Math 214 - Multivariable Calculus - Prep

Task 33.1

Polar coordinates is one type of new coordinate system, in particular this coordinate systems helps compute double integrals for planar regions that have some kind of rotational symmetry. We've actually been using change-of-coordinates since first semester calculus, every time we perform a substitution to complete an integral. What we'll do in this task is look at two changes-of-coordinates, one which stretches lengths, and the other which stretches areas.

Consider the integral $\ds\int_{-1}^{2} e^{-3x}\, dx$.
1. To complete this integral we use the substitution $u=-3x$. Solve for $x$ and compute the differential $dx$.
2. Perform the substitution, filling in the missing parts of $$\int_{x=-1}^{x=2} e^{-3x}\, dx = \int_{u=?}^{u=?} e^{u}? du.$$ Note that when a definite integral ends with $du$, the bounds should be in terms of $u$. To find the $u$ bounds, just ask, "If $x=-1$, then $u=?$" Don't spend any time completing the integral, rather just focus on completing the substitution above.
3. The $x$ values range from $-1$ to $2$. This is a directed interval whose width is 3 units, pointing from left to right along the $x$-axis (shown below). Our substitution $u=-3x$ transforms this directed interval into a different directed interval along the $u$-axis. Draw the transformed interval on the $u$-axis below.
4. How long is the new interval along the $u$-axis? What does your differential equation $dx=-\frac{1}{3}du$ have to do with this problem? What does the negative sign do?

Above we showed that the differential equation $dx = \frac{dx}{du}du$ tells us how to relate lengths along the $u$-axis to lengths along the $x$-axis. We could think of the number $\frac{dx}{du}$ as a length stretch factor. Let's now examine a two dimensional change-of-coordinates, and connect areas in the $uv$-plane to areas in the $xy$-plane.

Consider the change-of-coordinates $x=u-v$ and $y=u+v$, which we could also write as the coordinate transformation $\vec T(u,v) = (u-v,u+v)$.
1. In the table below, you're given several $(u,v)$ points. Find the corresponding $(x,y)$ pair. $$ \begin{array}{|c|c|c|} \text{Name}&(u,v)&(x,y)\\\hline A&(0,0)&(0,0)\\ B&(1,0)&(1-0,1+0) = (1,1)\\ C&(0,1)&(0-1,0+1) = (-1,1)\\ D&(1,1)&\\ E&(3,3)&\\ F&(2,4)&\\ G&(-2,4)&\\ \end{array}$$
2. There are two graphs below. One is a plot in the $uv$-plane of the points from the table, along with the parabola $v=u^2$, the line $v=u$, and the shaded box whose corners are the first four points. Complete a similar plot in the $xy$-plane by adding the remaining points, and then connect the points in your $xy$ plot to show how the parabola, line, and shaded box (done for you) transform because of this change-of-coordinates. How would you describe what this change-of-coordinates is doing?
  - The following Mathematica code allows you to draw any region from the $uv$-plane in the $xy$ plane. You can use it to check your work (as well as adapt the code to work with ANY new coordinate system).
```
R1 = ParametricRegion[{{u - v, u + v}, 0 <= u <= 1 && 0 <= v <= 1}, {u, v}];
Region[R1, Axes -> True, AxesLabel -> {x, y}]

R2 = ParametricRegion[{{u - v, u + v}, v == u^2}, {{u, -2, 2}, v}];
Region[R2, Axes -> True, AxesLabel -> {x, y}]
```
3. The differential of $x$ is $dx = du-dv$. Obtain a similar formula for the differential $dy$. Then write your answer as the vector equation (linear combination) $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} 1\\ ?\end{pmatrix}du+\begin{pmatrix}-1\\?\end{pmatrix}dv.$$
4. How are areas in the $uv$-plane related to areas in the $xy$-plane? In particular, the area of the 1 by 1 box in the $uv$ plane is 1. What is the area of the shaded region in the $xy$-plane. By how much are we stretching areas when we change from $uv$ to $xy$ coordinates (what's the "area stretch factor")?

The area stretch factor above is called a "Jacobian". We'll soon have quick way to compute a Jacobian, which will result from finding the area of a parallelogram (the next task).

Task 33.2

We need a quick way to compute the area of a parallelogram. In this task, we'll prove the following theorem and then use it in a few specific examples.

