## WebWork 18.2 #7

This problem is given in differential form. It is stated as follows:

Problem: Evaluate the integral $\int_C 5ydx+7xdy$ where $C$ is the straight line path from $(1,4)$ to $(3,6)$.

Hints: First, you should parameterize the curve. One possibility is $x=1+2t$ and $y=4+2t$ where $0\le t\le 1$. Since we now know what $x$ and $y$ are, the integral can be rewritten

$\int_C 5ydx+7xdy=\int_C 5(4+2t)dx+7(1+2t)dy.$

Now, we should convert this integral to be in terms of $t$ and $dt$, instead of having the $dx$ and $dy$. That’s easy, since $dx=2dt$ and $dy=2dt$ (differentiate $x$ and $y$ with respect to $t$). Now, we’re in this awkward looking situation:

$\int_0^1 5(4+2t)2dt+7(1+2t)2dt$

But, this is just simplified as

$\int_0^1 54+48t dt.$

Now, solve that.

## Homework 9

Due Friday, November 11

17.5: 26, 36

18.1: 32, 42

18.2: 30, 34

## WebWork 17.4 Problems

I’ve marked correct the problems in 17.4 that ask you to use a graph and a table in an unclear way. WebWork appears to have been wanting you to draw the 9 displacement vectors given in each table, and move those around the grid together until they matched somewhere with the vector field.

## Homework 8 Vector Field Plots

Some have asked about the vector field plots and how to do them. Here is an example using Sage: 2D Vector Field

## Assignment 7 Solutions

16.5 #34: Evaluate the integral

$\int_0^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2-z^2}}^{\sqrt{1-x^2-z^2}}\frac{1}{(x^2+y^2+z^2)^{1/2}}\,dy\,dz\,dx.$

Solution: The bounds on this integral tell us that we’re integrating over half the sphere of radius 1 centered at the origin (specifically the hemisphere where $x>0$).  Since $\rho$ represents distance from the origin, our function becomes

$\frac{1}{(x^2+y^2+z^2)^{1/2}}=\frac{1}{\rho}.$

Thus, our integral in spherical coordinates becomes

$\int_{-\pi/2}^{\pi/2}\int_0^\pi\int_0^1 \frac{1}{\rho}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$

Now, we can use our trick of viewing $\sin\phi$ as a constant in $\rho$, and so on. We can immediately rewrite the integral as

$\int_{-\pi/2}^{\pi/2}\,d\theta\int_0^\pi\sin\phi\,d\phi\int_0^1\rho\,d\rho.$

Evaluating the integral gives us

$\pi\left(-\cos\theta\Big|_0^\pi\right)\left(\frac{1}{2}\rho^2\Big|_0^1\right)=\pi.$

16.5 #48: The density of a solid sphere at any point is proportional to the square of the distance of the point to the center of the sphere. What is the ratio of the mass of a sphere of radius 1 to a sphere of radius 2?

Solution: There is an easy and a hard way to solve this. In spherical coordinates, the mass of a sphere of radius $r$ with the given density property (density is given by $\rho^2$ where $0\le\rho\le r$) is

$\int_0^{2\pi}\int_0^\pi\int_0^r \rho^2\rho^2\sin\phi\,d\rho d\phi d\theta = \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^r \rho^4 d\rho.$

What we want to do is to take the ratio of a sphere of radius 1 and a sphere of radius 2 using this formula. We get

$\frac{\displaystyle \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^1 \rho^4 d\rho}{\displaystyle \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^2 \rho^4 d\rho} = \frac{\displaystyle \int_0^1 \rho^4 d\rho}{\displaystyle \int_0^2 \rho^4 d\rho}.$

This is nice because now we only need to take a single antiderivative, and evaluate it on two intervals. This becomes

$\frac{\displaystyle \frac{1}{5}\rho^5\Big|_0^1}{\displaystyle \frac{1}{5}\rho^5\Big|_0^2}=\frac{\displaystyle\rho^5\Big|_0^1}{\displaystyle \rho^5\Big|_0^2}=\frac{1}{2^5}=\frac{1}{32}.$

17.1 #58: For $t>0$, a particle moves along the curve $x=a+b\sin kt$, $y=a+b\cos kt$ where $a,b,k$ are positive constants.

(a) Describe the motion in words.

Solution: The particle revolves around a circle located at $(a,a)$ having radius $b$ and period $2\pi/k$. Also, the particle starts at $(a,a+b)$ and revolves clockwise.

(b) What is the effect on the curve of the following changes? (i) Increasing $b$. (ii) Increasing $a$. (iii) Increasing $k$. (iv) Setting $a$ and $b$ equal.

Solution: Increasing $b$ will increase the radius of the circle. Increasing $a$ will move the center of the circle along the line $y=x$ farther from the origin. Increasing $k$ will make the particle travel faster around the circle. Setting $a$ and $b$ equal means that the radius of the circle is equal to the horizontal and vertical distance from the origin, meaning that the circle will rest on the $x$ and $y$-axes.

17.1 #64: Consider the line $x=5-2t$, $y=3+7t$, $z=4t$ and the plane $ax+by+cz=d$. All of the following questions have many possible answers. Find values of $a$, $b$, $c$, and $d$ such that

(a) The plane is perpendicular to the line.

