A two-dimmensional vector field is plotted relative to x and y coordinate axes,
diff --git a/ptx/sec_vvf.ptx b/ptx/sec_vvf.ptx
index 5b7671aaa..50ee1a456 100644
--- a/ptx/sec_vvf.ptx
+++ b/ptx/sec_vvf.ptx
@@ -62,8 +62,10 @@
- Graph of the vector \vec r(-2) =\la 4,1\ra.
- This vector starts at the origin and ends at the point (4,1).
+
+ Graph of the vector \vec r(-2) =\la 4,1\ra.
+ This vector starts at the origin and ends at the point (4,1).
+
Graph of a vector whose terminal point corresponds to a point on the curve.
@@ -93,12 +95,14 @@
- The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra.
- The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t).
- The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise.
- The function is plotted for t approximately between -2.5 and 1.5.
- The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5.
- The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases.
+
+ The image contains the plot of the vector valued function \vec r(t) = \la t^2,t^2+t-1\ra.
+ The image also contains the graph of the vector \vec r(-2) =\la 4,1\ra, which ends at a point on the function \vec r(t).
+ The function \vec r(t) resembles a parabola which has been rotated about 45 degrees clockwise.
+ The function is plotted for t approximately between -2.5 and 1.5.
+ The function begins near the point (5,2) and then slopes down crossing the x-axis at approximately x=2.5.
+ The slanted parabola then curves upwards around the point (0,-1), from which point it begins increasing in both x and y as t increases.
+
Graph of the vector and vector valued function from the example.
@@ -206,13 +210,15 @@
- The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2.
- The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t).
- The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12.
- From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1.
- The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop.
- After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2.
- The function is also symmetric about the y-axis.
+
+ The image contains the plot of the vector valued function \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra for -2\leq t\leq 2.
+ The image also contains the vectors \vec r(-1) =\la 0,\frac12 \ra and \vec r(2) =\la 6,\frac15 \ra, which both end at a point on the function \vec r(t).
+ The function \vec r(t) begins at the point (-6,\frac15) corresponding to the lower bound of t=-2, from which it slowly slopes upwards until crossing the y-axis at y=\frac12.
+ From here, the function slopes upwards and slightly outwards away from the y-axis until curving back and crossing the y-axis once again at y=1.
+ The function then slopes downwards and slightly away from the y-axis until curving back and crossing the y-axis again at y=\frac12, completing a loop.
+ After crossing the y-axis, the function continues slowly sloping downwards until reaching the point (6,\frac15) which corresponds to the upper bound of t=2.
+ The function is also symmetric about the y-axis.
+
Graph of two vectors and vector valued function from the example.
@@ -279,13 +285,15 @@
- The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi.
- The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t).
- The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0.
- Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane.
- The addition of the z coordinate given by t then makes this curve linearly increase at t increases.
- Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions.
- The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi).
+
+ The image contains the plot of the vector valued function \vec r(t) = \la \cos(t) ,\sin(t) ,t\ra for 0\leq t\leq 4\pi.
+ The image also contains the vector \vec r(7\pi/4)\approx (0.707,-0.707,5.498), which corresponds to a point on the function \vec r(t).
+ The function \vec r(t) begins at the point (1,0,0) corresponding to the lower bound of t=0.
+ Ignoring the z coordinate of the vector valued function, the curve is simply a circle centered at the origin in the xy plane.
+ The addition of the z coordinate given by t then makes this curve linearly increase at t increases.
+ Accounting for the z coordinate, the function \vec r(t) resembles a linearly increasing circular spiral, which completes two full revolutions.
+ The spiral ends at the same x and y coordinates as it started, but is 4\pi units above in the z direction, ending at the point (1,0,4\pi).
+
Graph of a vector and the three-dimensional vector valued function from the example.
@@ -432,9 +440,11 @@
- The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10.
- The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3).
- The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph.
+
+ The image contains the plot of the vector valued functions \vec r_1(t) = \la 0.2t,0.3t \ra, \vec r_2(t) = \la \cos(t) ,\sin(t) \ra on -10\leq t\leq10.
+ The function \vec r_1(t) = \la 0.2t,0.3t \ra is a line which begins at the point (-2,-3) corresponding to t=-10 and ends when t=10 at the point (2,3).
+ The second function \vec r_2(t) = \la \cos(t) ,\sin(t) \ra looks like a circle of radius 1, but as -10\leq t\leq10 the function completes \frac{20}{2\pi} full circular rotations, which cannot be seen in the graph.
+
Graph of the two vector valued functions prior to adding them together.
@@ -463,13 +473,15 @@
- The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t).
- The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10.
- The function begins in the third quadrant, near the point (-3,-2).
- Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards.
- The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate.
- The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant.
- The function nearly completes a third loop, before ending near the point (3,1.5).
+
+ The image contains the plot of the vector valued function \vec r(t) = \vec r_1(t)+\vec r_2(t).
+ The function is given by \vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra on -10\leq t\leq10.
+ The function begins in the third quadrant, near the point (-3,-2).