Area of a Parallelogram in the Plane

The area $A$ of a parallelogram whose edges are the two vectors $\vec u= (a,b)$ and $\vec v = (c,d)$ is given by $A=|ad-bc|$.

Suppose a parallelogram has edges that are parallel to the vectors $\vec u=(a,b)$ and $\vec v=(c,d)$. Prove that the area of this parallelogram is given by $|ad-bc|$. If you want some help, here are some steps you can follow:
1. Draw a generic parallelogram. Label one corner the origin, and then the two connecting edges are the vectors $\vec u = (a,b)$ and $\vec v=(c,d)$.
2. Add to your picture (1) the projection of $\vec u$ onto $\vec v$ (so $\vec u_{\parallel \vec v}$) and (2) the component of $\vec u$ that is orthogonal to $\vec v$ (so $\vec u_{\perp \vec v}$).
3. Explain why the area is $A = |\vec v||\vec u_{\perp \vec v}|$.
4. Use Mathematica to perform the computations, and then explain why the result given is equal to $|ad-bc|$.
```
u = {a, b};
v = {c, d};
uPerpv = u - Projection[u, v];
Norm[uPerpv]*Norm[v] // FullSimplify
```
Whether you are able to complete the proof above or not, let's practice using the result to compute areas of parallelogram (and triangles). Use the area formula $|ad-bc|$ to compute the requested areas below.
1. A parallelogram has vertices $(0,0)$, $(-2,5)$, $(3,4)$, and $(1,9)$. Find its area.
2. Find the area of the triangle with vertices $(0,0)$, $(-2,5)$, and $(3,4)$.
3. Find the area of the triangle with vertices $(-3,1)$, $(-2,5)$, and $(3,4)$. [You'll need to give vectors $\vec u$ and $\vec v$ that form the edges of a parallelogram.]

Task 33.3

Recall that the gradient of a function $f$ is the quantity $$\vec \nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right) = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)f ,$$ where in the last expression we let $\vec \nabla = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)$ and then treat $\vec \nabla f$ as a "vector" $\vec \nabla$ times a scalar $f$. The quantity $\vec \nabla = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)$ is an example of something we call an "operator," something that operates on functions.

Operator

An operator is a function whose input is a function itself. This allows us to say "operator on functions" instead of "function of functions."

We've already encountered several operators before this class. For example, the derivative operator $\frac{d}{dx}$ from first semester calculus takes a function such as $f(x) = x^2$ and returns a new function $\frac{d}{dx}f = 2x$. The integral operator $\int_a^b f dx$ takes a function $f$ and return a real number. The gradient operator takes a function $f$ and returns a vector of functions $\vec \nabla f = (f_x,f_y,f_z)$. This is just 3 examples of operators. Here are two more.

Divergence and Curl of a vector field

Consider the vector field $\vec F(x,y,z) = (M,N,P)$, where $M$, $N$, and $P$ are functions of $x$, $y$, and $z$.

The divergence of $\vec F$ is the scalar quantity $$ \begin{align*} \text{div}(\vec F) &= \vec \nabla \cdot \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\cdot \left(M,N,P\right) \\ &= \frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}\\ &= M_x+N_y+P_z . \end{align*} $$
The curl of $\vec F$ is the vector quantity $$ \begin{align*} \text{curl}(\vec F) &= \vec \nabla \times \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\times \left(M,N,P\right) \\ &= \left(\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z},\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x},\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) \\ &= \left(P_y-N_z,M_z-P_x,N_x-M_y\right) . \end{align*} $$

We will often say "del dot F" for the divergence of $\vec F$, and "del cross F" for the curl of $\vec F$.

Use the definitions above to compute the divergence and curl of each vector field below. Then compute the derivative (which will be a 3 by 3 matrix). Finish by checking your work with Mathematica (code for doing so is provided below). If you see any relationships worth mentioning, articulate them in your work.

$\vec F = (2x, 3y^2, e^z)$
$\vec F = (-3y, 3x, 5z)$
$\vec F = (z-3y, 3x, -x)$

In class, we'll talk about the physical meaning of each result above.

The code below computes the derivative, divergence, and curl of the vector field $\vec F = (x^2 y, x y z, 3 x + 4 z)$.

F = {x^2 y, x y z, 3 x + 4 z }
D[F, {{x, y, z}}] // MatrixForm
Div[F, {x, y, z}]
Curl[F, {x, y, z}]

Task 33.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.

