diff --git a/Old source files/S_mult_ints_double_ints_param_surf.ptx b/Old source files/S_mult_ints_double_ints_param_surf.ptx index e350b6438..4b23ca397 100644 --- a/Old source files/S_mult_ints_double_ints_param_surf.ptx +++ b/Old source files/S_mult_ints_double_ints_param_surf.ptx @@ -167,7 +167,7 @@ where (s,t) varies over the entire domain of f. Therefore, any familiar surface that we have studied so far can be generated as a parametric surface. But what is more powerful is that there are surfaces that cannot be generated by a single function z = f(x,y) (such as the unit sphere), but that can be represented parametrically. We now consider an important example.

- +

Consider the @@ -254,7 +254,7 @@ - +

In this activity, we seek a @@ -417,7 +417,7 @@ - +

Consider the cylinder with radius a and height h defined parametrically by diff --git a/Old source files/ez-S_multi_ints_double_ints_param_surf.ptx b/Old source files/ez-S_multi_ints_double_ints_param_surf.ptx new file mode 100644 index 000000000..36678c8c0 --- /dev/null +++ b/Old source files/ez-S_multi_ints_double_ints_param_surf.ptx @@ -0,0 +1,357 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ Consider the ellipsoid given by the equation + + \frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} = 1. + +

+ +

+ In Activity, we found that a parameterization of the sphere S of radius R centered at the origin is + + x(r,s) = R\cos(s) \cos(t), \ y(s,t) = R \cos(s) \sin(t), \ \text{ and } \ z(s,t) = R\sin(s) + + for -\frac{\pi}{2} \leq s \leq \frac{\pi}{2} and 0 \leq t \leq 2\pi. +

    +
  1. +

    + Let (x,y,z) be a point on the ellipsoid and let X = \frac{x}{4}, Y = \frac{y}{5}, and Z = \frac{z}{3}. Show that (X,Y,Z) lies on the sphere S. Hence, find a parameterization of S in terms of X, Y, and Z as functions of s and t. +

    +
    2. + +
  2. +

    + Use the result of part (a) to find a parameterization of the ellipse in terms of x, y, and z as functions of s and t. Check your parametrization by substituting x, y, and z into the equation of the ellipsoid. Then check your work by plotting the surface defined by your parameterization. +

    +
    3. +
+

+ + +

+

    +
  1. +

    + If (x,y,z) lies on the ellipse and X = \frac{x}{4}, + Y = \frac{y}{5}, and Z = \frac{z}{3}, then + + X^2+Y^2+Z^2 \amp = \left(\frac{x}{4}\right)^2 + \left(\frac{y}{5}\right)^2 + \left(\frac{z}{3}\right)^2 + \amp = \frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} + \amp = 1 + . + So (X,Y,Z) lies on the sphere S and S has parameterization + + X(r,s) = \cos(s) \cos(t), \ Y(s,t) = \cos(s) \sin(t), \ \text{ and } \ Z(s,t) = \sin(s) + + for -\frac{\pi}{2} \leq s \leq \frac{\pi}{2} and 0 \leq t \leq 2\pi. +

    +
    2. +
  2. +

    + Since x = 4X, y = 5Y, and z = 3Z we have + + x(r,s) = 4\cos(s) \cos(t), \ y(s,t) = 5\cos(s) \sin(t), \ \text{ and } \ z(s,t) = 3\sin(s) + + for -\frac{\pi}{2} \leq s \leq \frac{\pi}{2} and 0 \leq t \leq 2\pi. + Note that + + \frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} \amp = \cos^2(s)\cos^2(t) + \cos^2(s) \sin^2(t) + \sin^2(s) + \amp = \cos^2(s)[\cos^2(t) + \sin^2(t)] + \sin^2(s) + \amp = \cos^2(s) + \sin^2(s) + \amp = 1 + . + So x(r,s) = 4\cos(s) \cos(t), + y(s,t) = 5\cos(s) \sin(t), + and z(s,t) = 3sin(s) is a parametrization of the ellipse. +

    +
    3. +
+

+ + + + +

+ In this exercise, we explore how to use a parametrization and iterated integral to determine the surface area of a sphere. +

    +
  1. +

    + Set up an iterated integral whose value is the portion of the surface area of a sphere of radius R that lies in the first octant (see the parameterization you developed in Activity). +

    +
    2. + +
  2. +

    + Then, evaluate the integral to calculate the surface area of this portion of the sphere. +

    +
    3. + +
  3. +

    + By what constant must you multiply the value determined in (b) in order to find the total surface area of the entire sphere. +

    +
    4. + +
  4. +

    + Finally, compare your result to the standard formula for the surface area of sphere. +

    +
    5. +
+

+ + +

+

    +
  1. +

    + From Activity + we know that a parameterization of a sphere of radius R centered at the origin is + + x(s,t) = R \cos(s) \cos(t), \ y(s,t) = R\sin(s) \cos(t), \ \text{ and } \ z(s,t) = R \sin(t) + + for -\frac{\pi}{2} \leq s \leq \frac{\pi}{2} and t in [0, 2 \pi]. + By symmetry, + we can just calculate the surface area of the sphere in the first octant, + then multiply by 8. + So we can assume 0 \leq s, t \leq \frac{\pi}{2}. +

    +
    2. +
  2. +

    + With \vr(s,t) = \langle R \cos(s) \cos(t), R\sin(s) \cos(t), R \sin(t) \rangle we have + + \vr_s(s,t) \amp = \langle -R\sin(s) \cos(t), R\cos(s)\cos(t), 0 \rangle, + \vr_t(s,t) \amp = \langle -R\cos(s) \sin(t), -R\sin(s)\sin(t), -R\cos(t) \rangle + , + and so + + \lvert \vr_s(s,t) \times \vr_t(s,t) \rvert \amp = \lvert \langle -R\sin(s) \cos(t), R\cos(s)\cos(t), 0 \rangle \times \langle -R\cos(s) \sin(t), -R\sin(s)\sin(t), -R\cos(t) \rangle \rvert + \amp = \lvert \langle -R^2\cos(s)\cos^2(t), -R^2\sin(s)\cos^2(t), R^2\cos(t)\sin(t) \rangle \rvert + \amp = \sqrt{R^4\cos^2(s)\cos^4(t) + R^4\sin^2(s)\cos^4(t) + R^4\cos^2(t)\sin^2(t)} + \amp = R^2\sqrt{\cos^2(t)} + \amp = R^2\cos(t) + . + Therefore, the surface area of the first octant portion of the sphere is + + \int_0^{\pi/2} \int_0^{\pi/2} R^2 \cos(t) \, dt \, ds \amp = 8R^2 \int_0^{\pi/2} \sin(t) \restrict0^{\pi/2} \, ds + \amp = R^2 \int_0^{\pi/2} 1 \, ds + \amp = \frac{1}{2}\pi R^2 + . +

    +
    3. +
  3. +

    + We multiply the previous result by 8 to reproduce the surface area calculation in each of the eight octants. + So the surface area of a sphere of radius R is 4 \pi R^2. +

    +
    4. +
  4. +

    + The result of part (c) is the standard formula for the surface area of a sphere of radius R. +

    +
    5. +
+

+ + + + +

+ Consider the plane generated by z = f(x,y) = 24 - 2x - 3y over the region D = [0,2]\times[0,3]. +

    +
  1. +

    + Sketch a picture of the overall solid generated by the plane over the given domain. +

    +
    2. + +
  2. +

    + Determine a parameterization \vr(s,t) for the plane over the domain D. +

    +
    3. + +
  3. +

    + Use Equation to determine the surface area generated by f over the domain D. +

    +
    4. + +
  4. +

    + Observe that the vector \vu = \langle 2, 0, -4 \rangle points from (0,0,24) to (2,0,20) along one side of the surface generated by the plane f over D. Find the vector \vv such that \vu and \vv together span the parallelogram that represents the surface defined by f over D, and hence compute | \vu \times \vv |. What do you observe about the value you find? +

    +
    5. +
+

+ + +

+

    +
  1. +

    + The solid looks like a box with a slanted top. +

    +
    2. +
  2. +

    + If we let x(s,t) = s and y(s,t) = t, + then z(s,t) = 24-2s-3t provides a parameterization of the plane. + So we can let \vr(s,t) = s \vi + t \vj + (24-2s-3t) \vk. +

    +
    3. +
  3. +

    + In our situation we have \vr_s(s,t) = \vi - 2 \vk and \vr_t(s,t) = \vj - 3 \vk. + So \vr_s(s,t) \times \vr_t(s,t) = 2 \vi + 3\vj + \vk. + So the surface area generated by f over the domain D is found by + + \int_0^{3} \int_0^{2} | \langle 2,3,1 \rangle | \, dt \, ds \amp = \sqrt{14} \int_0^{3} \int_0^{2} \, dt \, ds + \amp = 6 \sqrt{14} + . +

    +
    4. +
  4. +

    + Take \vv = \langle 0, 2, -9 \rangle to be the vector that points from (0,0,24) to + (0,3,15) along one side of the surface generated by the plane f over D. + Then \vu and \vv together span the parallelogram that represents the surface defined by f over D. + Here we have + + | \vu \times \vv | = | \langle 12, 18, 6 \rangle | = \sqrt{504} = 6 \sqrt{14} + . + Since | \vu \times \vv | is the area of the parallelogram determined by \vu and \vv, + we should expect this area to be the same as the surface area of the surface defined by the plane f over D. +

    +
    5. +
+

+ + + + +

+ A cone with base radius a and height h can be realized as the surface defined by z = \frac{h}{a} \sqrt{x^2+y^2}, where a and h are positive. +

    +
  1. +

    + Find a parameterization of the cone described by z = \frac{h}{a} \sqrt{x^2+y^2}. (Hint: Compare to the parameterization of a cylinder as seen in Activity.) +

    +
    2. + +
  2. +

    + Set up an iterated integral to determine the surface area of this cone. +

    +
    3. + +
  3. +

    + Evaluate the iterated integral to find a formula for the lateral surface area of a cone of height h and base a. +

    +
    4. +
+

+ + +

+

    +
  1. +

    + Let t run along the positive z-axis from 0 to h. + By similar triangles, + the cross section of the cone at height t parallel to the x-y plane has radius \frac{h}{a}t. + We can think of the cone as made of circles with these varying radii, + so a parameterization of the cone is + + \vr(s,t) = \frac{h}{a}t \cos(s) \vi + \frac{h}{a}t \sin(s) \vj + t \vk + , + for 0 \leq s \leq 2 \pi and 0 \leq t \leq a. +

    +
    2. +
  2. +

    + We have + + \vr_s(s,t) = -\frac{h}{a}t\sin(s) \vi + \frac{h}{a}t\cos(s) \vj \ \text{ and } \ \vr_t(s,t) = \frac{h}{a} \cos(s) \vi + \frac{h}{a} \sin(s) \vj + \vk + , + and + + \vr_s(s,t) \times \vr_t(s,t) = -\frac{h}{a}t\cos(s) \vi + \frac{h}{a}t\sin(s) \vj + \left(\frac{a}{h}\right)^2 t \vk + . + By symmetry, + we calculate the area of the cone in the first octant and multiply by 4 to obtain the area of the surface of the cone as + + 4\int_0^{\pi/2} \int_0^a |\vr_s \times \vr_t| \, dt \, ds = 4\int_0^{\pi/2} \int_0^a \frac{a}{h^2} \sqrt{h^2+a^2} t \, dt \, ds + . +

    +
    3. +
  3. +

    + Evaluating the iterated integral yields + + 4\int_0^{\pi/2} \int_0^a \frac{a}{h^2} \sqrt{h^2+a^2} t \, dt \, ds \amp = 4\frac{a}{h^2} \sqrt{h^2+a^2} \int_0^{\pi/2} \frac{1}{2}t^2 \restrict0^h \, ds + \amp = 2 a \sqrt{h^2+a^2} \int_{0}^{\pi/2} \, ds + \amp = \pi a \sqrt{h^2+a^2} + . +

    +
    4. +
+

+ + + diff --git a/project.ptx b/project.ptx index 78e6c0f7b..fd871396c 100644 --- a/project.ptx +++ b/project.ptx @@ -1,6 +1,6 @@ - + diff --git a/source/C-mvprecalc.ptx b/source/C-mvprecalc.ptx index 3a362826c..7f759652b 100644 --- a/source/C-mvprecalc.ptx +++ b/source/C-mvprecalc.ptx @@ -26,7 +26,7 @@ More generally, we will need to work with functions and expressions that depend on more than one input or involve outputs more complicated than a single scalar. For instance, when trying to find the optimal price for a product, you will need to consider fixed costs like a building lease, labor costs, and the varying costs of all the materials that are used in the product. Another example would be trying to measure the atmospheric temperature, which will vary over a three-dimensional space and will vary over time of day as well. If we wanted to use wind to help in modeling weather phenomena, we will need new tools since wind has both a direction and a strength associated with it, and thus cannot be measured by a single number. For each of the cases mentioned above, it is not entirely clear how to relate changes in the input variables to changes in the different kinds of outputs. A long-term goal of our work in this book will be to carefully define the relationship between changes of inputs and outputs for many new types of functions.

- With our new expanded view of the world around us, we will also need to describe a variety of new examples of mathematical objects that are algebraically nice and exhibit different geometric features. You spent a few math classes to understand and apply to the calculus of one-variable in and one-variable out, we will expand our tools in much less time because of the depth and variety of ideas we have already encountered. Before we look at the calculus of these new kinds of functions, we will need to do a bit of precalculus again. + With our new expanded view of the world around us, we will also need to describe a variety of new examples of mathematical objects that are algebraically nice and exhibit different geometric features. Most students took several math classes to understand and apply to the calculus of functions with one-variable in and one-variable out; In this text, we will expand our tools in much less time because of the depth and variety of ideas that you have already encountered. Before we look at the calculus of these new kinds of functions, we will need to do a bit of precalculus again.

diff --git a/source/C-vvf.ptx b/source/C-vvf.ptx index 410212b31..c80109bfa 100644 --- a/source/C-vvf.ptx +++ b/source/C-vvf.ptx @@ -81,7 +81,7 @@ plot2+=text("$f(a)-f(a-h)$",(-0.25,-0.05),color="orange",fontsize=20) plot2 -

ADD ALT TEXT TO THIS IMAGE

 +

A plot of a downward facing parabola with a blue line going between two points on the parabola. The blue line corresponds to more horizontal change than vertical change. The right point on the parabola is labeled P:(a,f(a)). The change in the vertical coordinates between these two points is represented with a vertical segment in yellow and is labeled f(a)-f(a-h). The change in the horizontal coordinates between two points is represented by a green line segment that is labeled h.

var('t,x') @@ -101,7 +101,7 @@ plot3+=text("$f(a)-f(a-h)$",(0.10,-0.05),color="orange",fontsize=18) plot3 -

ADD ALT TEXT TO THIS IMAGE

 +

A plot of a downward facing parabola with a blue line going between two points on the parabola. The blue line in this plot is steeper than in the previous plot and the points are closer together on the parabola. The right point on the parabola is labeled P:(a,f(a)). The change in the vertical coordinates between these two points is represented with a vertical segment in yellow and is labeled f(a)-f(a-h). The change in the horizontal coordinates between two points is represented by a green line segment that is labeled h.

 var('t,x') @@ -121,7 +121,7 @@ plot4+=text("$f(a)-f(a-h)$",(0.20,-0.05),color="orange",fontsize=20) plot4 -

ADD ALT TEXT TO THIS IMAGE

 +

A plot of a downward facing parabola with a blue line going between two points on the parabola. The blue line in this plot is slightly steeper than in the previous plot and the two points are close together on the parabola. The right point on the parabola is labeled P:(a,f(a)). The change in the vertical coordinates between these two points is represented with a vertical segment in yellow and is labeled f(a)-f(a-h). The change in the horizontal coordinates between two points is represented by a green line segment that is labeled h.

