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\textbf{College of the Holy Cross, Spring 2014} \\
\textbf{Solutions for Math 136, Midterm Exam 3} \\
\textbf{May 2}
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%\noindent\textbf{Your Name}: $\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$
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%\noindent\textbf{Your Section}: 

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%\leftline{Little (8:00am) $\underline{\qquad}$ Ballantine (9:00am) $\underline{\qquad}$} \leftline{DeStefano 
%(10:00am) $\underline{\qquad}$ DeStefano (noon) $\underline{\qquad}$}

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%\noindent\textbf{Instructions}:  For full credit, you must show 
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\noindent
\noindent
I.  Let $R$ be the region bounded by $y = \sin(x)$, the $x$-axis, and 
$0 \le x \le \pi/3$.  
\begin{enumerate}
\item[A.] [10~points] Write down (but {\it do not try to evaluate}) the integral
that would compute the arc-length of the top edge of $R$.
\item[] {\it Solution:}  The top is the graph $y = \sin(x)$, so $\frac{dy}{dx} = \cos(x)$,
and the arclength is 
$$\int_0^{\pi/3} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx = \int_0^{\pi/3} \sqrt{1 + \cos^2(x)}\ dx.$$
\item[B.] [15~points] A thin metal plate of constant density has 
the shape of $R$.  Find the $x$-coordinate of the center of mass of 
the plate.
\item[] {\it Solution:}  The $x$-coordinate of the center of mass is 
$$\frac{\int_0^{\pi/3} x\sin(x)\ dx}{\int_0^{\pi/3} \sin(x)\ dx}$$
The denominator integral is simpler here:
$$\int_0^{\pi/3} \sin(x)\ dx = \left. -\cos(x)\right|_0^{\pi/3} = 1 - \frac{1}{2} = \frac{1}{2}.$$
For the numerator, we need to integrate by parts (with $u = x$, $dv = \sin(x)\ dx$, so
$du = dx$ and $v = -\cos(x)$):
\begin{eqnarray*} 
\int_0^{\pi/3} x\sin(x)\ dx & = & \left. -x\cos(x)\right|_0^{\pi/3} + \int_0^{\pi/3} \cos(x)\ dx\\
&=& -\frac{\pi}{6} + \left.\sin(x)\right|_0^{\pi/3}\\
&=& \frac{\sqrt{3}}{2} - \frac{\pi}{6}.
\end{eqnarray*}
So the $x$-coordinate of the center of mass is 
$$\overline{x} = \sqrt{3} - \frac{\pi}{3} \doteq .6849$$
\end{enumerate}

