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.

Due Friday, November 11

17.5: 26, 36

18.1: 32, 42

18.2: 30, 34

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.

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

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.

 

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 $y(t)=be^{dt}$.

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

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

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).