@@ -152,7 +152,11 @@ -

ADD ALT TEXT TO THIS IMAGE

+

A sequence of three images shows different Riemann sum converging to the defintie integral. + The left image shows a sinusoidal graph in blue that goes above and below the the horizontal axis. From left to right, points with inputs a, b, c, and d are labeled on a horizontal segment. The blue graph is above the horizontal axis between inputs a nd b and between c and d. The blue graph is below the horizontal axis between b and c. Eleven green rectangles extend vertically from the horizontal axis to the blue graph on the region of inputs from a to d. + The center image shows a sinusoidal graph in blue that goes above and below the the horizontal axis. From left to right, points with inputs a, b, c, and d are labeled on a horizontal segment. The blue graph is above the horizontal axis between inputs a nd b and between c and d. The blue graph is below the horizontal axis between b and c. Twenty two green rectangles extend vertically from the horizontal axis to the blue graph on the region of inputs from a to d. These green rectangles better approximate the area under the blue graph, as compared to the left image. + The right image shows a sinusoidal graph in blue that goes above and below the the horizontal axis. From left to right, points with inputs a, b, c, and d are labeled on a horizontal segment. The blue graph is above the horizontal axis between inputs a nd b and between c and d. The blue graph is below the horizontal axis between b and c. The area below the blue curve and above the horizontal axis for inputs from a to b is shaded in blue and labeled A_1. The area above the blue curve and below the horizontal axis for inputs from b to c is shaded in red and labeled A_2. The area below the blue curve and above the horizontal axis for inputs from c to d is shaded in blue and labeled A_3. +

diff --git a/source/S_Precalc_3Dcoord.ptx b/source/S_Precalc_3Dcoord.ptx index aa36bb0aa..c94f1f9a0 100644 --- a/source/S_Precalc_3Dcoord.ptx +++ b/source/S_Precalc_3Dcoord.ptx @@ -38,13 +38,22 @@ Notes to the Instructor and Dependencies

- This section only expects students to have familiarity with measuring distance in two dimensions. This can likely be covered in around 50 minutes and should offer opportunities for users to get used to the 3D interactable plots and the structure of in class activities. + This section only expects students to have familiarity with measuring distance in two dimensions. This can likely be covered in around 40 minutes and should offer opportunities for users to get used to the 3D interactable plots and the structure of in class activities. +

+

+ Some instructors may want to emphasize the idea of right and left handed coordinates more than others. In particular, many students in computer science have used left handed coordinate systems as used in many game engines. Similarly, many physics and engineering students will be expected to use an orientation similar to . +

+

+ While the discussion of how coordinates are measured in two dimensional rectangular coordinates is an idea that many students think they can skip past, we are using this idea to make sure everyone is on common ground with the particular geometric measurements that define coordinates. Many students struggle with later work on coordinates and coordinate systems because they do not understand the underlying geometric foundation. +

+

+ The discussion of octants is another place where some instructors may skip over this idea, but we have elected to use this as a way to generalize the very useful idea of quadrants from the Cartesian plane. Similarly, activities like ask student to attempt to draw in three dimensions. While this is usually very difficult, some instructors may use these kinds of activities as a step toward growing students ability to draw and communicate about figures in 3D, while other instructors may omit these tasks entirely. I personally (this is Nick talking) have found tremendous benefit to spending time and effort in getting students (and me) to work on drawing and understanding figures in three dimensions. Talking about surface and directional derivatives is significantly easier if everyone has some similar mental pictures and can represent those key ideas in their own hand. I think this argument applies to the descriptions of fundamental planes given in this section. The TL;DR version is that while you can save time by not doing most of this section, the cost is that many students will struggle with more difficult visualization and communication tasks.

Introduction

- The functions you are most familiar with have the form y=f(x) where x is the input or domain variable and y is viewed as the output variable. The graphs of these functions are plotted on a coordinate plane with the input expressed in terms of the horizontal coordinate and the output as the vertical coordinate. The following preview activity asks you to recall some algebraic and geometric ideas from before your single variable calculus course. + The functions you are most familiar with have the form y=f(x), where x is the input or domain variable and y is viewed as the output variable. The graphs of these functions are plotted on a coordinate plane with the input expressed in terms of the horizontal coordinate and the output as the vertical coordinate. The following preview activity asks you to recall some algebraic and geometric ideas from before your single variable calculus course.

@@ -75,6 +84,11 @@ (3,1)

+ +

+ (3,1) +

+ @@ -92,6 +106,21 @@ + +

+ + + plot99=point((0,1),color="black",size=20,gridlines=True,aspect_ratio=1,axes=true,axes_labels=["$x$","$y$"],xmin=-4,xmax=4,ymin=-4,ymax=4)+text("$P_1$",(0.25,1.25),fontsize=20,color="black") + plot99+=point((1,0),color="black",size=20)+text("$P_2$",(1.25,0.25),fontsize=20,color="black") + plot99+=point((2,-3),color="black",size=20)+text("$P_3$",(2.25,-3.25),fontsize=20,color="black") + plot99+=point((2,-3),color="black",size=20)+text("$P_3$",(2.25,-3.25),fontsize=20,color="black") + plot99+=point((3,-2),color="black",size=20)+text("$P_4$",(3.25,-2.25),fontsize=20,color="black") + plot99+=point((-3,2),color="black",size=20)+text("$P_5$",(-3.25,2.25),fontsize=20,color="black") + plot99 + + +

+ @@ -99,6 +128,11 @@ Give the coordinates of four points on the horizontal axis. What aspect do all of the points on the horizontal axis have in common? Use this idea to write an equation for the horizontal axis.

+ +

+ All points on the horizontal axis have a second coordinate of 0, so any points of the form (a,0) will work. Thus the equation for the horizontal axis is y=0. +

+ @@ -116,6 +150,20 @@ + + + + var('t') + plot9=parametric_plot((1,t),(t,-4,4),color="green",gridlines=True,aspect_ratio=1,axes=true,axes_labels=["$x$","$y$"],xmin=-4,xmax=4,ymin=-4,ymax=4) + plot9+=text("$x=1$",(1.75,1.75),color="green",fontsize=20) + plot9+=parametric_plot((t,-2),(t,-4,4),color="red") + plot9+=text("$y=-2$",(-1.75,-1.75),color="red",fontsize=20) + plot9+=parametric_plot((t,-t),(t,-4,4),color="blue") + plot9+=text("$-x=y$",(-2.75,1.75),color="blue",fontsize=20) + plot9 + + + @@ -123,6 +171,18 @@ Draw a plot of the points (-1,2) and (7,-4). On your plot, draw the line segment that measures the distance between the given points and the segments that measure the horizontal and vertical changes. Your plotted segments should make a right triangle.

+ + + + var('t') + plot9=line([(-1,2),(7,-4)],color="black",gridlines=True,aspect_ratio=1,axes=true,axes_labels=["$x$","$y$"]) + plot9+=parametric_plot((t,2),(t,-1,7),color="red") + plot9+=parametric_plot((7,t),(t,-4,2),color="blue") + plot9+=point([(-1,2),(7,-4)],color="black",size=20) + plot9 + + + @@ -130,6 +190,14 @@ Find the length of each of line segments in the plot that is your answer to . Explain how the right triangle idea you drew can be generalized to find the distance between two points (x_1,y_1) and (x_2,y_2).

+ +

+ The length of the vertical segment, shown in blue, is 6. The length of the horizontal segment, shown in red, is 8. The length of the diagonal segment, shown in black, is 10 and can be calculated with a pythagorean theorem like argument. +

+

+ In general the length of the vertical segment between (x_1,y_1) and (x_2,y_2) will be \abs{y_2-y_1}. The length of the horizontal segment between (x_1,y_1) and (x_2,y_2) will be \abs{x_2-x_1}. The length of the diagonal segment between (x_1,y_1) and (x_2,y_2) will be \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. +

+ @@ -309,7 +377,7 @@ A plot of the point (1,-2,3)

A plot with perpendicular arrows labeled x, y, and z. A point is labeled P with coordinates (1,-2,3) and a vertical green segment from the point to the xy-plane. A red segment from the intersection of the green segment with the xy-plane and at the y-axis is shown and labeled with 1. A blue segment from the intersection of the green segment with the xy-plane and the x-axis is shown and labeled with a -2.

 - + from sage.plot.plot3d.plot3d import axes var('t,s') @interact @@ -369,6 +437,11 @@ (6,-2,5)

+ +

+ (6,-2,5) +

+ @@ -376,11 +449,25 @@ Draw the point A (that is, your answer to ()) on a set of three-dimensional axes and include the line segments that show that coordinate's points as a set of directions from the origin (like in with the Show Segments option).

- -

- -

- + + + + from sage.plot.plot3d.plot3d import axes + var('t,s') + scale=7 + plot14=axes(4,2,color="black",frame=False,aspect_ratio=1)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + plot14+=line([(scale,0,0),(0,0,0),(0,-scale,0)],color="gray",thickness=2) + plot14+=point3d((6,-2,5),size=50,color="black") + plot14+=text3d("(6,-2,5)",(6.25,-3.25,3.25),color="black",fontsize=25) + plot14+=parametric_plot3d((t,-2,0),(t,0,6),thickness=3,color="red")+parametric_plot3d((6,-t,0),(t,0,2),color="blue",thickness=3) + plot14+=parametric_plot3d((6,-2,t),(t,0,5),color="green",thickness=3) + plot14+=text3d("-2",(6.5,-1,0),color="blue",fontsize=25) + plot14+=text3d("6",(3.5,-1.5,0),color="red",fontsize=25) + plot14+=text3d("5",(1.25,-2.25,2.5),color="green",fontsize=25) + show(plot14) + + + @@ -393,11 +480,37 @@ (-1,0,-2)

+ +

+ (-1,0,-2) +

+ -

- Draw the point B (that is, your answer to ()) on a set of three-dimensional axes and include the line segments that show that coordinate's points as a set of directions from the origin (like in with the Show Segments option). -

+ +

+ Draw the point B (that is, your answer to ()) on a set of three-dimensional axes and include the line segments that show that coordinate's points as a set of directions from the origin (like in with the Show Segments option). +

+ + + + + from sage.plot.plot3d.plot3d import axes + var('t,s') + scale=3 + plot14=axes(4,2,color="black",frame=False,aspect_ratio=1)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + plot14+=line([(-scale,0,0),(0,0,0),(0,0,-scale)],color="gray",thickness=2) + plot14+=point3d((-1,0,-2),size=50,color="black") + plot14+=text3d("(-1,0,-2)",(-1.25,0.25,-2.25),color="black",fontsize=25) + plot14+=parametric_plot3d((-t,0,0),(t,0,1),thickness=3,color="red") + plot14+=parametric_plot3d((-1,0,-t),(t,0,2),color="green",thickness=3) + plot14+=text3d("0",(-1.5,-1,0),color="blue",fontsize=25) + plot14+=text3d("-1",(-1,-0.5,0),color="red",fontsize=25) + plot14+=text3d("-2",(-1.25,-0.25,-1.5),color="green",fontsize=25) + show(plot14) + + + @@ -429,8 +542,8 @@ For instance, the graph of x=2 in two dimensions will be a vertical line with x-intercept of 2, because the points of the form (2,y) satisfy the equation x=2 (for any choice of y \in \mathbb{R}) as shown in . In the next activity we consider the three-dimensional analogs of these kinds equations.

- - + +

@@ -482,17 +595,16 @@

- - +

illustrates that equations where one of the rectangular coordinates is held constant lead to planes parallel to the coordinate planes. When we make the constant 0, we get the coordinate planes themselves coordinate planeplanecoordinate. The xy-plane satisfies z=0, the xz-plane satisfies y=0, and the yz-plane satisfies x=0 (see ). Planes of the form x=a, y=b, z=c are called fundamental planesfundamental planeplanefundamental are useful in understanding and building structures in three dimensions. You can see how the intersection of fundamental planes x=1, y=-2, and z=3 corresponds to the point (1,-2,3) in . As always, you should rotate and change the viewpoint of the figure to make sure you understand how each fundamental plane and the point P are oriented relative to the axes.

A grid of fundamental planes in three dimensions -

An interactive plot in three dimensions that shows a plane in red one unit to the left of the y and z axes, a plane in blue that is two units behind the x and z axes, and a plane in green that is three units above the x and y axes. All three planes go through a point labels as P and with coordinates (1,-2,3).

- +

An interactive plot in three dimensions that shows a plane in red one unit to the left of the y and z axes, a plane in blue that is two units behind the x and z axes, and a plane in green that is three units above the x and y axes. All three planes go through a point labels as P and with coordinates (1,-2,3).

+ @@ -503,8 +615,8 @@
A grid of fundamental planes in three dimensions -

A grid of planes that are parallel to the xy-plane in green, parallel to the xz-plane in blue, and parallel to the yz-plane in red.

- +

A grid of planes that are parallel to the xy-plane in green, parallel to the xz-plane in blue, and parallel to the yz-plane in red.

 + from sage.plot.plot3d.plot3d import axes var('t,s') @interact @@ -521,25 +633,15 @@
+ + + Distance in Three Dimensions +

+ We conclude this section by using our knowledge of how to measure straight-line distance in \R^2 to find a formula for distance in three dimensions. On a related note, we define a circle in \R^2 as the set of all points equidistant from a fixed point. In \R^3, we call the set of all points equidistant from a fixed point a spherespheredefinition. To find the equation of a sphere, we need to understand how to calculate the distance between two points in three-space, and we explore this idea in the next activity. +

- - - - - Distance in Three Dimensions -

- We conclude this section by using our knowledge of how to measure straight-line distance in \R^2 to find a formula for distance in three dimensions. On a related note, we define a circle in \R^2 as the set of all points equidistant from a fixed point. In \R^3, we call the set of all points equidistant from a fixed point a spherespheredefinition. To find the equation of a sphere, we need to understand how to calculate the distance between two points in three-space, and we explore this idea in the next activity. -

- - - + +

Remember that in two dimensions, the distance between P=(x_0, y_0) and Q=(x_1, y_1) is @@ -570,10 +672,6 @@

- @@ -652,45 +750,44 @@ - -

- The method used in does not depend on anything but the coordinates between the two points, so we can use the last result to measure the distance between any two points in \mathbb{R}^3. -

- - + +

+ The method used in does not depend on anything but the coordinates between the two points, so we can use the last result to measure the distance between any two points in \mathbb{R}^3. +

+ The distance between points in three dimensions -

- The distance between points P=(x_0, y_0, z_0) and Q=(x_1, y_1, z_1) (denoted as |PQ|) in \R^3 is given by the formula +

+ The distance between points P=(x_0, y_0, z_0) and Q=(x_1, y_1, z_1) (denoted as |PQ|) in \R^3 is given by the formula \vert{PQ}\vert = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2 + (z_1-z_0)^2} -

- - -

- As showed, the distance in two or three (or more!) dimensions depends on the change in each coordinate from one point to the other. Note that the distance does not depend on whether we consider P to Q or Q to P.