+ Near this point, the function is concave up, starting with a downward slope and later beginning to slope upwards.
+ The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate.
+ The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant.
+ The function nearly completes a third loop, before ending near the point (3,1.5).
+
Graph of the vector valued function coming from adding the circle and line vector valued functions.
@@ -497,15 +509,17 @@
- The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t).
- The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10.
- Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function.
- The function begins in the third quadrant, near the point (-15,-10).
- Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards.
- The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate.
- The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant.
- The function nearly completes a third loop, before ending near the point (15,7.5).
- The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t).
+
+ The image contains the plot of the vector valued function 5\vec r(t) = 5\vec r_1(t)+5\vec r_2(t).
+ The function is now given by \vec r(t) = \la\,5\cos(t) + t,5\sin(t) +1.5t\,\ra on -10\leq t\leq10.
+ Compared to the unscaled function \vec r(t), the function 5\vec r(t) is exactly 5 times larger than the original function.
+ The function begins in the third quadrant, near the point (-15,-10).
+ Like the unscaled function, near this point 5\vec r(t) is concave up, starting with a downwards slope and later beginning to slope upwards.
+ The function then changes to concave down, until it circles back on itself and continues in the same concave-up trajectory as when it started but in a slightly increased x and y coordinate.
+ The function creates a total of two of these loops, with the first loop being in the third quadrant, while the second is in the first quadrant.
+ The function nearly completes a third loop, before ending near the point (15,7.5).
+ The entirety of the original function \vec r(t) can be seen to fit between the two loops of the scaled function 5\vec r(t).
+
Graph of the vector valued function coming from scaling the vector valued function from the example.
@@ -564,15 +578,17 @@
Tracing a cycloid
- The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface.
- The point p is initially at the top of the circle of radius 1.
- Once the circle rolls, the point p is tracked to create a graph of a cycloid.
- The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on.
- After this point, the graph begins to increase.
- The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease.
- The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops.
- Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle.
- The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles.
+
+ The image shows the cycloid which comes from tracking a point p on a rolling circle of radius 1 on a flat surface.
+ The point p is initially at the top of the circle of radius 1.
+ Once the circle rolls, the point p is tracked to create a graph of a cycloid.
+ The graph coming from tracking the point p first begins to decrease, until the point p is at the bottom of the circle, at which point the point p touches the surface the ball is rolling on.
+ After this point, the graph begins to increase.
+ The circle continues to roll, with the point p once again becoming the top of the circle, after which the graph begins to decrease.
+ The circle continues rolling, with the point p becoming the lowest point on the circle two more times, after which the graph stops.
+ Between the starting point and the point at which p is at the bottom of the circle, the graph resembles a slightly stretched quarter of a circle.
+ The part of the graph that comes from the remaining two full revolutions of the circle resembles two horizontally stretched semi-circles.
+
Image showing the cycloid which comes from tracking a point on a rolling circle.
@@ -637,10 +653,12 @@
- The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface.
- The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0).
- After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0).
- The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before.
+
+ The graph shows the function \vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra, which is the graph of a cycloid which comes from tracking a point p on a rolling a circle of radius 1 on a flat surface.
+ The curve begins near the point (0,2), after which it begins to decrease until it reaches its lowest point near the point (2,0).
+ After this point, the curve begins increasing, until it reaches its highest point when p is at the top of the circle, after which it decreases until reaching another minimum near the point (8,0).
+ The curve continues in the same fashion, reaching another minimum near the point (14,0), after which it continues for a slight duration in the same fashion as before.
+
Graph showing the cycloid which comes from tracking a point on a rolling circle.
@@ -737,9 +755,11 @@
- Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1.
- The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1).
- The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1).
+
+ Graph of the function \vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra on -1\leq t\leq 1.
+ The function \vec r(t) is the right half of the unit circle, beginning at the point (0,-1), crossing the x-axis at the point (1,0), and ending at the point (0,-1).
+ The graph also contains the displacement vector, which begins at the start of the curve at the point (0,-1) and heads directly upwards until reaching the endpoint of the curve at the point (0,1).
+
Graph of the semicircle coming from plotting the vector valued function from the example.
@@ -977,9 +997,11 @@
- Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2.
- The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3).
- The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3).
+
+ Graph of the function \vec r(t) = \la t^2,t^2-1\ra for -2\leq t\leq 2.
+ The graph of the function \vec r(t) is a line which begins at the point (0,-1) and ends at the point (4,3).
+ The function begins at the point (4,3) when t=-2, linearly decreases until the point (0,-1) and once again follows the same linear path until ending at the point (4,3).
+
Graph of the vector valued function from the example.
@@ -1020,10 +1042,12 @@
- Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2.
- The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin.
- After reaching the origin, the function is concave and ends when it reaches the point (4,8).
- The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other.
+
+ Graph of the function \vec r(t) = \la t^2,t^3\ra, for -2\leq t\leq 2.
+ The graph of the function \vec r(t) begins at the point (4,-8), and upwards and to the left in a concave down fashion until reaching the origin.
+ After reaching the origin, the function is concave and ends when it reaches the point (4,8).