Solution: The line contains the point $(5,3,0)$, and more importantly we know that

$\frac{dx}{dt}=-2,\qquad\frac{dy}{dt}=7,\qquad \frac{dz}{dt}=4.$

That is, the vector $-2\vec{i}+7\vec{j}+4\vec{k}$ must be perpendicular to every displacement vector in the plane. Alternatively, this vector is a normal vector to the plane. Since we’re not wishing to orient the plane in any specific location, we may use the origin as a point in the plane. We have that the plane can be represented as

$-2(x-0)+7(y-0)+4(z-0)=0$

and so we have $-2x+7y+4z=0$, giving $a=-2$, $b=7$, $c=4$ and $d=0$.

(b) The plane is perpendicular to the line and through the point $(5,3,0)$.

Solution: This is a bit more complicated, but we can set up the initial equation as

$-2(x-5)+7(y-3)+4(z-0)=0.$

This tells us that $-2x+10+7y-21+4z=0$, and hence $-2x+7y+4z=11$. Thus, we take our previous answer and change $d$ to $11$.

(c) The line lies in the plane.

## WebWork 17.4 #2

Problem: Consider the vector field $\vec{v}=-4x\vec{i}+4y\vec{j}$. (a) Find the system of differential equations associated with this vector field. (b) Solve the system you found above to find the flow.

Ideas: For part (a), you’re just going to take the derivatives as we did in class. E.g., for my numbers I would have $x’=-4x$ and $y’=4y$. This makes part (b) a bit problematic, and you’re going to need to remember some of your differential equations tricks. The idea is that I want a function $x(t)$ such that $x’=-4x$. Notice that the rate of change of $x$ is a function containing $x$, and so you should automatically be thinking that $x$ is an exponential function (and the same is true for $y$). In fact, if we set $x(t)=ae^{ct}$ (for constants $a$ and $c$) then we have

$x'=-4x=-4ae^{ct}$

Of course, if $x=ae^{ct}$ then $x’=cae^{ct}$, and so $ca=-4a$ and $c=-4$. Thus, we have $x(t)=ae^{-4t}$, where $a$ simply provides an initial condition (a specific curve in the slope field). Similarly, we can find $y(t)$ given that $y’=4y$ and $y(t)=be^{dt}$.

## WebWork 17.3

I’m marking the following problems correct for all students in Section 17.3 of WebWork: #3, #4, #6, #8.

## Quiz from Friday, October 28

Problem: Find positive numbers $a$ and $b$ such that the change of variables $s=ax$ and $t=by$ transforms the integral $\int\int_R\,dx\,dy$ into

$\int\int_T\left|\frac{\partial(x,y)}{\partial(s,t)}\right|\,ds\,dt$

for the region $R$, the elliptical region $x^2/25+y^2/36\le 1$ and the region $T$, the circle $s^2+t^2\le 1$.

Solution: This is very nearly an example that we did in class. It’s also taken from a webwork problem (16.7#4), which is taken from your test (16.7#11). Recall that the equation of an ellipse is given by

$\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$

where the $a$ and $b$ give intercepts on the $x$ and $y$-axes, respectively. What we want to do is bring those intercepts back to the unit circle in coordinates $s$ and $t$. Thus, we really need to set $s=ax$ where $a=1/5$ and $t=by$ where $b=1/6$. This gives us the function $x(s,t)=5s$ and $y(s,t)=6t$, which we need in order to find the Jacobian:

$\left|\frac{\partial(x,y)}{\partial(s,t)}\right|=\left|\begin{array}{cc}5 & 0\\ 0 & 6\end{array}\right|=30.$

Replacement Question: In a sentence, explain why one gets sunburned faster in Colorado.

Solution: There is less atmosphere to block the sun’s rays.

Due November 2

17.2: 28, 32

17.3: 26, 34

17.4: 18, 20

## Quiz from Wednesday, October 26

Problem: Let $\theta$ represent the usual angle in the $xy$-plane emanating from the origin counter-clockwise from the $x$-axis, and let $\phi$ represent the angle from the $z$-axis to the $xy$-plane. If $\rho$ is distance from the origin, find the volume of the object created by rotating the region where $1\le\rho\le 3$ and $\pi/6\le\phi\le\pi/2$ in a full circle around the $z$-axis.

Solution: In spherical coordinates, we compute as follows.

\begin{align*} \int_0^{2\pi}\int_{\pi/6}^{\pi/2}\int_1^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta &= \int_0^{2\pi}\,d\theta\int_{\pi/6}^{\pi/2}\sin\phi\,d\phi\int_1^3\rho^2\,d\rho \\ &= 2\pi\left(-\cos\phi\right)\Big|_{\phi=\pi/6}^{\phi=\pi/2}\left(\frac{\rho^3}{3}\right)\Big|_{\rho=1}^{\rho=3}\\ &= 2\pi\cdot\frac{\sqrt{3}}{2}\left(\frac{27}{3}-\frac{1}{3}\right)\\ &= \frac{26\pi\sqrt{3}}{3}\end{align*}

Replacement Question: How many states share a border (including corners) with Colorado?

Answer: 7: Kansas, Nebraska, Wyoming, Utah, Arizona, New Mexico, and Oklahoma. There are only two states that share more borders with neighboring states than Colorado does: Tennessee and Missouri both share borders with 8 states (including each other).