+ -

- Equation can be used to derive the equation for a spheresphereequation of centered at a point (x_0,y_0,z_0) with radius R. Since the distance from any point (x,y,z) on such a sphere to the center (x_0,y_0,z_0) is R, the point (x,y,z) must satisfy the equation - - \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} = R - -

+

+ As showed, the distance in two or three (or more!) dimensions depends on the change in each coordinate from one point to the other. Note that the distance does not depend on whether we consider P to Q or Q to P. +

-

- Squaring both sides, we come to the standard equation for a sphere. -

+

+ Equation can be used to derive the equation for a spheresphereequation of centered at a point (x_0,y_0,z_0) with radius R. Since the distance from any point (x,y,z) on such a sphere to the center (x_0,y_0,z_0) is R, the point (x,y,z) must satisfy the equation + + \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} = R + +

+ +

+ Squaring both sides, we come to the standard equation for a sphere. +

- + The equation of a sphere -

- The equation of a sphere with center (x_0,y_0,z_0) and radius R is +

+ The equation of a sphere with center (x_0,y_0,z_0) and radius R is (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = R^2 -

- +

+

We see a strong similarity when we compare this equation to its two-dimensional analogue, the equation of a circle of radius R in the plane centered at (x_0,y_0): @@ -719,9 +816,7 @@

- - diff --git a/source/S_Vector_FluxIntegrals.ptx b/source/S_Vector_FluxIntegrals.ptx index 461d522aa..22ba71473 100644 --- a/source/S_Vector_FluxIntegrals.ptx +++ b/source/S_Vector_FluxIntegrals.ptx @@ -38,7 +38,7 @@ \vr_s=\frac{\partial \vr}{\partial s}=\langle{f_s,g_s,h_s}\rangle which measures the direction and magnitude of change in the coordinates of the surface when only s is varied. Similarly, the vector in yellow is \vr_t=\frac{\partial \vr}{\partial t}=\langle{f_t,g_t,h_t}\rangle which measures the direction and magnitude of change in the coordinates of the surface when only t is varied. We also plot the parallelogram that is formed by \vr_s and \vr_t, which is tangent to the surface. The area of this parallelogram offers an approximation for the surface area of a patch of the surface.

- A three-dimensional plot of + A three-dimensional plot showing how a parallelogram can be used to approximate a section of a curved surface

An interactive three-dimensional plot of a curved surface that is parabolic in one coordinate direction and cubic in the other. A grid of magenta and yellow traces along the surface are drawn. Two sliders at the top of the figure allow the user to change the location of a highlighted point on the surface. At the highlighted point, a vector is shown in magenta that is tilted to be tangent to the magenta trace through the highlighted point. A yellow vector is also drawn tangent to the yellow trace through the highlighted point. A yellow parallelogram is drawn with sides given by the magenta and yellow vectors.

@@ -296,7 +296,7 @@

- As with understanding line integrals of vector-valued functions in , we don't care about the output of the vector field at points away from the surface. We would really would like to examine the output vectors for the points on our surface. To do this, we will look at , which plots the output of our vector field at an array of points on our surface. + As with understanding line integrals of vector-valued functions in , we don't care about the output of the vector field at points away from the surface. We would really would like to examine the output vectors for the points on our surface. To do this, we will look at , which plots the output of our vector field at an array of points on our surface.

A three-dimensional vector field evaluated along a surface diff --git a/source/S_Vector_param_surf.ptx b/source/S_Vector_param_surf.ptx index 4e2ea4007..ce3634346 100644 --- a/source/S_Vector_param_surf.ptx +++ b/source/S_Vector_param_surf.ptx @@ -23,7 +23,7 @@

  • - How can we use parameterization of curved surfaces to compute surface area? + How can parameterizations of curved surfaces be used to compute surface area?

    • @@ -31,19 +31,19 @@ Notes to the Instructor and Dependencies

    - + This section was previously discussed in the chapter on multiple integration, but was not used until surface integrals were defined in Chapter 13.

    Introduction

    - Our work in and used parameterizations of curves in two or three dimensions to create and use many tools to describe points and other properties of the curves. Remember that a parameterization of a curve, C, is a vector-valued function of one variable, \vr(t) = \langle x(t), y(t), z(t) \rangle, such that the terminal point of the output vectors of \vr(t). Parameterizations of a curve also need to include bounds on the parameter t that describe where the curve begins and ends, if applicable. + Our work in and used parameterizations of curves in two or three dimensions to create and use tools describing points and other properties of the curves. Remember that a parameterization of a curve, C, is a vector-valued function of one variable, \vr(t) = \langle x(t), y(t), z(t) \rangle, such that the terminal points of the output vectors of \vr(t) will trace out a given curve in space. Parameterizations of a curve must include bounds on the parameter t that describe where the curve begins and ends, when applicable. The graph of a vector-valued function of one variable often does not include plots of the position vectors, but will plot only the terminal points of the output vectors as a curve in space.

    - Parameterizations of curves enabled us to easily generate points, use vector and calculus tools (like direction of travel, direction of turning, arc length, curvature, etc.), and describe motion along the curve. In each of these uses, the parameterization reduced our curve, as a set of points in space, to a function of one variable and we have a LOT of tools for functions of one variable. In particular, curves like circles, ellipses, hyperbolas, and parabolas are often not able to be expressed with one variable as a function of the others, but these curves can be studied with 1-variable calculus through our parameterizations. + Parameterizations of curves enabled us to easily generate points, use vector and calculus tools (like direction of travel, direction of turning, arc length, curvature, etc.), and describe motion along the curve. In each of these uses, the parameterization reduced our curve, as a set of points in space, to a function of one variable, which is so valuable because we have a LOT of tools for functions of one variable. In particular, curves like circles, ellipses, hyperbolas, and parabolas cannot be expressed with one variable as a function of the others, but these curves can be studied with 1-variable calculus through our use of parameterizations.

    - One of the central goals of this section is to extend the idea of parameterizations to curved surfaces in an attempt to apply the many multivariable tools of and . In the following preview activity, we will look at how we can use some geometric knowledge of a cone and the parameterization of a circle to create a vector-valued function that will describe the points on the surface of a cone. + One of the central goals of this section is to extend the idea of parameterizations to curved surfaces in an attempt to apply the many multivariable tools of and . In the following preview activity, we will look at how we can use geometric knowledge of right circular cylinders and cones to create a vector-valued function that will describe the points on the surface of a cone.

    -     Parameterizations of Surfaces

    - In , we concluded with a description of a cone in terms of measurements corresponding to the angle around the z-axis and the height above the xy-plane. + In , we finished with a description of a cone in terms of measurements corresponding to the angle around the z-axis and the height above the xy-plane. We used geometric intuition to come up with the coordinate relationships of our two surfaces in . We will spend some time in this subsection talking about how to use algebraic and geometric tools to give parameterizations of curved surfaces. The next subsection will talk about some of the important geometric measurements that can be done with a parameterization of a curved surface. In the last subsection, we will look at how parameterizations are useful in computing the surface area of a curved surface in three dimensions.

    - In a single-variable setting, any function may have its graph expressed parametrically. For instance, given y = g(x), by considering the parameterization \langle t, g(t) \rangle (where t belongs to the domain of g), we generate the same curve. What is more important is that certain curves that are not functions may be represented parametrically; for instance, the circle (which cannot be represented by a single function) can be parameterized by \langle \cos(t), \sin(t) \rangle, where 0 \le t \le 2\pi. + In a single-variable setting, any function may have its graph expressed parametrically. For instance, if we look the graph of y = g(x), we can consider the parameterization \langle t, g(t) \rangle (where t belongs to the domain of g), which will generate the same curve. Certain curves that are not expressible with y as a function of x may be represented parametrically; for instance, the circle (which cannot be represented with either the x or y-coordinate written as a function of the other) can be parameterized by \langle \cos(t), \sin(t) \rangle, where 0 \leq t \lt 2\pi.

    - In the same way, in a two-variable setting, the surface z = f(x,y) may be expressed parametrically by considering + When we look at a surface of the form z = f(x,y), we can express the (x,y,z) points on the surface parametrically by \langle x(s,t), y(s,t), z(s,t) \rangle = \langle s, t, f(s,t) \rangle, - where (s,t) varies over the entire domain of f. Therefore, any familiar surface that we have studied so far can be generated as a parametric surface. But what is more powerful is that there are surfaces that cannot be generated by a single function z = f(x,y) (such as the unit sphere), but that can be represented parametrically. We now consider an important example. + where (s,t) varies over the entire domain of f. In the parameterization above, we can think of s as acting like x and t as acting like y. Therefore, any familiar surface expressed as z=f(x,y) that we have studied so far can be generated as a parametric surface. The greater power of parameterizations is realized when dealing with surfaces that cannot be expressed by a single function z = f(x,y) (such as the unit sphere), but can be represented parametrically.

    - - +

    + For surfaces where we cannot express one coordinate as a function of the other two, like ellipsoids or hyperboloids, the best strategy is to use geometric knowledge of these surfaces to express each of the x-, y-, and z-coordinates in terms of two measurements/parameters. In , we use parameters that measures the rotational coordinate (around the z-axis) and a coordinate that was related to the height above or below the xy-plane (along with the radial coordinate in the cone). In the next example, we will look at how to parameterize a torus (think of the surface of a doughnut) using this same strategy. +

    + + +

    - Consider the - torus (or doughnut) shown in Figure. + In this example, we want to parameterize the torus shown in Figure.

    -
    + +
    + The surface of a torus shown centered at the origin with hole oriented around the z-axis + + +

    An 3D plot with axes labeled x, y, and z. A torus is drawn in light blue center at the origin and its inner opening oriented along the z-axis. Circular slices oriented radially from the z-axis are highlighted in green and circular slices centered around the z-axis are highlighted in red.

    +     + + from sage.plot.plot3d.plot3d import axes + var('s,t') + scale=4 + plot1=axes(scale,3,color="black",frame=False)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + a=1; b=2; + param(s,t)=((b+a*cos(s))*cos(t), (b+a*cos(s))*sin(t), a*sin(s)) + #set up grid n is number of steps in s m is number of steps in t + #a1 is min s a2 is max s b1 is min t b2 is max t + n=13;m=13;a1=0;a2=2*pi;b1=0;b2=2*pi; + #c1 is surface color o1 is surface opacity + #c2 and o2 are for constant t value traces + #c3 and o3 are for constant s value traces + c1="lightblue"; c2="red"; c3="green"; o1=0.7; o2=0.3; o3=0.3; + plot1+=parametric_plot3d(param,(s,a1,a2),(t,b1,b2),color=c1,opacity=o1) + for i in range(n+1): + sv=a1+i/n*(a2-a1) + plot1+=parametric_plot3d(param(sv,t),(t,b1,b2),color=c2,opacity=o2) + for i in range(m+1): + tv=b1+i/m*(b2-b1) + plot1+=parametric_plot3d(param(s,tv),(s,a1,a2),color=c3,opacity=o3) + show(plot1,aspect_ratio=1,viewpoint=[[-0.4047,-0.4447,-0.799],107.96],camera_position=[23.3495,2.6514,12.7857]) + +
    -

    - To find a parametrization of this torus, we recall our work in Preview - Activity. There, we saw that a circle of radius r - that has its center at the point (0,0,z_0) and is contained in the - horizontal plane z = z_0, as shown in Figure, can be - parametrized using the vector-valued function \vr defined by - - \vr(t) = r\cos(t)\vi + r\sin(t)\vj + z_0\vk - - where 0\leq t\leq 2\pi. + We need to be able to describe each (x,y,z) point on the torus in terms of two parameters. We can think of the torus as being generated by rotating a circle or radius a around the z-axis. If you look at a radial slice out from the z-axis out, as shown by the gray plane in , we generate the torus by revolving the points on the green circle around the z-axis, as shown by the red circles. This means that we can describe each point on the torus in terms of where each point corresponds to its location along the green circle and its location along a red circle (rotation around the z-axis).

    - -
    - A circle in a horizontal plane centered at (0,0,z_0). - -

    A three dimensional plot with axes labeled x, y, and z. A grid is shown on the xy-plane. A circle parallel to the xy-plane and centered on the z-axis is shown in blue. Line segments from the z-axis to the circle and from the circle to the xy-plane are shown in gray and labeled r and z, respectively.

     - +
    + A slice of a torus shown as the revolution of a circle (shown in green) around the z-axis. + + +

    An 3D plot with axes labeled x, y, and z. A gray plan extends from the z-axis in the direction of positive x and y coordinates. On this gray plane is a green circle, centered on the xy-plane, with 13 points on the circle shown in black. For each of the black points on the green circle is rotated around the z-axis as shown by circles drawn in red.

    +     + + from sage.plot.plot3d.plot3d import axes + var('s,t') + scale=4 + plot1=axes(scale,3,color="black",frame=False)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + a=1; b=2;sc=2*pi/3; tc=pi/4 + param(s,t)=((b+a*cos(s))*cos(t), (b+a*cos(s))*sin(t), a*sin(s)) + #plot1+=parametric_plot3d(param(sc,t),(t,0,2*pi),color='red',thickness=3) + n=13;m=13;a1=0;a2=2*pi;b1=0;b2=2*pi; + #c1 is surface color o1 is surface opacity + #c2 and o2 are for constant t value traces + #c3 and o3 are for constant s value traces + c1="lightblue"; c2="red"; c3="green"; o1=0.7; o2=0.3; o3=0.3; + #plot1+=parametric_plot3d(param,(s,a1,a2),(t,b1,b2),color=c1,opacity=o1) + for i in range(n+1): + sv=a1+i/n*(a2-a1) + plot1+=parametric_plot3d(param(sv,t),(t,b1,b2),color=c2,opacity=o2) + plot1+=sphere(param(sv,tc),color='black',size=0.1) + plot1+=parametric_plot3d(param(s,tc),(s,0,2*pi),color='green',thickness=3) + plot1+=parametric_plot3d((s,s,t),(s,0,3),(t,-2,2),color="gray",opacity=0.4) + plot1 + +
    -

    - To obtain the torus in Figure, we begin with a circle of radius a in the - xz-plane centered at (b,0), as shown - on the left of Figure. We may parametrize - the points on this circle, using the parameter s, by using the equations - - x(s) = b + a\cos(s)\ \mbox{and} \ - z(s) = a\sin(s), - - where 0\leq s \leq 2\pi. + We can think of the green slice of the torus as occurring on a constant value of the cylindrical coordinate \theta. We will use the t parameter to describe our position in terms of rotation around the z-axis; in other words, our parameter t will act as the \theta-coordinate from cylindrical coordinates. We will use our other parameter, s, to describe the location of a point on the green circle.

    - -
    - Revolving a circle to obtain a torus. - - -

    Revolving a circle to obtain a torus.

     - - -

    Revolving a circle to obtain a torus.

     - -     -
    -

    - Let's focus our attention on one point on this circle, such as the - indicated point, which has coordinates (x(s), 0, z(s)) - for a fixed value of the - parameter s. When this point is revolved about the z-axis, we - obtain a circle contained in a horizontal plane centered at - (0,0,z(s)) and having radius x(s), as shown on the right of Figure. If we let t be the new parameter that generates the circle for the rotation about the z-axis, this circle may be - parametrized by - - \vr(s,t) = x(s)\cos(t)\vi + x(s)\sin(t)\vj + z(s)\vk. - + Looking at the slice of the torus shown by the grey plane in , we get a two dimensional plot as shown in . We need to describe points in r and z in terms of the parameter s. Our green circle in the rz-plane will be described by r(s)=b+a \cos(s) and z(s)=\sin(s) with 0\leq s \leq 2\pi. This is a typical parameterization of a circle with a horizontal translation by a.