+ The function is also symmetric about the x-axis, meaning that the upper and lower half of the curves are mirror images of each other.
+
Graph of the vector valued function from the example.
@@ -1066,10 +1090,12 @@
- Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2.
- The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2.
- At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2.
- The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12.
+
+ Graph of the function \vec r(t) = \la 1/t,1/t^2\ra, for -2\leq t\leq 2.
+ The graph of the function \vec r(t) begins at the point (-\frac12,\frac14) corresponding to t=2, heads upwards and to the left as t increases, following the path of the left side of the parabola y=x^2.
+ At t=2 the function starts at the point (\frac12,\frac14) and as t decreases follows the path of the right side of the parabola given by y=x^2.
+ The function is also mirrored about y-axis, and is equivalent to the parabola y=x^2 outside of the region -\frac12 \leq x \frac12.
+
Graph of the vector valued function from the example.
@@ -1112,15 +1138,17 @@
- Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi.
- The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi.
- As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease.
- The function continues to decrease, crossing the xaxis at x=1.
- After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin.
- The function now increases upwards and to the right until reaching a maximum near point (0.25,1).
- After this point, the curve decreases, crossing the x-axis once again at x=1.
- The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0).
- The function is also symmetric about the x-axis.
+
+ Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi.
+ The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi.
+ As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease.
+ The function continues to decrease, crossing the xaxis at x=1.
+ After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin.
+ The function now increases upwards and to the right until reaching a maximum near point (0.25,1).
+ After this point, the curve decreases, crossing the x-axis once again at x=1.
+ The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0).
+ The function is also symmetric about the x-axis.
+
Graph of the vector valued function from the example.
@@ -1161,15 +1189,17 @@
- Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi.
- The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi.
- As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease.
- The function continues to decrease, crossing the xaxis at x=1.
- After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin.
- The function now increases upwards and to the right until reaching a maximum near point (0.25,1).
- After this point, the curve decreases, crossing the x-axis once again at x=1.
- The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0).
- The function is also symmetric about the x-axis.
+
+ Graph of the function \vec r(t) = \la \frac1{10}t^2,\sin(t) \ra, for -2\pi\leq t\leq 2\pi.
+ The graph of the function \vec r(t) begins at the point (\frac{(2\pi)^2}{10},0) corresponding to t=-2\pi.
+ As t increases, the function curves upwards and to left, until reaching a maximum near the point (2.25,1), after which it begins to decrease.
+ The function continues to decrease, crossing the xaxis at x=1.
+ After this, the function continues to decrease until reaching a minimum near the point (0.25,-1), after which it curves upwards until it reaches the origin.
+ The function now increases upwards and to the right until reaching a maximum near point (0.25,1).
+ After this point, the curve decreases, crossing the x-axis once again at x=1.
+ The function now continues decreasing until it reaches a minimum near the point (2.25,-1) after which it begins to increase, until ending at its starting point of (\frac{(2\pi)^2}{10},0).
+ The function is also symmetric about the x-axis.
+
Graph of the vector valued function from the example.
@@ -1211,11 +1241,13 @@
- Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2].
- The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin.
- The rightmost point of the oval is the point (3,0).
- The leftmost point of the oval is the point (-3,0).
- The highest point of the oval is the point (0,2), while the lowest is the point (0,-2).
+
+ Graph of the function \vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra, on [0,2].
+ The graph of the function \vec r(t) is an oval having a horizontal width of 6 and a height of 4 centered at the origin.
+ The rightmost point of the oval is the point (3,0).
+ The leftmost point of the oval is the point (-3,0).
+ The highest point of the oval is the point (0,2), while the lowest is the point (0,-2).
+
Graph of the vector valued function from the example.
@@ -1257,15 +1289,17 @@
- Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi].
- The graph of the function \vec r(t) resembles the symbol \infty.
- The curve begins at the point (3,0) which corresponds to t=0.
- The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2).
- From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2).
- The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right.
- The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin.
- Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0).
- The curve is symmetric about both the x and y axes.
+
+ Graph of the function \vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra, on [0,2\pi].
+ The graph of the function \vec r(t) resembles the symbol \infty.
+ The curve begins at the point (3,0) which corresponds to t=0.
+ The curve then begins going upwards and to the left, until it reaches a maximum near the point (2,2).
+ From here the curve begins going downwards and to the left, passing through the origin, until reaching a minimum near (-2,-2).
+ The curve then begins going upwards to the left, passing through the point (-3,0), after which it begins going to the right.
+ The curve reaches a maximum near the point (-2,2), after which it begins decreasing and going to the right, once again crossing the origin.
+ Finally, the curve reaches another minimum near the point (2,-2), after which it begins going upwards to the right, until reaching its starting point given by (3,0).
+ The curve is symmetric about both the x and y axes.
+
Graph of the vector valued function from the example.
@@ -1307,13 +1341,15 @@
- Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi].
- The graph of the function is symmetric about both the x and y axes.
- The curve crosses the x-axis at the point x=2 corresponding to t=\pi.
- After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion.
- Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left.
- As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2.