\vskip 10pt
\noindent
II.  Both parts of this problem deal with the differential equation $\frac{dy}{dx} = xy$.
\begin{enumerate}
\item[A.] [15~points] Find the general solution $y(x)$ of the equation 
by separating variables and integrating.
\item[] {\it Solution:} This is separable:
\begin{eqnarray*}
\frac{dy}{dx} &=& xy\\
\int \frac{dy}{y} &=& \int x\ dx\\
\ln|y| &=& \frac{x^2}{2} + c\\
y &=& k e^{x^2/2} \quad \text{where}\quad k = \pm e^c.
\end{eqnarray*}
\item[B.] [5~points] Find the particular solution $y(x)$ satisfying the initial 
condition $y(0) = 4$ and compute the exact value of $y(2)$.  
\item[] {\it Solution:}  With $y(0) = 4$, we get $4 = k e^0$, so $k = 4$.  
The particular solution is $y = 4 e^{x^2/2}$.   With $x = 2$, we have $y = 4 e^2 \doteq 29.56$. 
\end{enumerate}
\vskip 10pt
\noindent
III. [10~points]  A drug is administered to a patient intravenously at a constant
rate of $10$mg per hour.  The patient's body breaks down the drug and removes
it from the bloodstream at a rate proportional to the amount present, with some
proportionality constant $k$.  Write a 
differential equation for the function $Q(t) = $ amount of the drug present (in mg) 
in the bloodstream at time $t$ (in hours) that describes this situation.  
Note: You \emph{do not need to solve} the equation.
\vskip 10pt
\noindent
{\it Solution:}  The differential equation comes from analyzing both contributions
to the change of the amount of drug in the bloodstream:
$$\frac{dQ}{dt} = kQ + 10.$$
Here $k < 0$ is the proportionality constant giving the rate
at which the drug is broken down and removed from the bloodstream.
The term $+10$ represents the constant intravenous dose of $10$mg per hour.  
\vskip 10pt 
\noindent 
IV. 
\begin{enumerate}
\item[A.]  [10~points] Does the geometric series 
$\ds \sum_{n=0}^\infty \cdot \frac{2^n}{\pi^n}$ converge or diverge?  
If it is convergent, say why and find the 
sum; if it is not convergent say why not.
\item[] {\it Solution:}  This is a geometric series with $a = 1$ and 
$r = \frac{2}{\pi}$.  Since $|r| < 1$, it converges to the sum 
$$\sum_{n=0}^\infty \frac{2^n}{\pi^n} = \frac{1}{1 - \frac{2}{\pi}} = \frac{\pi}{\pi - 2}
\doteq 2.752$$
\item[B.] [10~points] Explain why the Integral Test can be applied to the series
$\ds \sum_{n=1}^\infty \frac{1}{n^2 + 4}$ and use it to determine if the series
converges or diverges.
\item[] {\it Solution:}  The $n$th term of the series is $\frac{1}{n^2 + 4} = f(n)$,
for the function $f(x) = \frac{1}{x^2 + 4}$.  This function is continuous at all $x > 1$, 
it is decreasing (since $f'(x) = \frac{-2x}{(x^2 + 4)^2} < 0$ for all $x > 1$), and 
$\lim_{x \to \infty} \frac{1}{x^2 + 4} = 0$.  Hence the Integral Test applies and 
\begin{eqnarray*}
\int_1^\infty \frac{dx}{x^2 + 4} &=& \lim_{b \to \infty} \int_1^b \frac{dx}{x^2 + 4}\\
&=& \lim_{b\to\infty} \left. \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right|_1^b\\
&=& \lim_{b\to\infty} \frac{1}{2}\tan^{-1}\left(\frac{b}{2}\right) - \frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)\\
&=& \frac{\pi}{4} - \frac{\tan^{-1}\left(\frac{1}{2}\right)}{2}.
\end{eqnarray*}
Since the improper integral converges (is finite), the same is true for the infinite
series.  
\end{enumerate}
\vskip 10pt
\noindent
V.  All parts of this question refer to the power series 
\[
\sum_{n=1}^\infty \frac{(x - 1)^n}{n 3^n}.
\]
\begin{enumerate}
\item[A.] [15~points] Use the Ratio Test to determine the radius of convergence.
\item[] {\it Solution:}  
Applying the Ratio Test, 
\begin{eqnarray*}
\lim_{n\to\infty}\left|\frac{(x-1)^{n+1}}{(n+1) 3^{n+1}}\cdot\frac{n 3^n}{(x-1)^n}\right|
&=& \lim_{n\to\infty} \frac{n}{3(n+1)} |x - 1|\\
&=& \frac{1}{3} |x - 1|.
\end{eqnarray*}
For absolute convergence, we need $\frac{1}{3} |x - 1| < 1$, so $|x - 1| < 3$, or
$-2 < x < 4$.
  
The series is centered at $a = 1$, so the radius of convergence is $3$.

\item[B.] [10~points] Test convergence at the endpoints of the interval from part A to determine 
the interval of convergence.  Explain your conclusions.

\item[] {\it Solution:}  The inequality $|x - 1| < 3$ is equivalent to $-3 < x - 1 < 3$, 
or $-2 < x < 4$.  At $x = 4$, we substitute and obtain
\[
\sum_{n=1}^\infty \frac{1}{n}.
\]
This is the harmonic series which \emph{diverges}. At $x = -2$, we
have 
\[
\sum_{n=1}^\infty \frac{(-1)^n}{n}.
\]
The Alternating Series Test implies that this \emph{converges}.  
So the interval of convergence is $[-2,4)$, or all $x$ with $-2 \le x < 4$.  

\end{enumerate}
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