    - +
    + + + + var('x,t') + a=1; b=2; + sc=2*pi/3; + param(t)=(b+a*cos(t),a*sin(t)) + plot2=parametric_plot((b+a*cos(t),a*sin(t)),(t,0,2*pi),color='green',thickness=2,axes_labels=['','$z$'],ticks=[[],[]],xmin=-1,xmax=4,ymin=-1.5,ymax=1.5) + plot2+=line([(0,a*sin(sc)),param(sc),(b,0)],color="gray",thickness=2) + plot2+=point(param(sc),color="red",faceted=True,zorder=3,size=25) + plot2+=text('$r$',(0.75,1.1),color="gray",fontsize=20) + plot2+=text('$a$',(1.9,0.5),color="gray",fontsize=20) + plot2+=line([(b,0.1),(b,-0.1)],color='gray') + plot2+=text('$b$',(2,-.25),color="gray",fontsize=20) + plot2 + + +

    - Now using our earlier parametric equations for x(s) and z(s) for the original smaller circle, we have an overall parameterization of the torus given by - - \vr(s,t) =(b+a\cos(s))\cos(t)\vi + - (b+a\cos(s))\sin(t)\vj + - a\sin(s)\vk. - + We can use our cylindrical coordinate transformations to write the x, y, and z-coordinates in terms of our parameters s and t. The cylindrical coordinates of our points can be written (in terms of s and t) as + \theta = t \quad r = b+a\cos(s) \quad z=a\sin(s) + So the rectangular coordinates of the points on the torus can be written (in terms of s and t) as + + x \amp= r \cos(\theta) = (b+a \cos(s))*\cos(t) + y \amp= r \sin(\theta) = (b+a \cos(s))*\sin(t) + z \amp= z = a \sin(s) +

    -

    - To trace out the entire torus, we require that the parameters vary through the values 0\leq s\leq 2\pi and - 0\leq t\leq 2\pi. + Just as in the case of parameterizing a curve in space with a vector valued function of one variable, we need to specify bounds on the parameters used to describe our surface. Remember that for a point on the torus, the t-parameter describes the location in terms of rotation around the z-axis and the s-parameter describes the point's location around the circular slice (in the rz-plane). Both the s and t parameters will be bounded below by 0 and above by 2 \pi. We can organize our parametric equations for the torus using a vector valued function of two variables such as + + \vr (s,t) =& \langle x(s,t),y(s,t),z(s,t)\rangle + =& \langle \left(b+a \cos(s)\right)\cos(t), \left(b+a \cos(s)\right)\sin(t), a \sin(s) \rangle + + for s,t \in [0,2 \pi).

    + - +

    + In our next activity, we will look at how to parameterize a sphere centered at the origin two ways: 1) we think of the sphere as being a surface of revolution around the z-axis and 2) we use spherical coordinates. +

    - - + +

    - In this activity, we seek a - parametrization of the sphere of radius R centered at the origin, as - shown on the left in Figure. - Notice that this sphere may be obtained by revolving a half-circle - contained in the xz-plane about the z-axis, as shown on the right. + In this activity, we want to give a parametrization of the sphere of radius R centered at the origin, as shown in . You can check the box at the top of the plot of to highlight what a semicircular slice of the sphere looks like along a slice with a constant \theta coordinate. Notice that this slice shows how the sphere can be made by rotating this half circle around the z-axis.

    - -
    - A sphere obtained by revolving a half-circle. - -

    A sphere obtained by revolving a half-circle.

    -     -

    A sphere obtained by revolving a half-circle.

    -     -     +
    + A plot of a sphere of radius R centered at the origin + + + + + + +

    An interactive three dimensional plot that shows axes labeled x, y, and z. A sphere centered at the origin is shown in light blue. A checkbox above the interactive plot will toggle a half circle plotted along the sphere from the positive z-axis to the negative z-axis.

    +
    - -
      -
    1. -

      - Begin by writing a parametrization of this half-circle using - the parameter s: - - - x(s) = \ldots \ \ \ \ \ \ \ \ \ - z(s) = \ldots. - - Be sure to state the domain of the parameter s. -

      -
      2. - -
    2. -

      - By revolving the points on this half-circle about the - z-axis, obtain a parametrization \vr(s,t) of the points on the - sphere of radius R. Be sure to include the domain of both - parameters s and t. (Hint: What is the radius of the circle obtained when revolving a point on the half-circle around the z axis?) -

      -
      3. - -
    3. -

      - Draw the surface defined by your parameterization with - appropriate technology. -

      -
      4. -
    + + + +

    + Our first approach to parameterizing the sphere of radius R will be similar to our approach to parameterizing the torus as a surface of revolution around the z-axis. In other words, we will let the parameter t represent the location as a rotation around the z-axis. So we will be able to parameterize the sphere with + x(s,t)= r(s) \cos(t) , \quad y(s,t)=r(s) \sin(t), \quad z(s,t)= z(s) + where r(s) and z(s) describe the other cylindrical coordinates in terms of another parameter. +

    +
    + A plot of a half circle in the rz-plane representing the slice of the sphere we will be rotation around the z-axis + + + var('t') + param(t)=(2*cos(t),2*sin(t)) + plot2=parametric_plot((2*cos(t),2*sin(t)),(t,-pi/2,pi/2),color='green',thickness=2,axes_labels=['$r$','$z$'],ticks=[[],[]]) + plot2+=text("$R$",(2.1,0.1),color="black",fontsize=15) + plot2.show(xmin=-0.5,xmax=3,ymin=-2.5,ymax=2.5) + plot2 + + +

    A two dimensional plot with horizontal axis labeled r and vertical axis labeled z. The right half of a circle centered at the origin is shown in green and its radius is labeled R.

    +     + +
    +

    + We want to describe the points on the green half circle shown in in terms of our parameter s. Remember that there is not a unique way to parameterize a curve. Give a parameterization of the green half circle in as a function of the parameter s. Be sure to state the bounds on your parameter. +

    - + +

    + We will use a common parameterization of a circle with r(s)=R \cos(s) and z(s)=R\sin(s) where -\frac{\pi}{2}\leq s \leq \frac{\pi}{2}. +

    +     +     + + +

    + We want to combine your work for the previous task with Equation to generate a parameterization of the sphere of radius R. Plug in your answer for the previous task into Equation and state your answer as parametric functions for x, y, and z. Be sure to state the bounds on your parameters s and t. + + \amp x(s,t) \amp= + \amp y(s,t) \amp= + \amp z(s,t) \amp= + \amp\leq s \amp \leq + \amp\leq t \amp \leq + +

    +     +     + + +

    + Use to verify that your parameterization for the previous task will plot the sphere of radius 3 centered at the origin. Remember to adjust the upper and lower bounds of s and t to match your parameterization. +

    +
    + An interactive plot where users can input their parameterization and bounds + + + var('s,t') + @interact + def paramplot3d2(param_array=input_grid(1,3, default=['(2.5+cos(s))*cos(t)','(2.5+cos(s))*sin(t)','sin(s)'],label='\\( \\langle x(s,t),y(s,t),z(s,t) \\rangle=\\)',width=20), svalues=input_grid(1,2,default=[0,pi],label='Lower and Upper bound on \\( s\\)'), tvalues=input_grid(1,2,default=[0,pi],label='Lower and Upper bound on \\( t\\)')): + from sage.plot.plot3d.plot3d import axes + scale=4.5 + plot1=axes(scale,1,color="black")+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + param(s,t)=(param_array[0][0],param_array[0][1],param_array[0][2]) + #set up grid n is number of steps in s m is number of steps in t + #a1 is min s a2 is max s b1 is min t b2 is max t + n=15;m=11;a1=svalues[0][0];a2=svalues[0][1];b1=tvalues[0][0];b2=tvalues[0][1]; + #c1 is surface color o1 is surface opacity + #c2 and o2 are for constant t value traces + #c3 and o3 are for constant s value traces + c1="lightblue"; c2="gray"; c3="gray"; o1=0.9; o2=0.5; o3=0.5; + plot1+=parametric_plot3d(param(s,t),(s,a1,a2),(t,b1,b2),color=c1,opacity=o1) + for i in range(n+1): + sv=a1+i/n*(a2-a1) + plot1+=parametric_plot3d(param(sv,t),(t,b1,b2),color=c2,opacity=o2) + for i in range(m+1): + tv=b1+i/m*(b2-b1) + plot1+=parametric_plot3d(param(s,tv),(s,a1,a2),color=c3,opacity=o3) + show(plot1,aspect_ratio=1,viewpoint=[[-0.4047,-0.4447,-0.799],107.96],camera_position=[23.3495,2.6514,12.7857]) + + +
    +     +     + + +

    + We now want to shift to using spherical coordinates in order to create a parameterization of the sphere of radius R. In particular, we want to express the rectangular coordinate equation x^2+y^2+z^2=R^2 in spherical coordinates. State an equation in spherical coordinates that describes the same set of points as x^2+y^2+z^2=R^2. +

    +     + +

    + This equation is of the form \rho= \text{constant} +

    +     +     + + +

    + For surfaces of the form z=f(x,y) were easily parameterized because we could write each of the x, y, and z-coordinates in terms of two variables, namely x and y. We can apply a similar idea here because in terms of spherical coordinates, our surface has one spherical coordinate fixed and the other two spherical coordinates can be used as our parameters. Write a couple of sentences to describe which spherical coordinate you want s to act as and which you would like t to act as. You should include the bounds on these parameters as part of your descriptions. +

    +     +     + + +

    + Let's put the last couple of tasks together to come up with a spherical coordinate parameterization of x^2+y^2+z^2=R^2. Specifically, you will need to plug in the constant value for the appropriate spherical coordinate and s or t (as dictated by your choice in the previous task) for the other two spherical coordinates into the following transformation equations. + x=\rho \cos(\theta) \sin(\phi) \quad \quad y=\rho \sin(\theta) \sin(\phi) \quad \quad z=\rho \cos(\phi) + This should give you equations for x, y, and z in terms of the parameters s and t, as well as bounds on both s and t. + x(s,t) = \hspace{1.5in} y(s,t) = \hspace{1.5in} z(s,t)= + with the bounds on s and t given by + \hspace{1.0in} \leq s \leq \hspace{1.0in} \text{ and } \hspace{1.0in} \leq t \leq \hspace{1.0in} +

    +     +     + + +

    + Use to verify that your parameterization of the sphere of radius 3 does parameterize the correct surface. +

    +     +     + + + +

    + As you can see in the previous activity, there are many ways to parameterize the same surface. The parameterizations you generated for the sphere depended on whether you viewed the sphere as a surface of revolution or whether you utilized spherical coordinates. There are many other ways to parameterize a sphere, but these are likely the two most familiar and common ideas. +

    +

    + In the exercises of this section, you will be asked to generate and understand the parameterization of the quadric surfaces and cylinder surfaces described in . Additionally, you will need to understand how shifts and stretches can be easily applied to each rectangular coordinate separately. +

    + + + The Geometry of a Parametric Surface +

    + In this subsection, we will go over several useful ideas about how to view geometric information given by the parameterization of a curved surface. For this section, we will examine a surface S_1, given by a parameterization \vr(s,t)=\langle x(s,t), y(s,t), z(s,t) \rangle with a \leq s \leq b and c \leq t \leq d. +

    +

    + In , we saw how holding one of the input coordinates constant allowed us to restrict the graph of z=f(x,y) to a one variable slice which we called the trace. In particular, we could look at the trace along x=a which is given by z=f(a,y), for example . Restricting our focus to one slice allowed us to define partial derivatives and many other tools in . +

    +

    + What happens when we hold one of our parameters constant and allow the other to vary? We will get a slice of the parametric surface but this will not be the direction of one coordinate (x, y, or z), but will describe the surface in a constant direction of one of our parameters, s or t. The trace given by holding s constant will yield \vr(a,t), a curve parameterized as a vector valued function of one variable (in this case, t). We can use all of our tools from to understand this curve. Specifically, the derivative, \frac{d\vr}{dt} (a,t)=\langle \frac{dx}{dt}(a,t),\frac{dy}{dt}(a,t),\frac{dz}{dt}(a,t)\rangle will give a vector that is tangent to the curve given by \vr(a,t). +

    +

    + Similarly, we can look at the trace generated in the direction of a constant value of t. We will get a trace parameterized by \vr(s,b) with a tangent vector given by \frac{d\vr}{ds}(s,b). +

    +

    + In , you can see plot of our parametric surface given by \vr(s,t) for s_1 \leq s \leq s_2 and t_1 \leq t \leq t_2 plotted as a light blue surface. There are five constant values of t plotted along the surface in shown by the magenta curves and five constant values of s plotted along the surface as shown by the yellow curves. The tangent vectors given by \frac{d\vr}{ds} (s,b) and \frac{d\vr}{dt} (a,t) are shown at a particular point on the surface. You can use the sliders at the top of to change the location at which the tangent vectors are shown. +

    +
    + A three-dimensional plot of a parametric surface with traces corresponding to constant s values highlighted in yellow and constant t values highlighted in magenta. The tangent vector corresponding to \frac{d \vr}{dt} (a,t) is shown with the yellow arrow and the tangent vector corresponding to \frac{d \vr}{ds} (s,b) is shown with the red arrow. + + + var('t,s,u,v,z,y,z') + f=-x^2/2-3*y^3/3 + @interact + def _(a=slider(-1,1,.5,0,label="s value",display_value=False),b=slider(-1,1,.5,.5,label="t value")): + plot1=plot3d(f,(x,-1,1),(y,-1,1),frame=False,opacity=.3)+sphere((a,b,f(x=a,y=b)),size=.05,color="black") + for i in range(0,5): + plot1+=parametric_plot3d((s,-1+i/2,f(x=s,y=-1+i/2)),(s,-1,1),color="#D81B60") + plot1+=parametric_plot3d((-1+i/2,s,f(y=s,x=-1+i/2)),(s,-1,1),color="#FFC107") + f_val=f(x=a,y=b) + pos=(a,b,f_val) + fx=diff(f,x)(x=a,y=b) + fy=diff(f,y)(x=a,y=b) + vt=(1/2+a,b,1/2*fx+f_val) + vs=(a,b+1/2,1/2*fy+f_val) + cr=(a-fx/2,b-fy/2,1/2+f_val) + vectt=arrow3d(pos,vt,color="#D81B60") + vects=arrow3d(pos,vs,color="#FFC107") + plot1+=vectt+vects + plot1+=parametric_plot3d((a+t/2,b+s/2,f_val+t/2*fx+s/2*fy),(t,0,1),(s,0,1),color="orange",opacity=.6) + plot1+=arrow3d(pos,cr,color="#004D40") + show(plot1) + + +

    An interactive three-dimensional plot of a curved surface that is parabolic in one coordinate direction and cubic in the other. A grid of magenta and yellow traces along the surface are drawn. Two sliders at the top of the figure allow the user to change the location of a highlighted point on the surface. At the highlighted point, a vector is shown in magenta that is tilted to be tangent to the magenta trace through the highlighted point. A yellow vector is also drawn tangent to the yellow trace through the highlighted point. A yellow parallelogram is drawn with sides given by the magenta and yellow vectors.