- From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left.
+
+ Graph of the function \vec r(t) = \la 2\sec(t) ,\tan(t) \ra, on [-\pi,\pi].
+ The graph of the function is symmetric about both the x and y axes.
+ The curve crosses the x-axis at the point x=2 corresponding to t=\pi.
+ After this point the upper part of the curve goes upwards and to the right in a mostly linear fashion.
+ Since the curve is symmetric about the y-axis, the lower part of the curve is the mirrored version of the upper part, meaning that it goes downwards and to the left.
+ As the curve is symmetric about the x-axis, the left side of the curve is the mirrored version of the right part, meaning that it crosses the x-axis at x=-2.
+ From this point, the upper part of the left side of the curve goes upwards and to the left, while the lower part goes downwards and to the left.
+
Graph of the vector valued function from the example.
@@ -1365,10 +1401,12 @@
- Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi].
- The graph of the function is a circular spiral centered about the y-axis.
- Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane.
- Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0).
+
+ Graph of the function \vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra, on [0,2\pi].
+ The graph of the function is a circular spiral centered about the y-axis.
+ Ignoring the y-axis, the curve is simply a singular circle of radius 2 in the xz-plane.
+ Incorporating the y coordinate then creates the linearly increasing spiral, which begins at the point (2,0,0), completes precisely one full revolution and ends at the point (2,2\pi,0).
+
Graph of the vector valued function from the example.
@@ -1417,10 +1455,12 @@
- Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi].
- The graph of the function is an oval-shaped spiral centered about the z-axis.
- Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane.
- Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2).
+
+ Graph of the function \vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra on [0,2\pi].
+ The graph of the function is an oval-shaped spiral centered about the z-axis.
+ Ignoring the z-axis, the curve is simply an oval having a horizontal width of 6 and a height of 2 in the xy-plane.
+ Incorporating the z coordinate then creates the linearly increasing oval spiral, which begins at the point (3,0,0), completes precisely one full revolution and ends at the point (3,0,2).
+
Graph of the vector valued function from the example.
@@ -1468,13 +1508,15 @@
- Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi].
- The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis.
- The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1.
- Ignoring the z coordinate, the curve is a unit circle in the xy plane.
- Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane.
- If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane.
- This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes.
+
+ Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra on [0,2\pi].
+ The graph of the function is an oval lying in the plane coming from rotating the xy plane 45 degrees towards the z-axis.
+ The oval lying in this plane has a horizontal width of \sqrt{2} and a height of 1.
+ Ignoring the z coordinate, the curve is a unit circle in the xy plane.
+ Similarly ignoring the y coordinate, the curve is a unit circle in the xz plane.
+ If we now ignore the x coordinate, the resulting curve is a diagonal line given by z=y in the yz plane.
+ This line turns back on itself, which can be seen in the image of the oval when considering all three coordinate axes.
+
Graph of the vector valued function from the example.
@@ -1522,17 +1564,19 @@
- Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi].
- The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis.
- The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane.
- Ignoring the z coordinate, the curve is a unit circle in the xy plane.
- Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively.
- We now describe the z coordinate with respect to travelling along the unit circle in the xy plane.
- Starting at t=, the function begins at the point (1,0,0).
- As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1.
- Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}.
- Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}.
- Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0).
+
+ Graph of the function \vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra on [0,2\pi].
+ The graph of the function resembles a saddle centered at the origin whose height is defined by the z-axis.
+ The two sides of the saddle that taper off fall into negative z and lie in the second and third quadrants in the xy plane.
+ Ignoring the z coordinate, the curve is a unit circle in the xy plane.
+ Ignoring the x or y coordinates individually, the curve looks like the \infty symbol in the yz and the xz planes, respectively.
+ We now describe the z coordinate with respect to travelling along the unit circle in the xy plane.
+ Starting at t=, the function begins at the point (1,0,0).
+ As t increases and we travel along the unit circle in the x and y coordinates, z increases until we get to t=\frac{\pi}{2} at which z=1.
+ Then, continuing along the unit circle, z decreases until it reaches a minimum of z=-1 when t=\frac{3\pi}{4}.
+ Continuing along the circle, z begins to increase once again, reaching one more maximum of z=1 when t=\frac{5\pi}{4}.
+ Finally, z begins to decrease, reaching its last minimum of z=1 when t=\frac{7\pi}{4}, after which z increases, and the curve ends where it began, at the point (1,0,0).
+
Graph of the vector valued function from the example.
diff --git a/ptx/sec_vvf_calc.ptx b/ptx/sec_vvf_calc.ptx
index 66bf615b8..e63c127f2 100644
--- a/ptx/sec_vvf_calc.ptx
+++ b/ptx/sec_vvf_calc.ptx
@@ -226,12 +226,14 @@
- Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1].
- The function \vec r is a small part of a concave down circular arc.
- The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r .
- The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1).
- The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1).
- The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r.
+
+ Graph of an arbitrary vector-valued \vec r on the interval which includes [t_0,t_1].
+ The function \vec r is a small part of a concave down circular arc.