Big View
Prep Day 33 Task 33.1 Polar coordinates is one type of new coordinate system, in particular this coordinate systems helps compute double integrals for planar regions that have some kind of rotational symmetry. We've actually been using change-of-coordinates since first semester calculus, every time we perform a substitution to complete an integral. What we'll do in this task is look at two changes-of-coordinates, one which stretches lengths, and the other which stretches areas. Consider the integral $\ds\int_{-1}^{2} e^{-3x}\, dx$. To complete this integral we use the substitution $u=-3x$. Solve for $x$ and compute the differential $dx$. Perform the substitution, filling in the missing parts of $$\int_{x=-1}^{x=2} e^{-3x}\, dx = \int_{u=?}^{u=?} e^{u}? du.$$ Note that when a definite integral ends with $du$, the bounds should be in terms of $u$. To find the $u$ bounds, just ask, "If $x=-1$, then $u=?$" Don't spend any time completing the integral, rather just focus on completing the substitution above. The $x$ values range from $-1$ to $2$. This is a directed interval whose width is 3 units, pointing from left to right along the $x$-axis (shown below). Our substitution $u=-3x$ transforms this directed interval into a different directed interval along the $u$-axis. Draw the transformed interval on the $u$-axis below. How long is the new interval along the $u$-axis? What does your differential equation $dx=-\frac{1}{3}du$ have to do with this problem? What does the negative sign do? Above we showed that the differential equation $dx = \frac{dx}{du}du$ tells us how to relate lengths along the $u$-axis to lengths along the $x$-axis. We could think of the number $\frac{dx}{du}$ as a length stretch factor. Let's now examine a two dimensional change-of-coordinates, and connect areas in the $uv$-plane to areas in the $xy$-plane. Consider the change-of-coordinates $x=u-v$ and $y=u+v$, which we could also write as the coordinate transformation $\vec T(u,v) = (u-v,u+v)$. In the table below, you're given several $(u,v)$ points. Find the corresponding $(x,y)$ pair. $$ \begin{array}{\|c\|c\|c\|} \text{Name}&(u,v)&(x,y)\\\hline A&(0,0)&(0,0)\\ B&(1,0)&(1-0,1+0) = (1,1)\\ C&(0,1)&(0-1,0+1) = (-1,1)\\ D&(1,1)&\\ E&(3,3)&\\ F&(2,4)&\\ G&(-2,4)&\\ \end{array}$$ There are two graphs below. One is a plot in the $uv$-plane of the points from the table, along with the parabola $v=u^2$, the line $v=u$, and the shaded box whose corners are the first four points. Complete a similar plot in the $xy$-plane by adding the remaining points, and then connect the points in your $xy$ plot to show how the parabola, line, and shaded box (done for you) transform because of this change-of-coordinates. How would you describe what this change-of-coordinates is doing? The following Mathematica code allows you to draw any region from the $uv$-plane in the $xy$ plane. You can use it to check your work (as well as adapt the code to work with ANY new coordinate system). R1 = ParametricRegion[{{u - v, u + v}, 0 <= u <= 1 && 0 <= v <= 1}, {u, v}]; Region[R1, Axes -> True, AxesLabel -> {x, y}] R2 = ParametricRegion[{{u - v, u + v}, v == u^2}, {{u, -2, 2}, v}]; Region[R2, Axes -> True, AxesLabel -> {x, y}] The differential of $x$ is $dx = du-dv$. Obtain a similar formula for the differential $dy$. Then write your answer as the vector equation (linear combination) $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} 1\\ ?\end{pmatrix}du+\begin{pmatrix}-1\\?\end{pmatrix}dv.$$ How are areas in the $uv$-plane related to areas in the $xy$-plane? In particular, the area of the 1 by 1 box in the $uv$ plane is 1. What is the area of the shaded region in the $xy$-plane. By how much are we stretching areas when we change from $uv$ to $xy$ coordinates (what's the "area stretch factor")? The area stretch factor above is called a "Jacobian". We'll soon have quick way to compute a Jacobian, which will result from finding the area of a parallelogram (the next task). Task 33.2 We need a quick way to compute the area of a parallelogram. In this task, we'll prove the following theorem and then use it in a few specific examples. Area of a Parallelogram in the Plane The area $A$ of a parallelogram whose edges are the two vectors $\vec u= (a,b)$ and $\vec v = (c,d)$ is given by $A=\|ad-bc\|$. Suppose a parallelogram has edges that are parallel to the vectors $\vec u=(a,b)$ and $\vec v=(c,d)$. Prove