    +
    +

    + We can use the red and yellow tangent vectors at the highlighted point (shown in black) to create the tangent plane at the highlighted point. A parallelogram of the tangent plane is shown in orange. Note that the orange parallelogram will move and tilt to estimate the curved surface near the highlighted point. We can use the properties of the cross product to both compute a vector that will be orthogonal to the curved surface at the highlighted point and find the area of the orange parallelogram ( and ). +

    +

    + We take a few moments now to summarize the ideas above, but state our derivatives of \vr(s,t) as partial derivatives with respect to s or t, since those will correspond to holding the t and s parameters constant (respectively). +

      +
    • +

      + Curves in the direction of change of s and t are given by vector valued functions \vr(s,b) and \vr(a,t), respectively. +

      +
      • +
    • +

      + Vectors tangent to the parametric surface given by \vr(s,t) through the point given by \vr(a,b) will be given by \frac{\partial \vr}{\partial s}(a,b)= \vr_s(a,b) and \frac{\partial \vr}{\partial t}(a,b)= \vr_t(a,b). +

      +
      • +
    • +

      + The vector given by (\vr_s \times \vr_t)(a,b) will be orthogonal to surface at the point \vr(a,b) and \vecmag{(\vr_s \times \vr_t)(a,b)} will give the area of the the parallelogram along the tangent plane to the surface with edges \vr_s(a,b) and \vr_t(a,b). +

      +
      • +
    +

    + +

    + In this example, we will look at each of the properties above as they apply to the parameterization of the torus from . The points on the torus will be given by the parameterization: + + \vr (s,t) =& \langle x(s,t),y(s,t),z(s,t)\rangle + =& \langle (b+a \cos(s)) \cos(t), (b+a \cos(s)) \sin(t), a \sin(s) \rangle + + for s,t \in [0,2 \pi). +

    +
    + + + + var('t,s,u,v,z,y,z') + import numpy as np + @interact + def _(a=slider(0,2*pi,pi/6,default=pi/3,label="s value"),b=slider(0,2*pi,pi/6,default=pi/3,label="t value")): + from sage.plot.plot3d.plot3d import axes + scale=4 + plot1=axes(scale,1,color="black",frame=False)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + a1=1; b1=2;sc=2*pi/3; tc=pi/4 + param(s,t)=((b1+a1*cos(s))*cos(t), (b1+a1*cos(s))*sin(t), a1*sin(s)) + plot1+=parametric_plot3d(param(s,t),(s,0,2*pi),(t,0,2*pi),frame=False,opacity=.3)+sphere(param(a,b),size=.05,color="black") + for i in range(0,12): + plot1+=parametric_plot3d(param(s,2*pi*i/12),(s,0,2*pi),color="#D81B60",opacity=0.5) + plot1+=parametric_plot3d(param(2*pi*i/12,s),(s,0,2*pi),color="#FFC107",opacity=0.5) + pos=param(a,b) + rs=diff(param,s)(s=a,t=b)*pi/6 + rt=diff(param,t)(s=a,t=b)*pi/6 + cr=np.cross(rt,rs) + vectt=arrow3d(pos,rs+pos,color="#D81B60",size=3) + vects=arrow3d(pos,rt+pos,color="#FFC107") + plot1+=vectt+vects + tp(s,t)=t*rt+s*rs+pos + plot1+=parametric_plot3d(tp,(t,0,1),(s,0,1),color="orange",opacity=.8) + plot1+=arrow3d(pos,cr+pos,color="#004D40") + show(plot1) + + +
    +
      +
    • +

      + The traces with constant values of t are shown in magenta on and are parameterized by \vr(s,t_0)= \langle (b+a \cos(s)) \cos(t_0), (b+a \cos(s)) \sin(t_0), a \sin(s) \rangle when t=t_0. +

      +
      • +
    • +

      + The traces with constant values of s are shown in yellow on and are parameterized by \vr(s_0,t)= \langle (b+a \cos(s_0)) \cos(t), (b+a \cos(s_0)) \sin(t), a \sin(s_0) \rangle when s=s_0. +

      +
      • +
    • +

      + The vector tangent to the surface along the trace with constant t-value is given by \vr_s=\langle (-a\sin(s))\cos(t), (-a\sin(s))\sin(t),0\rangle and is shown with the red arrow on . +

      +
      • +
    • +

      + The vector tangent to the surface along the trace with constant s-value is given by \vr_t=\langle (a+b\cos(s))(-\sin(t)),(a+b\sin(s))(\cos(t)),0 \rangle and is shown with the yellow arrow on . +

      +
      • +
    • +

      + The green vector is calculated by \vr_s \times \vr_t, is orthogonal to the surface at the highlighted point, and the length of the green vector corresponds to the area of the orange parallelogram. Note that as the length of the yellow vector increases, the parallelogram's area and the length of the green vector both increase correspondingly. +

      +
      • +
    +     - The Surface Area of Parametrically Defined Surfaces

    - Recall that a differentiable function is locally linear that is, if we zoom in on the surface around a point, the surface looks like its tangent plane. We now exploit this idea in order to determine the surface area generated by a parametrization \langle x(s,t), y(s,t), z(s,t) \rangle. The basic idea is a familiar one: we will subdivide the surface into small pieces, in the approximate shape of small parallelograms, and thus estimate the entire the surface area by adding the areas of these approximation parallelograms. Ultimately, we use an integral to sum these approximations and determine the exact surface area. + In this subsection, we will use our new tool of parameterizing curved surfaces and a classic calculus approach to look at how to compute the surface area of a curved surface. For step one of our classic calculus approach, we will break our parametrized surface into pieces that correspond to a grid of s and t steps. On each of these pieces of our st-grid, we will approximate the surface area of our curved surface with the area of a parallelogram tangent to the surface at that point. The sum of these parallelogram areas will give us the approximation we seek. For step two of our classic calculus approach, we will look at how smaller step sizes in s and t appears as part of the sum used in the approximation. The third step of our classic calculus approach will be to use a limit of the surface area approximation to get a Riemann sum definition for a double integral.

    -

    - Let + For this development, we will use a parametrized surface given by \vr(s,t) = x(s,t) \vi + y(s,t) \vj + z(s,t) \vk - define a surface over a rectangular domain a \leq s \leq b and c \leq t \leq d. As a function of two variables, s and t, it is natural to consider the two partial derivatives of the vector-valued function \vr, which we define by + with bounds on our parameters given by a \leq s \leq b and c \leq t \leq d. In this case, we are using a rectangular domain in the st-plane but in general, this same approach will work for non-rectangular domains. The partial derivatives of this parameterization, \vr_s(s,t) \amp = x_s(s,t) \vi + y_s(s,t) \vj + z_s(s,t) \vk - \vr_t(s,t) \amp = x_t(s,t) \vi + y_t(s,t) \vj + z_t(s,t) \vk. + \vr_t(s,t) \amp = x_t(s,t) \vi + y_t(s,t) \vj + z_t(s,t) \vk + will give vectors tangent to the surface in the direction of changes in s and t, respectively.

    -

    - In the usual way, we slice the domain into small rectangles. In particular, we partition the interval [a,b] into m subintervals of length \Delta s = \frac{b-a}{n} and let s_0, s_1, \ldots, s_m be the endpoints of these subintervals, where a = s_0\lt s_1\lt s_2 \lt \cdots \lt s_m = b. Also partition the interval [c,d] into n subintervals of equal length \Delta t = \frac{d-c}{n} and let t_0, t_1, \ldots, t_n be the endpoints of these subintervals, where c = t_0\lt t_1\lt t_2 \lt \cdots \lt t_n = d. These two partitions create a partition of the rectangle R = [a,b] \times [c,d] in st-coordinates into mn sub-rectangles R_{ij} with opposite vertices (s_{i-1},t_{j-1}) and (s_i, t_j) for i between 1 and m and j between 1 and n. These rectangles all have equal area \Delta A = \Delta s \cdot \Delta t. + We will break our section of the curved surface into steps in s and t. In particular, we partition the interval of s-values [a,b] into m subintervals of length \Delta s = \frac{b-a}{m} and let s_0, s_1, \ldots, s_m be the endpoints of these subintervals, where a = s_0\lt s_1\lt s_2 \lt \cdots \lt s_m = b. We also partition the interval of t-values [c,d] into n subintervals of equal length \Delta t = \frac{d-c}{n} and let t_0, t_1, \ldots, t_n be the endpoints of these subintervals, where c = t_0\lt t_1\lt t_2 \lt \cdots \lt t_n = d. +

    +

    + These subintervals partition the rectangle R = [a,b] \times [c,d] in st-coordinates into mn sub-rectangles R_{ij} with opposite vertices (s_{i-1},t_{j-1}) and (s_i, t_j) for i between 1 and m and j between 1 and n. These rectangles all have equal area \Delta A = \Delta s \cdot \Delta t.

    -

    - Now we want to think about the small piece of area on the surface itself that lies above one of these small rectangles in the domain. - Observe that if we increase s by a small amount \Delta s from the point - (s_{i-1},t_{j-1}) in the domain, then \vr changes by approximately - \vr_s(s_{i-1},t_{j-1}) \Delta s. Similarly, if we increase t by a - small amount \Delta t from the point (s_{i-1},t_{j-1}), then \vr - changes by approximately \vr_t(s_{i-1},t_{j-1}) \Delta t. So we can - approximate the surface defined by \vr on the st-rectangle - [s_{i-1},s_i] \times [t_{j-1}, t_{j}] with the - parallelogram determined by the vectors \vr_s(s_{i-1},t_{j-1}) \Delta - s and \vr_t(s_{i-1},t_{j-1}) \Delta t, as seen in Figure. + Each of these rectangles from the st-domain will split our curved surface into a m by n grid of pieces, but these pieces of the curved surface will not all be the same size (in surface area). Looking at the surface shown in , we can see the traces given for five equally spaced values of s and t breaks our curved surface into a four by four grid of pieces. Note that these pieces of the curved surface do not have the same surface area. So we will need to estimate the surface area for each particular piece of our grid using our tools from the parameterization.

    - +

    + Each of the pieces of the curved surface correspond to increasing s by a small amount \Delta s or increasing t by a small amount \Delta t from the point (s_{i-1},t_{j-1}) in the st-parameter plane. We want to find the vectors corresponding to each of the sides of the orange parallelogram in . In particular, the vector \vr_t \Delta t will be the vector that corresponds to the side of the parallelogram along a constant value of s (shown as a yellow vector in ) and the vector \vr_s \Delta s will be the vector that corresponds to the side of the parallelogram along a constant value of t (shown as a red vector in ). +

    +

    + The area of the parallelogram will then be S_{ij}=\vecmag{(\vr_s \Delta s) \times (\vr_t \Delta t)} evaluated at the point given by (s_{i-1},t_{j-1}). This will be the approximation of the surface area for a piece of our grid of the curved surface. We can simplify this magnitude of a cross product to be S_{ij}=\vecmag{\vr_s \times \vr_t} \Delta s \Delta t, which means that our approximation over all of these pieces of the grid will sum to + SA\approx \sum_{i=1}^m \sum_{j=1}^n S_{ij}= \sum_{i=1}^m \sum_{j=1}^n \vecmag{\vr_s(s_{i-1},t_{j-1}) \times \vr_t(s_{i-1},t_{j-1})} \Delta s \Delta t. +

    +

    + This has satisfied both the first and second steps of the classic calculus approach because we have approximated the surface area and quantified how that approximation will change with a smaller scale of steps (in s and t). Note that this approximation takes the form of a Riemann sum corresponding to the setup of a double integral. In particular, if we take the limit as \Delta s and \Delta t go to zero, our Riemann sum will be come a double integral of \vecmag{\vr_s \times \vr_t} over the rectangular region of integration with a \leq s \leq b and c \leq t \leq d, which gives the following formula for calculating the surface area. +

    + + + Surface area +

    + Let \vr(s,t) = \langle x(s,t), y(s,t), z(s,t) \rangle be a parameterization of a smooth surface over a domain D. The area of the surfacesurface area defined by \vr on D is given by + + S = \iint_D \vecmag{\vr_s \times \vr_t } \ dA. + +

    +     +

    - Say that the small parallelogram has area S_{ij}. If we can find its area, then all that remains is to sum the areas of all of the generated parallelograms and take a limit. Recall from our earlier work in the course that given two vectors \vu and \vv, the area of the parallelogram spanned by \vu and \vv is given by the magnitude of their cross product, \vecmag{ \vu \times \vv}. In the present context, it follows that the area, S_{ij}, of the parallelogram determined by the vectors \vr_s(s_{i-1},t_{j-1}) \Delta s and \vr_t(s_{i-1},t_{j-1}) \Delta t is - - - S_{ij} \amp = \vecmag{(\vr_s(s_{i-1},t_{j-1}) \Delta s) \times (\vr_t(s_{i-1},t_{j-1}) \Delta t)} - - - \amp = \vecmag{\vr_s(s_{i-1},t_{j-1}) \times \vr_t(s_{i-1},t_{j-1})} \Delta s \Delta t, - - - where the latter equality holds from standard properties of the cross product and length. + In this example, we will use our parameterizations for the sphere of radius R from to calculate the surface area of a sphere. Our first approach to parameterizing the sphere was to consider the sphere as a surface of revolution around the z-axis. This gave the parameterization of + \vr(s,t) = \langle \sqrt{R^2-s^2}\cos(t), \sqrt{R^2-s^2}\sin(t), s \rangle + with s acting as the z-coordinate, going from R to -R, and t acting as the \theta, going from 0 to 2\pi.

    -

    - We sum the surface area approximations from Equation over all sub-rectangles to obtain an estimate for the total surface area, S, given by - - S \approx \sum_{i=1}^m \sum_{j=1}^n \vecmag{\vr_s(s_{i-1},t_{j-1}) \times \vr_t(s_{i-1},t_{j-1})} \Delta s \Delta t. - + We can calculate \vr_s, \vr_t, \vecmag{\vr_s \times \vr_t} as follows: + + \vr_s \amp= \left\langle \frac{-2s}{\sqrt{R^2-s^2}}\cos(t), \frac{-2s}{\sqrt{R^2-s^2}}\sin(t), 1 \right\rangle + \vr_t \amp= \left\langle -\sqrt{R^2-s^2}\sin(t), \sqrt{R^2-s^2}\cos(t), 0 \right\rangle + \vr_s \times \vr_t \amp= \left\langle -\sqrt{R^2-s^2} \cos(t), -\sqrt{R^2-s} \sin(t), -\frac{2s}{\sqrt{R^2-s^2}} \right\rangle + \vecmag{\vr_s \times \vr_t} \amp= \sqrt{(R^2-s^2) \cos^2(t)+(R^2-s) \sin^2(t)+ \frac{4s^2}{R^2-s^2}} + + Unfortunately, our expression for \vecmag{\vr_s \times \vr_t} will not simplify much more and we will need to use a trig substitution to compute the appropriate double integral. In particular, the surface area double integral will be + \int_{-R}^R \int_0^{2\pi} \sqrt{(R^2-s^2)+ \frac{4s^2}{R^2-s^2}} dt \emskip ds . + We will omit the calculations that result from this and look at using our other parameterization to get a much easier calculation of the surface area of a sphere.