+ The graph includes the vectors \vec r (t_0) and \vec r (t_1), which begin from the origin and end at the corresponding point of the function \vec r .
+ The graph also includes the vector \vec r (t_1) - \vec r (t_0), which begins where \vec r (t_0) ends, and then terminates at the same termination point as \vec r (t_1).
+ The three vectors \vec r (t_0), \vec r (t_1) and \vec r (t_1) - \vec r (t_0) for a triangle, where following the path of \vec r (t_0) and \vec r (t_1) - \vec r (t_0) takes you to the same point as \vec r (t_1).
+ The vector \vec r (t_0) terminates on the left side of the circular arc, while \vec r (t_1) terminates further on the right side of the circular arc of given by the function \vec r.
+
Illustration of a vector-valued function on a given interval.
@@ -258,11 +260,13 @@
- Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image.
- This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} .
- The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point.
- The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image.
- However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0).
+
+ Graph of the same arbitrary vector-valued \vec r as well as the vectors \vec r (t_0) and \vec r (t_1) as described in the previous image.
+ This time the graph includes two additional vectors \vrp (t_0) and \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} .
+ The vector \vrp (t_0) begins at the termination point of \vec r (t_0), and is tangent to the function \vec r at this point.
+ The vector \frac{\vec r(t_1)-\vec r(t_0)}{t_1-t_0} also begins at the termination point of \vec r (t_0), and follows the path of the vector \vec r (t_1) - \vec r (t_0) from the previous image.
+ However, this vector does not end at the termination point of \vec r (t_1) but instead terminates at some point further away in the same direction as the vector \vec r (t_1) - \vec r (t_0).
+
Illustration of a vector-valued function on a given interval showcasing a derivative vector.
@@ -459,10 +463,12 @@
- Graph of the vector-valued function \vec r(t) = \la t^2,t\ra.
- The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis.
- The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra.
- The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes.
+
+ Graph of the vector-valued function \vec r(t) = \la t^2,t\ra.
+ The graph of the function \vec r(t) is simply the graph of the parabola y=x^2, but instead of opening towards the positive y-axis, the function \vec r(t) opens towards the positive x-axis.
+ The function \vec r(t) is plotted on the interval [-2,2], where \vec r(-2)= \la 4,-2\ra and \vec r(2)= \la 4,2\ra.
+ The graph also includes the derivative function \vrp(t) = \la 2t, 1\ra which takes the path of the horizontal line y=1 going from left to right in the standard coordinate axes.
+
Graph of the vector-valued function from the example and its derivative.
@@ -497,11 +503,13 @@
- Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image.
- The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra.
- The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image.
- The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra.
- The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t).
+
+ Graph of the vector-valued function \vec r(t) = \la t^2,t\ra described in the previous image.
+ The graph also includes two copies of the vector \vrp(1) = \la 2,1\ra.
+ The first copy of the vector \vrp(1) = \la 2,1\ra begins at the origin, and ends at the point (2,1), which is also a point on the derivative function \vrp(t) = \la 2t, 1\ra from the previous image.
+ The second copy of the vector \vrp(1) = \la 2,1\ra begins at the point (1,1), which corresponds to the termination point of \vec r(1) = \la 1,1 \ra.
+ The second copy of the vector \vrp(1) = \la 2,1\ra is tangent to the function \vec r(t) = \la t^2,t\ra at the point (1,1) corresponding to when t=1 in the function \vec r(t).
+
Graph of the vector-valued function from the example and its derivative.
@@ -562,14 +570,16 @@
Viewing a vector-valued function and its derivative at one point
- Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra.
- The function is a circular spiral which climbs the z-axis.
- Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane.
- Adding the z coordinate then creates the linearly increasing spiral.
- The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra
- The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1).
- The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2).
- The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t).
+
+ Graph of the vector-valued function \vec r(t) = \la \cos(t) , \sin(t) , t\ra.
+ The function is a circular spiral which climbs the z-axis.
+ Looking at the graph from above the z-axis, the function resembles a unit circle in the xy-plane.
+ Adding the z coordinate then creates the linearly increasing spiral.
+ The graph also includes two copies of the vector \vrp(\pi/2) = \la -1,0,1\ra
+ The first copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the origin, and ends at the point (-1,0,1).
+ The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra begins at the point (0,1,\pi/2), which corresponds to the termination point of \vec r(\pi/2).
+ The second copy of the vector \vrp(\pi/2) = \la -1,0,1\ra is also tangent to the function \vec r(t) at the point (0,1,\pi/2) corresponding to when t=\pi/2 in the function \vec r(t).
+
Graph of the vector-valued function from the example and its derivative at a point.
@@ -706,13 +716,15 @@
- Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5].
- The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate.
- The function then curves towards the origin.
- After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends.
- The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1).
- The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra.
- Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1).
+
+ Graph of the vector-valued function \vec r(t) = \la t,t^2,t^3\ra on [-1.5,1.5].
+ The function begins at the point (-1.5,2.25,-3.375), from which it begins to increase linearly in the x and z coordinates, and decrease in the y coordinate.
+ The function then curves towards the origin.