that the area of this parallelogram is given by $\|ad-bc\|$. If you want some help, here are some steps you can follow: Draw a generic parallelogram. Label one corner the origin, and then the two connecting edges are the vectors $\vec u = (a,b)$ and $\vec v=(c,d)$. Add to your picture (1) the projection of $\vec u$ onto $\vec v$ (so $\vec u_{\parallel \vec v}$) and (2) the component of $\vec u$ that is orthogonal to $\vec v$ (so $\vec u_{\perp \vec v}$). Explain why the area is $A = \|\vec v\|\|\vec u_{\perp \vec v}\|$. Use Mathematica to perform the computations, and then explain why the result given is equal to $\|ad-bc\|$. u = {a, b}; v = {c, d}; uPerpv = u - Projection[u, v]; Norm[uPerpv]Norm[v] // FullSimplify Whether you are able to complete the proof above or not, let's practice using the result to compute areas of parallelogram (and triangles). Use the area formula $\|ad-bc\|$ to compute the requested areas below. A parallelogram has vertices $(0,0)$, $(-2,5)$, $(3,4)$, and $(1,9)$. Find its area. Find the area of the triangle with vertices $(0,0)$, $(-2,5)$, and $(3,4)$. Find the area of the triangle with vertices $(-3,1)$, $(-2,5)$, and $(3,4)$. [You'll need to give vectors $\vec u$ and $\vec v$ that form the edges of a parallelogram.] Task 33.3 Recall that the gradient of a function $f$ is the quantity $$\vec \nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right) = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)f ,$$ where in the last expression we let $\vec \nabla = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)$ and then treat $\vec \nabla f$ as a "vector" $\vec \nabla$ times a scalar $f$. The quantity $\vec \nabla = \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)$ is an example of something we call an "operator," something that operates on functions. Operator* An operator is a function whose input is a function itself. This allows us to say "operator on functions" instead of "function of functions." We've already encountered several operators before this class. For example, the derivative operator $\frac{d}{dx}$ from first semester calculus takes a function such as $f(x) = x^2$ and returns a new function $\frac{d}{dx}f = 2x$. The integral operator $\int_a^b f dx$ takes a function $f$ and return a real number. The gradient operator takes a function $f$ and returns a vector of functions $\vec \nabla f = (f_x,f_y,f_z)$. This is just 3 examples of operators. Here are two more. Divergence and Curl of a vector field Consider the vector field $\vec F(x,y,z) = (M,N,P)$, where $M$, $N$, and $P$ are functions of $x$, $y$, and $z$. The divergence of $\vec F$ is the scalar quantity $$ \begin{align} \text{div}(\vec F) &= \vec \nabla \cdot \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\cdot \left(M,N,P\right) \\ &= \frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}\\ &= M_x+N_y+P_z . \end{align} $$ The curl of $\vec F$ is the vector quantity $$ \begin{align} \text{curl}(\vec F) &= \vec \nabla \times \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\times \left(M,N,P\right) \\ &= \left(\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z},\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x},\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) \\ &= \left(P_y-N_z,M_z-P_x,N_x-M_y\right) . \end{align} $$ We will often say "del dot F" for the divergence of $\vec F$, and "del cross F" for the curl of $\vec F$. Use the definitions above to compute the divergence and curl of each vector field below. Then compute the derivative (which will be a 3 by 3 matrix). Finish by checking your work with Mathematica (code for doing so is provided below). If you see any relationships worth mentioning, articulate them in your work. $\vec F = (2x, 3y^2, e^z)$ $\vec F = (-3y, 3x, 5z)$ $\vec F = (z-3y, 3x, -x)$ In class, we'll talk about the physical meaning of each result above. The code below computes the derivative, divergence, and curl of the vector field $\vec F = (x^2 y, x y z, 3 x + 4 z)$. F = {x^2 y, x y z, 3 x + 4 z } D[F, {{x, y, z}}] // MatrixForm Div[F, {x, y, z}] Curl[F, {x, y, z}] Task 33.4 Pick some problems related to the topics we are discussing from the Text Book Practice page. Edit Page History Source Attach File Backlinks List Group Page last modified on November 07, 2022, at 01:54 PM
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Day 33

Task 33.1

Task 33.2

Task 33.3

Task 33.4