    -

    - Taking the limit as m, n \to \infty shows that the surface area of the surface defined by \vr over the domain D is given as follows. + We can now use our parameterization of a sphere that came from using spherical coordinates. + \vr(s,t) = \langle R \sin(s) \cos(t), R \sin(s)\sin(t), R\cos(s) \rangle + with s acting as the \phi-coordinate, going from 0 to 2\pi, and t acting as the \theta, going from 0 to 2\pi. +

    +

    + We can calculate \vr_s, \vr_t, \vecmag{\vr_s \times \vr_t} as follows: + + \vr_s \amp= \langle R \cos(s) \cos(t), R \cos(s)\sin(t), -R\sin(s) \rangle + \vr_t \amp= \langle -R \sin(s) \sin(t), R \sin(s) \cos(t), 0 \rangle + \vr_s \times \vr_t \amp= \langle R^2 \sin^2(s) \cos(t), R^2 \sin^2(s) \sin(t), R^2 \cos^2(t) \cos(s)\sin(s)+R^2 \sin^2(t) \cos(s)\sin(s) \rangle + \amp= R^2\sin(s) \langle \sin(s) \cos(t), \sin(s) \sin(t), \cos(s) \rangle + \vecmag{\vr_s \times \vr_t} \amp= R^2\sin(s) + + You may recognize this expression from our work on the volume element in spherical coordinates, . In both this case and the volume element of spherical coordinates, we are measuring how warped our space is based on transformations using spherical coordinates.

    - - - Surface area

    - Let \vr(s,t) = \langle x(s,t), y(s,t), z(s,t) \rangle be a parameterization of a smooth surface over a domain D. The area of the surfacesurface area defined by \vr on D is given by - - S = \iint_D \vecmag{\vr_s \times \vr_t } \ dA. - + Our double integral will then be + \int_{0}^\pi \int_0^{2\pi} R^2\sin(s) dt \. ds + which will evaluate to + R^2 (2 \pi) (-\cos(s)\restrict{_{s=0}^{s=\pi}})= R^2 (2 \pi)(2)=4\pi R^2 + You may recognize this result as the formula you were likely handed (without explanation) for the surface area of a sphere.

    -     - +     - - -

    - Consider the cylinder with radius a and height h defined parametrically by + + +

    + In this activity, we will compute the surface area of a right circular cylinder. In particular, we will consider the cylinder with radius a and height h defined parametrically by \vr(s,t) = a\cos(s) \vi + a\sin(s) \vj + t \vk for 0 \leq s \leq 2\pi and 0 \leq t \leq h, as shown in Figure. -

    - +

    A cylinder.

    A cylinder.
    -

    -

      -
    1. -

      - Set up an iterated integral to determine the surface area of this cylinder. -

      -
      2. - -
    2. -

      - Evaluate the iterated integral. -

      -
      3. - -
    3. -

      - Recall that one way to think about the surface area of a cylinder is to cut the cylinder horizontally and find the perimeter of the resulting cross sectional circle, then multiply by the height. Calculate the surface area of the given cylinder using this alternate approach, and compare your work in (b). -

      -
      4. -
    -

    + + + +

    + Calculate \vr_s, \vr_t, \vecmag{\vr_s \times \vr_t} based on the parameterization given above. +

    -

      -
    1. -

      - We have - - \vr_s(s,t) = -a\sin(s) \vi + a\cos(s) \vj \ \text{ and } \ \vr_t(s,t) = \vk - , - so the area of the surface of the cylinder is - - \int \int_D \vecmag{\vr_s \times \vr_t} \, dA = \int \int_D a \, dA - . - In this case, dA = ds \ dt, - so an iterated integral that represents the area of the surface is - - \int \int_D \vecmag{\vr_s \times \vr_t} \, dA = \int_{0}^{h} \int_{0}^{2 \pi} a \, ds \, dt - . -

      -
      2. -
    2. -

      - We can find the surface area of the cylinder by multiplying the circumference of the circle of radius a by the height h to obtain 2 \pi a h. - Evaluating the iterated integral yields - - \int_{0}^{h} \int_{0}^{2 \pi} a^2 \, ds \, dt \amp = a \int_{0}^{h} s \restrict{0}^{2 \pi} \, dt - \amp = 2 \pi a \int_{0}^{h} \, dt - \amp = 2 \pi a t \restrict{0}^{h} - \amp = 2 \pi a h - . -

      -
      3. -
    3. -

      - If we slice the cylinder horizontally, - a cross section is a circle of radius a. - The circumference of this circle is 2 \pi a. - We multiply by the height of the cylinder to obtain the surface area as 2 \pi a h as expected. -

      -
      4. -
    + We have + + \vr_s(s,t) = -a\sin(s) \vi + a\cos(s) \vj \ \text{ and } \ \vr_t(s,t) = \vk + + so \vecmag{\vr_s \times \vr_t} will simplify to a

    -     + + + +

    + Use the calculations from the previous task to set up an iterated integral to determine the surface area of this cylinder. +

    +     + +

    + So the area of the surface of the cylinder is + + \int \int_D \vecmag{\vr_s \times \vr_t} \, dA = \int \int_D a \, dA + . + In this case, dA = ds \ dt, + so an iterated integral that represents the area of the surface is + + \int \int_D \vecmag{\vr_s \times \vr_t} \, dA = \int_{0}^{h} \int_{0}^{2 \pi} a \, ds \, dt + +

    +     +     + + +

    + Evaluate your iterated integral from the previous task. +

    +     + +

    + 2 \pi a h +

    +     +     + + +

    + One way to think about the surface area of a cylinder is to cut the cylinder horizontally and find the perimeter of the resulting cross sectional circle, then multiply by the height. Calculate the surface area of the given cylinder using this alternate approach, and compare your result the value from the previous task. +

    +     + +

    + If we slice the cylinder horizontally, a cross section is a circle of radius a. The circumference of this circle is 2 \pi a. We multiply by the height of the cylinder to obtain the surface area as 2 \pi a h as expected. +

    +     +     +     -

    - As we noted earlier, we can take any surface z = f(x,y) and generate a corresponding parameterization for the surface by writing \langle s, t, f(s,t) \rangle. Hence, we can use our recent work with parametrically defined surfaces to find the surface area that is generated by a function f = f(x,y) over a given domain. -

    +

    + As we noted earlier, we can take any surface z = f(x,y) and generate a corresponding parameterization for the surface by writing \langle s, t, f(s,t) \rangle. Hence, we can use our recent work with parametrically defined surfaces to find the surface area that is generated by a function f = f(x,y) over a given domain. +

    - + + +

    + Let z = f(x,y) define a smooth surface, and consider the corresponding parameterization \vr(s,t) = \langle s, t, f(s,t) \rangle. +

    +     +

    - Let z = f(x,y) define a smooth surface, and consider the corresponding parameterization \vr(s,t) = \langle s, t, f(s,t) \rangle. -

      -
    1. -

      - Let D be a region in the domain of f. Using Equation, show that the area, S, of the surface defined by the graph of f over D is + Let D be a region in the domain of f. Using Equation, show that the area, S, of the surface defined by the graph of f over D is S = \iint_D \sqrt{\left(f_x(x,y)\right)^2 + \left(f_y(x,y)\right)^2 + 1} \ dA. -

      -
      2. - -
    2. -

      - Use the formula developed in (a) to calculate the area of the surface defined by f(x,y) = \sqrt{4-x^2} over the rectangle D = [-2,2] \times [0,3]. -

      -
      3. - -
    3. -

      - Observe that the surface of the solid describe in (b) is half of a circular cylinder. Use the standard formula for the surface area of a cylinder to calculate the surface area in a different way, and compare your result from (b). -

      -
      4. -
    -

    +

    -

      -
    1. -

      - Recall that any surface defined by a function - f = f(x,y) in Cartesian coordinates can be thought of as a surface defined parametrically with parameters s and t by - - x(s,t)=s, \ \ \ \ y(s,t) = t, \ \ \ \ \text{ and } \ \ \ \ z = f(s,t) - . - The surface is then given by the vector-valued function \vr with \vr(s,t) = \langle s, t, f(s,t) \rangle. - In this case we have - - \vr_s(s,t) = \langle 1,0,f_s(s,t)\rangle \ \ \ \ \text{ and } \ \ \ \ \vr_t(s,t) = \langle 0,1,f_t(s,t)\rangle - . - So - - \lvert \vr_s \times \vr_t \rvert = \lvert \langle -f_s(s,t), -f_t(s,t), 1 \rangle \rvert = \sqrt{f_s(s,t)^2 + f_t(s,t)^2+1} - , - and the formula for surface area becomes - - \int \int_D \sqrt{\left(f_x(x,y)\right)^2 + \left(f_y(x,y)\right)^2 + 1} \ dA - . -

      -
      2. -
    2. -

      - Using the formula from part (a) we have that the surface area is - - \int_{-2}^2 \int_0^3 \sqrt{\left(-x(4-x^2)^{-1/2}\right)^2 + 0^2 + 1} \, dy \, dx \amp = \int_{-2}^2 \int_0^3 \sqrt{\frac{x^2}{4-x^2}+1} \, dy \, dx - \amp = \int_{-2}^2 \int_0^3 \frac{2}{\sqrt{4-x^2}} \, dy \, dx - \amp = \int_{-2}^2 \int_0^3 \frac{2}{\sqrt{4-x^2}}y\restrict0^3 \, dx - \amp = 6\int_{-2}^2 \frac{1}{\sqrt{4-x^2}} \, dx - \amp = 6 \arcsin{\frac{x}{2}}\restrict{-2}^2 - \amp = 6\left(\arcsin(1)-\arcsin(-1)\right) - \amp = 6\pi - . -

      -
      3. -
    3. -

      - The cross sections of the surface of this solid are semicircles with radius 2. - So the surface area of this solid should be 2\pi \times 3 = 6\pi. -

      -
      4. -
    + Recall that any surface defined by a function f = f(x,y) in Cartesian coordinates can be thought of as a surface defined parametrically with parameters s and t by + + x(s,t)=s, \ \ \ \ y(s,t) = t, \ \ \ \ \text{ and } \ \ \ \ z = f(s,t) + . + The surface is then given by the vector-valued function \vr with \vr(s,t) = \langle s, t, f(s,t) \rangle. In this case we have + + \vr_s(s,t) = \langle 1,0,f_s(s,t)\rangle \ \ \ \ \text{ and } \ \ \ \ \vr_t(s,t) = \langle 0,1,f_t(s,t)\rangle + . + So + + \lvert \vr_s \times \vr_t \rvert = \lvert \langle -f_s(s,t), -f_t(s,t), 1 \rangle \rvert = \sqrt{f_s(s,t)^2 + f_t(s,t)^2+1} + , + and the formula for surface area becomes + + \int \int_D \sqrt{\left(f_x(x,y)\right)^2 + \left(f_y(x,y)\right)^2 + 1} \ dA + .

    +     + + +

    + Use the formula developed in (a) to calculate the area of the surface defined by f(x,y) = \sqrt{4-x^2} over the rectangle D = [-2,2] \times [0,3]. +

    +     + +

    + Using the formula from part (a) we have that the surface area is + + \int_{-2}^2 \int_0^3 \sqrt{\left(-x(4-x^2)^{-1/2}\right)^2 + 0^2 + 1} \, dy \, dx \amp = \int_{-2}^2 \int_0^3 \sqrt{\frac{x^2}{4-x^2}+1} \, dy \, dx + \amp = \int_{-2}^2 \int_0^3 \frac{2}{\sqrt{4-x^2}} \, dy \, dx + \amp = \int_{-2}^2 \int_0^3 \frac{2}{\sqrt{4-x^2}}y\restrict0^3 \, dx + \amp = 6\int_{-2}^2 \frac{1}{\sqrt{4-x^2}} \, dx + \amp = 6 \arcsin{\frac{x}{2}}\restrict{-2}^2 + \amp = 6\left(\arcsin(1)-\arcsin(-1)\right) + \amp = 6\pi + . +

    +     +     + + +

    + Observe that the surface of the solid describe in (b) is half of a circular cylinder. Use the standard formula for the surface area of a cylinder to calculate the surface area in a different way, and compare your result from (b). +

    +     + +

    + The cross sections of the surface of this solid are semicircles with radius 2. So the surface area of this solid should be 2\pi \times 3 = 6\pi. +

    +     +     - - Summary +

    • @@ -617,7 +910,7 @@

    -     + diff --git a/source/S_mult_ints_double_ints_rects.ptx b/source/S_mult_ints_double_ints_rects.ptx index 53aa554ff..55828f8c1 100644 --- a/source/S_mult_ints_double_ints_rects.ptx +++ b/source/S_mult_ints_double_ints_rects.ptx @@ -46,7 +46,7 @@ In single-variable calculus, recall that we used the classic calculus approach to define the definite integral as the area under a graph. Specifically, we approximated the area under the graph of a positive function f on an interval [a,b] by adding areas of rectangles whose heights are determined by the curve. We then broke the interval [a,b] into smaller subintervals, constructing rectangles on each of these smaller intervals to approximate the region under the curve on that subinterval, then summing the areas of these rectangles to approximate the area under the curve. We defined the definite integral of f using the limit of this Riemann sum as the size of all of the subintervals goes to zero.

    - In the Preview Activity, we will review a few ideas from integrals in single variable calculus. The rest of this section will then be used to extend the ideas of integration and its interpretations to functions of two variables over a rectangular region. + In the Preview Activity, we will review a few ideas from integrals in single-variable calculus. The rest of this section will then be used to extend the ideas of integration and its interpretations to functions of two variables over a rectangular region.

    @@ -138,10 +138,10 @@ Double Riemann Sums over Rectangles

    - The motivating interpretation of a definite integral for a function of one variable was the area under a curve over a particular interval of input values. We will use the same motivating interpretation for the definite integral of function of two variables. The definite integral of a nonnegative function of two variables will measure the volume of the region beneath the graph of f over a region of input points. The definite integral for a function of two variables is called the double integral of the function f. We can show this geometrically as the volume of the solid below the surface given by z=f(x,y) as shown in . The surface given by z=f(x,y) is shown in blue and the region of inputs above which we want to find the volume is shown in red in the xy-plane. The double integral of f over the red region should measure the volume of the solid shaded in gray. + The motivating interpretation of a definite integral for a function of one variable was the area under a curve over a particular interval of input values. We will use the same motivating interpretation for the definite integral of a function of two variables. The definite integral of a nonnegative function of two variables will measure the volume of the region beneath the graph of f over a region of input points. The definite integral for a function of two variables is called the double integral of the function f. We can show this geometrically as the volume of the solid below the surface given by z=f(x,y) as shown in . The surface given by z=f(x,y) is shown in blue and the region of inputs above which we want to find the volume is shown in red in the xy-plane. The double integral of f over the red region should measure the volume of the solid shaded in gray.

    - A plot of the volume under as surface z=f(x,y) over a region R, in red + A plot of the volume under a surface z=f(x,y) over a region R, in red

    A plot of the volume under as surface z=f(x,y) over a region R, in red

     @@ -247,7 +247,7 @@

    - To understand the numerical calculations involved in the classic calculus approach for a double integral, it is most important to understand the region of integration. Thus, we will not look a graph of z=f(x,y). Instead, we will stay focused in the xy-plane. On the axes below, outline the rectangular region R that corresponds to the region of integration. + To understand the numerical calculations involved in the classic calculus approach for a double integral, it is most important to understand the region of integration. Thus, we will not look at a graph of z=f(x,y). Instead, we will stay focused in the xy-plane. On the axes below, outline the rectangular region R that corresponds to the region of integration.