+ After passing through the origin, the function begins to increase in all x,y and z coordinates until it reaches the point (1.5,2.25,3.375), at which it ends.
+ The graph also contains the line \ell(t) = \la -1,1,-1\ra + t\la 1,-2,3\ra, which is the tangent line to the function at \vec r(-1), which is the point (-1,1,-1).
+ The line can be described as the line which passes through the point (-1,1,-1), moving in the direction of the vector \la 1,-2,3\ra.
+ Additionally, the line is defined for all t, so it also moves in the opposite direction of the vector \la 1,-2,3\ra, or in other words in the direction of \la -1,2,-3\ra from the point (-1,1,-1).
+
Graph of the vector-valued function from the example and the tangent line to a point on the curve.
@@ -793,11 +805,13 @@
- Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra.
- The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin.
- After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates.
- The curve is also symmetric about the y-axis.
- The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1.
+
+ Graph of the vector-valued function \vec r(t) = \la t^3,t^2\ra.
+ The function begins near the point (-3,2), from which it is concave down and sloping downwards towards the origin.
+ After passing through the origin, the curve is concave down and begins increasing in both x and y coordinates.
+ The curve is also symmetric about the y-axis.
+ The graph also contains the line \ell(t) = \la -1,1\ra + t\la 3,-2\ra , which is tangent to the function \vec r(t) = \la t^3,t^2\ra at the point (-1,1) corresponding to when t=1.
+
Graph of the vector-valued function from the example and the tangent line to a point on the curve.
@@ -983,13 +997,15 @@
- Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2].
- The function looks like the parabola y=x^2 -1, which takes the path going towards positive x.
- The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3).
- The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t).
- The function \vec u(t) looks like the unit circle which is missing a piece of the top.
- The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle.
- The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle.
+
+ Graph of the vector-valued function \vec r(t) = \la t,t^2 -1\ra on [-2,2].
+ The function looks like the parabola y=x^2 -1, which takes the path going towards positive x.
+ The function begins at the point (-2,3), decreases until reaching the point (0,-1), and then increases until ending at the point (2,3).
+ The graph also contains the function \vec u(t) which is the unit vector that points in the direction of \vec r(t).
+ The function \vec u(t) looks like the unit circle which is missing a piece of the top.
+ The circular arc from the graph of \vec u(t) begins at the point where the vector \vec r(-2)=\la -2,3\ra crosses the unit circle.
+ The circular arc then goes counterclockwise following the path of the unit circle, until ending at the point where the vector\vec r(2)=\la 2,3\ra crosses the unit circle.
+
Graph of the vector-valued function with the unit vector function which points in the direction of the function.
@@ -1059,14 +1075,16 @@
- Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t).
- The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0).
- The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2.
- The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1)
- From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc.
- The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0).
- From here the vector points to the right, as it is tangent to the bottom point of the circular arc.
- If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t).
+
+ Graph of the function \vec u(t) which is the unit vector that points in the direction of \vec r(t).
+ The graph also contains three unit vectors, \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0).
+ The vector \vec u\,'(-2)\approx \la -0.320,-0.213\ra begins at the starting point of the function \vec u(t) and points in the direction tangent to the circular arc at t=-2.
+ The vector \vec u\,'(-1)= \la 0,-2\ra begins at the point (-1,0) corresponding to the termination point of \vec u(-1)
+ From here, the vector points straight down, as it is tangent to the leftmost point of the circular arc.
+ The vector \vec u\,'(0)= \la 1,0\ra begins at the point (0,-1) corresponding to the termination point of \vec u(0).
+ From here the vector points to the right, as it is tangent to the bottom point of the circular arc.
+ If the vectors \vec u\,'(-2), \vec u\,'(-1) and \vec u\,'(0) were extended to lines, all three lines would be tangent to a point on the nearly complete unit circle given by \vec u(t).
+
Graph of the circular unit vector function along with three derivative vectors of the function.
@@ -1782,11 +1800,12 @@
- Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra.
- The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise.
- The curve passes through the point (2,0), the origin, and the point (0,2) in the given order.
- The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1).
- The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1.
+ Graph of the function \vec r(t) = \la t^2+t, t^2-t\ra.
+ The function \vec r(t) resembles a standard parabola which has been rotated 45 degrees clockwise.
+ The curve passes through the point (2,0), the origin, and the point (0,2) in the given order.
+ The graph also contains \vrp(1) beggining at the point (2,0) corresponding to the termination point of \vec r(1).
+ The vector \vrp(1)= \la 3,1 \ra and is tangent to \vec r(t) = \la t^2+t, t^2-t\ra corresponding to the point where t=1.
+
Graph of the vector-valued function and its derivative at a point.
@@ -1830,13 +1849,15 @@
- Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra.
- The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
- After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
- From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
- The graph of the function is also symmetric about the y-axis.
- The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1).
- The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
+ Graph of the function \vec r(t) = \la t^2-2t+2,t^3-3t^2+2t\ra.
+ The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
+ After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
+ From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
+ The graph of the function is also symmetric about the y-axis.
+ The graph also contains \vrp(1) beginning at the point (1,0) corresponding to the termination point of \vec r(1).
+ The vector \vrp(1)= \la 0,-1 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
Graph of the vector-valued function and its derivative at a point.
@@ -1880,13 +1901,15 @@
- Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra.
- The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
- After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
- From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
- The graph of the function is also symmetric about the y-axis.
- The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1).
- The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
+ Graph of the function \vec r(t) = \la t^2+1,t^3-t\ra.
+ The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
+ After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
+ From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
+ The graph of the function is also symmetric about the y-axis.
+ The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1).
+ The vector \vrp(1)= \la 2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
Graph of the vector-valued function and its derivative at a point.
@@ -1930,13 +1953,15 @@
- Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra.
- The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
- After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
- From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
- The graph of the function is also symmetric about the y-axis.
- The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1).
- The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
+ Graph of the function \vec r(t) = \la t^2-4t+5,t^3-6t^2+11t-6\ra.
+ The function \vec r(t) begins in the fourth quadrant, and begins heading upwards and to the left towards the point (2,0).
+ After crossing this point, the curve loops up into the first quadrant and back down until it reaches the point (1,0).
+ From this point, the curve loops down into the fourth quadrant and back up until it once again reaches the point (2,0), from which it begins increasing in both x and y coordinates in the first quadrant.
+ The graph of the function is also symmetric about the y-axis.
+ The graph also contains \vrp(1) beginning at the point (2,0) corresponding to the termination point of \vec r(1).
+ The vector \vrp(1)= \la -2,2 \ra and is tangent to \vec r(t) corresponding to the point where t=1.
+
Graph of the vector-valued function and its derivative at a point.
diff --git a/ptx/sec_work.ptx b/ptx/sec_work.ptx
index 597785bcb..e92b03bfe 100644
--- a/ptx/sec_work.ptx
+++ b/ptx/sec_work.ptx
@@ -338,11 +338,13 @@
Illustrating the important aspects of stretching a spring in computing work in
- Image of a spring showing the spring both before and after streching.
- In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring.
- On the x-axis, the block that the spring is going to move is pictured to be centered at x=1.
- In the second illustration, we see the streched spring, with the same block still attached on the right side.
- This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right.
+
+ Image of a spring showing the spring both before and after streching.
+ In the first illustration, we see the unstreched spring with a block attached to it on its right side, with a rightward facing force vector labeled F, showing the direction of the force that is applied by the spring.
+ On the x-axis, the block that the spring is going to move is pictured to be centered at x=1.
+ In the second illustration, we see the streched spring, with the same block still attached on the right side.
+ This time, the block is centered at x=6, showing the spring moved the block a total of 5 inches to the right.
+
Image showing the important aspects of strenching a spring which are used in computing work.
@@ -533,11 +535,13 @@
- Image of a cylindrical storage tank with a radius of 10 and a height of 30ft.
- On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank.
- The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis.
- The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i.
- Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i.
+
+ Image of a cylindrical storage tank with a radius of 10 and a height of 30ft.
+ On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank.
+ The image also contains the ith subinterval of y, which corresponds to the shaded region between y_{i-1} and y_i on the y-axis.
+ The height of this subinterval of y given by y_{i}-y_{i-1} is also labeled \Delta y_i.
+ Additionally, the distance from the top of this subinterval, occuring at y_{i} and the point 5 feet above the tank is given on the y-axis as 35-y_i.
+
Illustration of a cylindrical water tank with measurements used to compute the work required to empty it.
@@ -649,11 +653,13 @@
- Image of the same cylindrical storage tank with a radius of 10 and a height of 30.
- On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank.
- Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank.
- At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y.
- Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y.
+
+ Image of the same cylindrical storage tank with a radius of 10 and a height of 30.
+ On the left side of the storage tank the y-axis is shown with measurements of y=0 at the base of the tank, y=30 and the top of the tank, and y=35 5 feet above the tank.
+ Now instead of containing the ith subinterval of y, the image contains an arbitrarily chosen point y which is contained in the tank.
+ At this point y, the volume is given as V(y)=100\pi dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y.
+ Additionally, the distance from the arbitrary point y and the point 5 feet above the tank is given on the y-axis as 35-y.
+
Illustration of a cylindrical water tank with simplified measurements used to compute the work required to empty it.
@@ -724,12 +730,14 @@
- Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2.
- This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases.
- On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level.
- The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank.
- At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y.
- Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y.
+
+ Image of a conical storage tank with its tip 10 below ground and the base at ground level with a radius of 2.
+ This means as we move further below ground level, the area of the circular cross-section of the conical storage tank decreases.
+ On the left side of the storage tank the y-axis is shown with measurements of y=3 which is 3 feet above the tank, y=0 which is the top of the tank, and y=-10 which is the tip of the cone, 10 feet below ground level.
+ The image contains an arbitrarily chosen point y between y=-10 and y=0 which is contained in the tank.
+ At this point y, the volume is given as V(y)=\pi (\frac{y}{5} + 2)^2 dy, where d is the arbitrarily short distance between the top and bottom of the thin slice of water in the tank at the point y.