    @@ -394,7 +394,7 @@

    Write a sentence about why the volume of each rectangular prism used for this approximation is f(x_i,y_j) \Delta A. - Write a couple sentences about how you would find an approximation of the volume under the surface z=f(x,y) over the region R. (Do not do calculation, but rather explain what calculation is being done.) + Remember that x_i and y_j are the points from parts c and d of this activity. Write a couple sentences about how you would find an approximation of the volume under the surface z=f(x,y) over the region R. (Do not do calculation, but rather explain what calculation is being done.)

    @@ -649,7 +649,7 @@ Signed Volume

    - Suppose that f(x,y) assumes both positive and negatives values on the rectangle R, as shown in Figure. When constructing a Riemann sum, for each i and j, the product f(x_{ij}^*, y_{ij}^*) \enspace \Delta A can be interpreted as a signed volume of a box with base area \Delta A and signed height f(x_{ij}^*, y_{ij}^*). Since f can have negative values, this height could be negative. The sum + Suppose that f(x,y) assumes both positive and negative values on the rectangle R, as shown in Figure. When constructing a Riemann sum, for each i and j, the product f(x_{ij}^*, y_{ij}^*) \enspace \Delta A can be interpreted as a signed volume of a box with base area \Delta A and signed height f(x_{ij}^*, y_{ij}^*). Since f can have negative values, this height could be negative. The sum \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \enspace \Delta A @@ -711,7 +711,7 @@

    - The double integral of f over the rectangle R will be positive because there is more volume above the xy-plane (shown in blue) thanvolume below the xy-plane, which is counted as negative and shown in red. + The double integral of f over the rectangle R will be positive because there is more volume above the xy-plane (shown in blue) than volume below the xy-plane, which is counted as negative and shown in red.

    diff --git a/source/S_mult_ints_triple_ints.ptx b/source/S_mult_ints_triple_ints.ptx index 1852d9822..a6aa10eaa 100644 --- a/source/S_mult_ints_triple_ints.ptx +++ b/source/S_mult_ints_triple_ints.ptx @@ -817,7 +817,6 @@
  • - The triple integral has several useful interpretations:

    For a solid region S in \R^3, the following are common and important interpretations of triple integrals over S:

      diff --git a/source/S_mv_deriv_directional_deriv.ptx b/source/S_mv_deriv_directional_deriv.ptx index d1295a743..e3d10fb8d 100644 --- a/source/S_mv_deriv_directional_deriv.ptx +++ b/source/S_mv_deriv_directional_deriv.ptx @@ -40,7 +40,7 @@ Introduction

      - The partial derivatives of a multivariable function tell us the instantaneous rate at which the function's output changes as we hold all but one input variable constant (allowing the remaining input variable to change). It is natural to wonder how we can measure the rate at which a function changes in a direction other than parallel to a coordinate axes. In this section, we investigate this question and will connect the rates of change in other directions to the rates of change given by the partial derivatives. The Preview Activity investigates these concepts in terms of a contour map representing elevation in terms of location. + The partial derivatives of a multivariable function tell us the instantaneous rate at which the function's output changes as we hold all but one input variable constant (allowing the remaining input variable to change). It is natural to wonder how we can measure the rate at which a function changes in a direction other than parallel to a coordinate axis. In this section, we investigate this question and will connect the rates of change in other directions to the rates of change given by the partial derivatives. The Preview Activity investigates these concepts in terms of a contour map representing elevation in terms of location.

      We can use cardinal directions to specify the direction of displacement vectors. These directions can be described by a compass rose. The compass rose given in is an example of a sixteen point compass rose. Directions of the form ESE are read as east-southeast and point in the direction halfway between east and southeast. @@ -267,7 +267,7 @@

      - We will look at an example that is algebraically and geometrically simple but allows us to use . We will consider the rate of change for the function f(x,y)=2.5-\frac{(x-1)^2}{2}-\frac{(y+1)^2}{9} at the input point (2,2) in several directions. Because f(2,2)=2.5-\frac{(2-1)^2}{2}-\frac{(2-2)^2}{9}=2.5-0.5-1=1, this point lies on the level curve with value 1 as shown in . + We will look at an example that is algebraically and geometrically simple but allows us to use . We will consider the rate of change for the function f(x,y)=2.5-\frac{(x-1)^2}{2}-\frac{(y+1)^2}{9} at the input point (2,2) in several directions. Because f(2,2)=2.5-\frac{(2-1)^2}{2}-\frac{(2+1)^2}{9}=2.5-0.5-1=1, this point lies on the level curve with value 1 as shown in .

      A contour plot of f(x,y)=2.5-\frac{(x-1)^2}{2}-\frac{(y+1)^2}{9} with the point (2,2) highlighted @@ -290,7 +290,7 @@ We will compute the directional derivative of f at the point (2,2) in the direction of \langle-2,-1\rangle. Since \langle-2,1\rangle is not a unit vector, we use \vu=\frac{1}{\sqrt{5}}\langle -2,1\rangle as the unit vector in this direction. To use equation, we will need the partial derivatives of f: f_x= -\frac{2(x-1)}{2} \quad \text{ and } \quad f_y=-\frac{2(y+1)}{9} At (2,2), we have f_x(2,2)=-1 and f_y(2,2)=-\frac{2}{3}. By equation, the directional derivative of f at (2,2) in the direction \vu is - Df_\vu (2,2)=f_x(2,2) u_1 + f_y(2,2) u_2 = (-1)(-\frac{2}{\sqrt{5}})+\left(-\frac{2}{3}\right)(\frac{1}{\sqrt{5}})=\frac{4}{3\sqrt{5}} . + Df_\vu (2,2)=f_x(2,2) u_1 + f_y(2,2) u_2 = \left(-1\right)\left(-\frac{2}{\sqrt{5}}\right)+\left(-\frac{2}{3}\right)\left(\frac{1}{\sqrt{5}}\right)=\frac{4}{3\sqrt{5}} . We can interpret this as saying that for a small step in the direction of \vu at (2,2), we would expect the output of f to increase by \frac{4}{3\sqrt{5}} times the step size.

      @@ -306,7 +306,7 @@

      - Remember that the directional derivative is a local measurement and only applies in a small neighborhood of a point and only in the direction \vu. In other words, Df_\vu(2,2) does not describe the the rate of change at other points along the blue curve, only the rate of change along the blue curve for a small step away from our base point in the red direction. With plots like , you may be tempted to use information far away from the point of interest to try to figure out the rate of change. However, like most calculus measurements, we must only use plots to provide information on what is happening near a specific point. + Remember that the directional derivative is a local measurement and only applies in a small neighborhood of a point and only in the direction \vu. In other words, Df_\vu(2,2) does not describe the rate of change at other points along the blue curve, only the rate of change along the blue curve for a small step away from our base point in the red direction. With plots like , you may be tempted to use information far away from the point of interest to try to figure out the rate of change. However, like most calculus measurements, we must only use plots to provide information on what is happening near a specific point.

      If we compute Df_\vj(2,2), the rate of change for the output of f in the direction \vj=\langle 0,1\rangle at (2,2), using equation, then we have @@ -713,8 +713,8 @@ The first statement explains why the gradient is perpendicular to the level curve through the point of interest. Because a level curve is the set of points for which the function has a particular output value, the output will not change along the level curve. Thus, the directional derivative in a direction tangent to the level curve must be zero. We can expand this explanation to the other statements as well. The output of f increases in any direction that makes an acute angle with the gradient vector and the output of f decreases in any direction that makes an obtuse angle with the gradient.

      - - + +

      For this activity, we will look at the value of the directional derivative for temperature as a function of location. When looking at a plot of thermoclines (locations with the same temperature), you notice that your town is exactly on the thermocline labeled as 70 degrees, shown as the point P on @@ -827,7 +827,9 @@

      - + + +

      We will look algebraically calculating and explaining the vector properties of the gradient using the function f(x,y)=xy-2y^2+\frac{x^2}{2}+2y-x at the input (1,1). @@ -867,6 +869,7 @@ +

      Note that the third and fifth bullets above are algebraic statements of the second and fourth bullets, respectively.

      @@ -956,7 +959,7 @@ A heat-seeking missile will always travel in the direction in which the temperature increases most rapidly; that is, it will always travel in the direction of the gradient \nabla T. If a missile is fired from the point (2,4), then its path will be that shown on the right in Figure.

      - This type strategy is sometimes called gradient ascent. The gradient decent method is used to find relative minimums for functions by taking steps in the direction opposite the gradient. Gradient decent methods are used extensively in economics and machine learning to find where the difference between predictions from a model and data are as small as possible (minimize error). + This type of strategy is sometimes called gradient ascent. The gradient decent method is used to find relative minimums for functions by taking steps in the direction opposite the gradient. Gradient decent methods are used extensively in economics and machine learning to find where the difference between predictions from a model and data are as small as possible (minimize error).

      diff --git a/source/S_mv_deriv_optimization.ptx b/source/S_mv_deriv_optimization.ptx index 9e202a9dd..82c3cdc7f 100644 --- a/source/S_mv_deriv_optimization.ptx +++ b/source/S_mv_deriv_optimization.ptx @@ -230,7 +230,7 @@

      - In the analogy of , these critical points correspond to locations where our level will measure no change in elevation for every direction. How can we ensure that directional derivative will be zero in every direction? If \nabla f =\vec{0}, then Df_{\vu} = \vec{0}\cdot\vu = 0 for every direction. Geometrically, if \nabla f =\vec{0} at a point (x_0,y_0), then the tangent plane at this point will be horizontal, meaning its equation has the form z=c for a constant c. + In the analogy of , these critical points correspond to locations where our level will measure no change in elevation for every direction. How can we ensure that the directional derivative will be zero in every direction? If \nabla f =\vec{0}, then D_{\vu} f = \vec{0}\cdot\vu = 0 for every direction. Geometrically, if \nabla f =\vec{0} at a point (x_0,y_0), then the tangent plane at this point will be horizontal, meaning its equation has the form z=c for a constant c.

      This means that we can find critical points of a function f by computing partial derivatives and identifying any values of (x,y) for which one of the partials doesn't exist or for which both partial derivatives are simultaneously zero. For the latter, note that we have to solve the system of equations @@ -651,10 +651,10 @@ We begin with the convenient directions in which to look at concavity: parallel to the x- and y-axes. Recall from that f_{xx} and f_{yy} measure the concavity of the traces in the x and y directions, respectively. If f_{xx} \gt 0 at a critical point, then in the direction parallel to the x-axis, the trace curve is concave up. If f_{xx} \lt 0 at a critical point, then in the direction parallel to the x-axis, the trace curve is concave down. Similar statements hold for f_{yy}.

      - Suppose that f_{xx} (a,b) \lt 0 and f_{yy} \gt 0 at a critical point (a,b). Because the concavity is different for two different direction, the critical point (a,b) must be a saddle point. Unfortunately, if f_{xx} (a,b) \gt 0 and f_{yy} \gt 0 at a critical point (a,b), then we do not necessarily have a local minimum, as shown in . + Suppose that f_{xx} \lt 0 and f_{yy} \gt 0 at a critical point (a,b). Because the concavity is different for two different direction, the critical point (a,b) must be a saddle point. Unfortunately, if f_{xx} \gt 0 and f_{yy} \gt 0 at a critical point (a,b), then we do not necessarily have a local minimum, as shown in .

      - A surface plot where f_{xx} (a,b) \gt 0 and f_{yy} \gt 0 at the origin. + A surface plot where f_{xx} \gt 0 and f_{yy} \gt 0 at the origin.

      A three dimensional plot with axes labeled x, y, and z and a grid on the xy-plane. A curved surface is shown in light blue. The surface curves up along the directions given by the x and y axes and down in directions where x=\pm y. The upward curved path of the surface above the x and y axes is highlighted in red.

      diff --git a/source/S_mv_functions.ptx b/source/S_mv_functions.ptx index 67a39a14a..a591e9f5b 100644 --- a/source/S_mv_functions.ptx +++ b/source/S_mv_functions.ptx @@ -622,7 +622,7 @@
      - + Traces

      When studying functions of several variables, we are often interested in how each individual variable affects the function while the other variable is fixed. In Preview Activity, we saw that the amount of money in an account depends on the amount initially invested and the duration of the investment. However, if we fix the initial investment, the amount of money in the account depends only on the duration of the investment, and if we fix the duration of the investment, then the amount of money in the account depends only on the initial investment. This idea of keeping one variable constant while we allow the other to change will be an important tool for us when studying functions of several variables. This will be the first of many times we will employ the following approach

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      diff --git a/source/acmv-index.ptx b/source/acmv-index.ptx index 99055ca7f..2932f1eec 100644 --- a/source/acmv-index.ptx +++ b/source/acmv-index.ptx @@ -183,7 +183,7 @@ Permission is granted to copy, distribute and/or modify this document under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. The work may be used for free by any party so long as attribution is given to the author(s), the work and its derivatives are used in the spirit of share and share alike; no party may sell this work or any of its derivatives for profit. All trademarks are the registered marks of their respective owners. The graphic -

      ADD ALT TEXT TO THIS IMAGE

       +

      Creative Commons logo

             that may appear in other locations in the text shows that the work is licensed with the Creative Commons and that the work may be used for free by any party so long as attribution is given to the author(s) and if the material is modified, the resulting contributions are distributed under the same license as this original. Full details may be found by visiting https://creativecommons.org/licenses/by-sa/4.0/ or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.       diff --git a/source/exercises/ez-S_multi_ints_double_ints_param_surf.ptx b/source/exercises/ez-S_multi_ints_double_ints_param_surf.ptx index abc5015fc..36678c8c0 100644 --- a/source/exercises/ez-S_multi_ints_double_ints_param_surf.ptx +++ b/source/exercises/ez-S_multi_ints_double_ints_param_surf.ptx @@ -60,7 +60,7 @@

      - In Activity, we found that a parameterization of the sphere S of radius R centered at the origin is + In Activity, we found that a parameterization of the sphere S of radius R centered at the origin is x(r,s) = R\cos(s) \cos(t), \ y(s,t) = R \cos(s) \sin(t), \ \text{ and } \ z(s,t) = R\sin(s) @@ -129,7 +129,7 @@

      1. - Set up an iterated integral whose value is the portion of the surface area of a sphere of radius R that lies in the first octant (see the parameterization you developed in Activity). + Set up an iterated integral whose value is the portion of the surface area of a sphere of radius R that lies in the first octant (see the parameterization you developed in Activity).

        2. @@ -158,7 +158,7 @@

        1. - From Activity + From Activity we know that a parameterization of a sphere of radius R centered at the origin is x(s,t) = R \cos(s) \cos(t), \ y(s,t) = R\sin(s) \cos(t), \ \text{ and } \ z(s,t) = R \sin(t) @@ -289,7 +289,7 @@

          1. - Find a parameterization of the cone described by z = \frac{h}{a} \sqrt{x^2+y^2}. (Hint: Compare to the parameterization of a cylinder as seen in Activity.) + Find a parameterization of the cone described by z = \frac{h}{a} \sqrt{x^2+y^2}. (Hint: Compare to the parameterization of a cylinder as seen in Activity.)

            2. diff --git a/source/exercises/ez-S_mv_DirectionalDerivatives.ptx b/source/exercises/ez-S_mv_DirectionalDerivatives.ptx index eb959d1e7..7bf6b270f 100644 --- a/source/exercises/ez-S_mv_DirectionalDerivatives.ptx +++ b/source/exercises/ez-S_mv_DirectionalDerivatives.ptx @@ -1,38 +1,22 @@ - - - - + - - - - + - - - - + - - - - - - - - + @@ -111,7 +95,7 @@
            A plot of f(x,y) with the measurements of change in input and output labeled for a change of inputs given by \langle a,b\rangle -

            A plot of f(x,y) with the measurements of change in input and output labeled for a change of inputs given by \langle a,b\rangle

            +

            A plot of a surface of the form z=f(x,y) is shown in light blue where the surface has a constant height of 2 above the y-axis and is decreasing as the y-coordinate moves away from 0. Additionally, the surface decreases as the x-coordinate gets closer to 0. The point on the surface above the input (-2,1) is highlighted in blue.

             @@ -174,7 +158,7 @@

          -

          ADD ALT TEXT TO THIS IMAGE

           +

          A plot of a surface of the form z=f(x,y) is shown in light blue where the surface has a constant height of 2 above the y-axis and is decreasing as the y-coordinate moves away from 0. Additionally, the surface decreases as the x-coordinate gets closer to 0.

          @@ -193,7 +177,7 @@

        -

        ADD ALT TEXT TO THIS IMAGE

         +

        A plot of a surface of the form z=f(x,y) is shown in light blue where the surface has a constant height of 2 above the y-axis and is decreasing as the y-coordinate moves away from 0. Additionally, the surface decreases as the x-coordinate gets closer to 0.

        @@ -491,7 +475,7 @@ An illustration of this direction at the point (1,2,f(1,2)) is shown in the figure below, with the gradient vector scaled appropriately to make it easier to plot. -

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         +

        The gradient of \nabla f is plotted

      2. @@ -500,7 +484,7 @@ The gradient is perpendicular to the level curves, and a picture of the unit vector in the direction of the gradient at (1,2) is shown in the figure below, with the gradient vector scaled appropriately to make it easier to plot. -

        ADD ALT TEXT TO THIS IMAGE

         +

        The gradient of \nabla f is plotted

      3. @@ -532,7 +516,7 @@ An illustration of this direction at the point (1,2,f(1,2)) is shown in the figure below, with the gradient vector scaled appropriately to make it easier to plot. -

        ADD ALT TEXT TO THIS IMAGE

         +

        The gradient of \nabla f is plotted

      4. @@ -541,7 +525,7 @@ The gradient is perpendicular to the level curves, and a picture of the unit vector in the direction of the gradient at (1,2) is shown in the figure below, with the gradient vector scaled appropriately to make it easier to plot. -

        ADD ALT TEXT TO THIS IMAGE

         +

        The gradient of \nabla f is plotted

      5. @@ -559,135 +543,6 @@ - - - -

        - The properties of the gradient that we have observed for functions of two variables also hold for functions of more variables. In this problem, we consider a situation where there are three independent variables. Suppose that the temperature in a region of space is described by - - T(x,y,z) = 100e^{-x^2-y^2-z^2} - - and that you are standing at the point (1,2,-1). -

          -
        1. -

          - Find the instantaneous rate of change of the temperature in the direction of - \vv=\langle 0, 1, 2\rangle at the point (1,2,-1). Remember that you should first find a - unit vector in the direction of \vv. -

          -
          2. - -
        2. -

          - In what direction from the point (1,2,-1) would you move to cause the temperature to - decrease as quickly as possible? -

          -
          3. - -
        3. -

          - How fast does the temperature decrease in this direction? -

          -
          4. - -
        4. -

          - Find a direction in which the temperature does - not change at (1,2,-1). -

          -
          5. -
        -

        -         - - -
          - -
        1. -

          - 0 -

          -
          2. -
        2. -

          - - -\nabla T(1,2,-1) = \left\langle 200e^{-6}, 400e^{-6}, -200e^{-6} \right\rangle \approx \langle 0.5, 1, -0.5 \rangle - -

          -
          3. -
        3. -

          - The rate of change of the temperature in this direction is - - -\left\vecmag{\nabla T(1,2,-1) \right} = -200 \sqrt{6}e^{-6} \approx -1.21 - - So the temperature decreases at approximately 1.21 units of temperature per one unit of distance in this direction. -

          -
          4. -
        4. -

          - possible answers: \langle -1,1,1 \rangle or in the direction of - \langle 0,1,2\rangle as shown in part (a) -

          -
          5. -
        -         - - -
          -
        1. -

          - A unit vector in the direction of \vv is \vu = \frac{1}{\sqrt{5}} \vv. - Since - - T_x(x,y,z) = -200xe^{-x^2-y^2-z^2} \ , T_y(x,y,z) = -200ye^{-x^2-y^2-z^2}, \ \text{ and } \ T_z(x,y,z) = -200ze^{-x^2-y^2-z^2} - , - the instantaneous rate of change of the temperature in the direction of \vv=\langle 0, 1, 2\rangle is then - - D_{\vu}T(1,2,1) \amp = \nabla T(1,2,-1) \cdot \vu - \amp = \langle T_x(1,2,-1), T_y(1,2,-1), T_z(1,2,-1) \rangle \cdot \vu - \amp =\left\langle -200e^{-6}, -400e^{-6}, 200e^{-6} \right\rangle \cdot \frac{1}{\sqrt{5}} \langle 0,1,2\rangle - \amp =-\frac{200}{\sqrt{5}}e^{-6}(0+4-4) - \amp = 0 - . -

          -
          2. -
        2. -

          - To cause the temperature to decrease as quickly as possible we move in the direction of - - -\nabla T(1,2,-1) = \left\langle 200e^{-6}, 400e^{-6}, -200e^{-6} \right\rangle \approx \langle 0.5, 1, -0.5 \rangle - . -

          -
          3. -
        3. -

          - The rate of change of the temperature in this direction is - - -\vecmag{\nabla T(1,2,-1) \right} = -\sqrt{\left(200e^{-6}\right)^2 + \left(400e^{-6}\right)^2 + \left(-200e^{-6}\right)^2} = -200 \sqrt{6}e^{-6} \approx -1.21 - . - So the temperature decreases at approximately 1.21 in this direction. -

          -
          4. -
        4. -

          - Let \vu = \langle u_1, u_2, u_3 \rangle be a vector in a direction in which the temperature does not change. - Then - - 0 \amp = \nabla T(1,2,-1) \cdot \vu - \amp = \left\langle -200e^{-6}, -400e^{-6}, 200e^{-6} \right\rangle \cdot \langle u_1, u_2, u_3 \rangle - \amp = -200e^{-6}(u_1+2u_2-u_3) - . - So a direction in which the temperature does not change is the direction of the vector - \langle -1,1,1 \rangle (or in the direction of - \langle 0,1,2\rangle as shown in part (a)). -

          -
          5. -
        -         - - -

        @@ -725,7 +580,7 @@ Since the gradient at a point is orthogonal to the contour at that point, the contours through the indicated points are drawn orthogonal to the gradients. This is illustrated in the figures below. -

        ADD ALT TEXT TO THIS IMAGE

        ADD ALT TEXT TO THIS IMAGE

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         +

        The gradient of \nabla f is plotted with curves through the highlighted points such that the curves are orthogonal to the vectors given by the gradient at each point.

        The gradient of \nabla f is plotted with curves through the highlighted points such that the curves are orthogonal to the vectors given by the gradient at each point.

        The gradient of \nabla f is plotted with curves through the highlighted points such that the curves are orthogonal to the vectors given by the gradient at each point.

      6. @@ -733,7 +588,7 @@ The gradient points in the direction of the greatest increase of the function, so a curve that moves in the direction opposite the gradient is a curve on which f is decreasing. This is illustrated in the figures below. -

        ADD ALT TEXT TO THIS IMAGE

        ADD ALT TEXT TO THIS IMAGE

        ADD ALT TEXT TO THIS IMAGE

         +

        The gradient of \nabla f is plotted

        The gradient of \nabla f is plotted

        The gradient of \nabla f is plotted

      @@ -746,7 +601,7 @@ Since the gradient at a point is orthogonal to the contour at that point, the contours through the indicated points are drawn orthogonal to the gradients. This is illustrated in the figures below. -

      ADD ALT TEXT TO THIS IMAGE

      ADD ALT TEXT TO THIS IMAGE

      ADD ALT TEXT TO THIS IMAGE

       +

      The gradient of \nabla f is plotted

      The gradient of \nabla f is plotted

      The gradient of \nabla f is plotted

    1. @@ -754,7 +609,7 @@ The gradient points in the direction of the greatest increase of the function, so a curve that moves in the direction opposite the gradient is a curve on which f is decreasing. This is illustrated in the figures below. -

      ADD ALT TEXT TO THIS IMAGE

      ADD ALT TEXT TO THIS IMAGE

      ADD ALT TEXT TO THIS IMAGE

       +

      The gradient of \nabla f is plotted

      The gradient of \nabla f is plotted

      The gradient of \nabla f is plotted

    diff --git a/source/exercises/ez-S_mv_HD.ptx b/source/exercises/ez-S_mv_HD.ptx index f2390268a..606ed022d 100644 --- a/source/exercises/ez-S_mv_HD.ptx +++ b/source/exercises/ez-S_mv_HD.ptx @@ -8,9 +8,157 @@ + + + + + + + + + + + + + + + + + + + + + + + + +

    + The properties of the gradient that we have observed for functions of two variables also hold for functions of more variables. In this problem, we consider a situation where there are three independent variables. Suppose that the temperature in a region of space is described by + + T(x,y,z) = 100e^{-x^2-y^2-z^2} + + and that you are standing at the point (1,2,-1). +

      +
    1. +

      + Find the instantaneous rate of change of the temperature in the direction of + \vv=\langle 0, 1, 2\rangle at the point (1,2,-1). Remember that you should first find a + unit vector in the direction of \vv. +

      +
      2. + +
    2. +

      + In what direction from the point (1,2,-1) would you move to cause the temperature to + decrease as quickly as possible? +

      +
      3. + +
    3. +

      + How fast does the temperature decrease in this direction? +

      +
      4. + +
    4. +

      + Find a direction in which the temperature does + not change at (1,2,-1). +

      +
      5. +
    +

    +     + + +
      + +
    1. +

      + 0 +

      +
      2. +
    2. +

      + + -\nabla T(1,2,-1) = \left\langle 200e^{-6}, 400e^{-6}, -200e^{-6} \right\rangle \approx \langle 0.5, 1, -0.5 \rangle + +

      +
      3. +
    3. +

      + The rate of change of the temperature in this direction is + + -\left\vecmag{\nabla T(1,2,-1) \right} = -200 \sqrt{6}e^{-6} \approx -1.21 + + So the temperature decreases at approximately 1.21 units of temperature per one unit of distance in this direction. +

      +
      4. +
    4. +

      + possible answers: \langle -1,1,1 \rangle or in the direction of + \langle 0,1,2\rangle as shown in part (a) +

      +
      5. +
    +     + + +
      +
    1. +

      + A unit vector in the direction of \vv is \vu = \frac{1}{\sqrt{5}} \vv. + Since + + T_x(x,y,z) = -200xe^{-x^2-y^2-z^2} \ , T_y(x,y,z) = -200ye^{-x^2-y^2-z^2}, \ \text{ and } \ T_z(x,y,z) = -200ze^{-x^2-y^2-z^2} + , + the instantaneous rate of change of the temperature in the direction of \vv=\langle 0, 1, 2\rangle is then + + D_{\vu}T(1,2,1) \amp = \nabla T(1,2,-1) \cdot \vu + \amp = \langle T_x(1,2,-1), T_y(1,2,-1), T_z(1,2,-1) \rangle \cdot \vu + \amp =\left\langle -200e^{-6}, -400e^{-6}, 200e^{-6} \right\rangle \cdot \frac{1}{\sqrt{5}} \langle 0,1,2\rangle + \amp =-\frac{200}{\sqrt{5}}e^{-6}(0+4-4) + \amp = 0 + . +

      +
      2. +
    2. +

      + To cause the temperature to decrease as quickly as possible we move in the direction of + + -\nabla T(1,2,-1) = \left\langle 200e^{-6}, 400e^{-6}, -200e^{-6} \right\rangle \approx \langle 0.5, 1, -0.5 \rangle + . +

      +
      3. +
    3. +

      + The rate of change of the temperature in this direction is + + -\vecmag{\nabla T(1,2,-1) \right} = -\sqrt{\left(200e^{-6}\right)^2 + \left(400e^{-6}\right)^2 + \left(-200e^{-6}\right)^2} = -200 \sqrt{6}e^{-6} \approx -1.21 + . + So the temperature decreases at approximately 1.21 in this direction. +

      +
      4. +
    4. +

      + Let \vu = \langle u_1, u_2, u_3 \rangle be a vector in a direction in which the temperature does not change. + Then + + 0 \amp = \nabla T(1,2,-1) \cdot \vu + \amp = \left\langle -200e^{-6}, -400e^{-6}, 200e^{-6} \right\rangle \cdot \langle u_1, u_2, u_3 \rangle + \amp = -200e^{-6}(u_1+2u_2-u_3) + . + So a direction in which the temperature does not change is the direction of the vector + \langle -1,1,1 \rangle (or in the direction of + \langle 0,1,2\rangle as shown in part (a)). +

      +
      5. +
    +     +

    Consider the surface x^2-y^2+z^2=4. diff --git a/source/interacts/surface-param.txt b/source/interacts/surface-param.txt new file mode 100644 index 000000000..e89813d74 --- /dev/null +++ b/source/interacts/surface-param.txt @@ -0,0 +1,24 @@ +@interact +def sphereplot(stb=checkbox(default=False,label="Show rotational slice:")): + from sage.plot.plot3d.plot3d import axes + var('s,t') + scale=3.5 + plot1=axes(scale,1,color="black",frame=False)+text3d("x",(scale+0.5,0,0),fontsize=30)+text3d("y",(0,scale+0.5,0),fontsize=30)+text3d("z",(0,0,scale+0.5),fontsize=30) + param(s,t)=(2*sin(s)*cos(t),2*sin(s)*sin(t),2*cos(s)) + #set up grid n is number of steps in s m is number of steps in t + #a1 is min s a2 is max s b1 is min t b2 is max t + n=9;m=9;a1=0;a2=pi;b1=0;b2=2*pi; + #c1 is surface color o1 is surface opacity + #c2 and o2 are for constant t value traces + #c3 and o3 are for constant s value traces + c1="lightblue"; c2="black"; c3="black"; o1=0.7; o2=0.2; o3=0.2; + plot1+=parametric_plot3d(param,(s,a1,a2),(t,b1,b2),color=c1,opacity=o1) + for i in range(n+1): + sv=a1+i/n*(a2-a1) + plot1+=parametric_plot3d(param(sv,t),(t,b1,b2),color=c2,opacity=o2) + for i in range(m+1): + tv=b1+i/m*(b2-b1) + plot1+=parametric_plot3d(param(s,tv),(s,a1,a2),color=c3,opacity=o3) + if stb: + plot1+=parametric_plot3d(param(s,pi/3),(s,0,pi),color="green",thickness=3) + show(plot1,aspect_ratio=1,viewpoint=[[-0.4047,-0.4447,-0.799],107.96],camera_position=[23.3495,2.6514,12.7857]) \ No newline at end of file