+ Additionally, the distance from the arbitrary point y and the point 3 feet above the tank is given on the y-axis as 3-y.
+
Illustration of a conical water tank with measurements used to compute the work required to empty it.
@@ -806,16 +814,18 @@
- Image of the swimming pool described in the example.
- The pool is 25 long.
- On the leftmost side of the image is the deep end of the pool, which is
- 10 long having a depth of 6.
- After the 10 mark,
- the bottom of the pool slopes up until it rises to a depth of 3
- a distance of 5 away in the x dimension.
- The shallow end of the pool has a depth of 3
- and length of 10.
- The end of the shallow end marks the end of the pool.
+
+ Image of the swimming pool described in the example.
+ The pool is 25 long.
+ On the leftmost side of the image is the deep end of the pool, which is
+ 10 long having a depth of 6.
+ After the 10 mark,
+ the bottom of the pool slopes up until it rises to a depth of 3
+ a distance of 5 away in the x dimension.
+ The shallow end of the pool has a depth of 3
+ and length of 10.
+ The end of the shallow end marks the end of the pool.
+
Illustration of the swimming pool from the example.
@@ -844,16 +854,18 @@
- Image of the same swimming pool described in the example.
- The image now contains the two coordinate axes x and y.
- On the y axis are five markings.
- The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool.
- The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water.
- The image also contains an arbitrary label y lying somewhere below the water level of the pool.
- On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end.
- The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool.
- On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool.
- The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively.
+
+ Image of the same swimming pool described in the example.
+ The image now contains the two coordinate axes x and y.
+ On the y axis are five markings.
+ The label y=0 marks the bottom of the deep end of the pool, the label y=3 marks the bottom of the shallow end of the pool.
+ The label y=6 marks the top of the water in the pool, and the label y=8 marks a point 2 above the top of the water.
+ The image also contains an arbitrary label y lying somewhere below the water level of the pool.
+ On the x-axis, the label x=0 marks the leftmost edge of the pool, x=10 marks the end of the deep end and after the pool slopes up to a depth of 3, the label x=15 marks the start of the shallow end.
+ The distance between the leftmost edge and rightmost edge of the pool when y is above y=3, which is the depth of shallow end, is constant, and is given by the length of the pool.
+ On the other hand, the distance between the same two edges of the pool when y is below y=3 is the length of the deep end, plus the additional distance in the x direction between the end of the deep end, x=10, and the point at which the arbitrary y level meets the sloped portion of the pool.
+ The sloped portion of the pool is given as the line which passes through the points (10,0) and (15,3), which are the end of the deep end, and the start of the shallow end, respectively.
+
Illustration of the swimming pool from the example showing necessary differential elements.
@@ -1516,11 +1528,13 @@
- Image of a fuel storage tank.
- The storage tank has a base that has a 2 length and 15 width.
- The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side.
- The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base.
- The depth of the tank, which is the distance between the top and bottom of the tank is 10.
+
+ Image of a fuel storage tank.
+ The storage tank has a base that has a 2 length and 15 width.
+ The forward, left and back walls of the tank are perpendicular to the base, while the right side of the tank slopes upward further away to the right side.
+ The top of the tank has a length of 5 due to the slope of the right wall of the fuel tank and the top same 15 width as the base.
+ The depth of the tank, which is the distance between the top and bottom of the tank is 10.
+
Illustration of the fuel storage tank from the problem.
@@ -1613,10 +1627,12 @@
- Image of a water tank in the shape of a circular truncated cone.
- The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base.
- The water tank has a circular base with a radius of 5.
- From the base, the tank has a height of 10, and a circular top with a radius of 2.
+
+ Image of a water tank in the shape of a circular truncated cone.
+ The water tank is laying on its base and is in the shape of a cone, from which a small portion including the top is sliced off paralell to the base.
+ The water tank has a circular base with a radius of 5.
+ From the base, the tank has a height of 10, and a circular top with a radius of 2.
+
Illustration of the water tank in the shape of a truncated cone.
@@ -1660,10 +1676,12 @@
- Image of the pyramidal water tank.
- The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank.
- The top of the water tank has a square top with a side length of 2.
- From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank.
+
+ Image of the pyramidal water tank.
+ The water tank is in the shape of a pyramid with a square base which is laying on its tip, meaning that the square base is the top of the tank.
+ The top of the water tank has a square top with a side length of 2.
+ From the top, the tank has a depth of 7, which is the shortest distance between the square top of the tank and the peak of the pyramid, which is the bottom of the tank.
+
Illustration of the water tank in the shape of an inverted pyramid.
@@ -1704,11 +1722,13 @@
+
Image of a water tank in the shape of a truncated pyramid.
The water tank has a square base with a side length of 2.
From the base, the pyramidal water tank expands on all sides until reaching the top of the tank.
The top of the water tank is a square with a side length of 5.
From the top, the tank has a depth of 9, which is the shortest distance between the bottom and top of the water tank.
+
Illustration of the water tank in the shape of a truncated pyramid.