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\centerline{Mathematics 136 -- Calculus 2}
\centerline{Exam 3 -- Sample Exam Questions -- Solutions}
\centerline{April 22, 2014}

\vskip 10pt

\begin{enumerate}
\item[I.]  
\begin{enumerate}
\item[(A)] The arclength is 
\begin{eqnarray*}
L &=& \int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^2 \sqrt{36t^2 + 36t^4}\ dt\\
&=& \int_0^2 6t\sqrt{1 + t^2}\ dt, \quad \text{ so let } u = 1 + t^2; \quad \text{form is } \int u^{1/2}\ du \\
&=& \left. 2 (1 + t^2)^{3/2}\right|_0^2\\
&=& 2(5\sqrt{5} - 1).
\end{eqnarray*}
\item[(B)]  The arclength is 
\begin{eqnarray*}
L &=& \int \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}\ dx\\
&=& \int_0^3 \sqrt{1 + \left(\frac{1}{4}(x^2 + 4)^{1/2}(2x)\right)^2}\ dx\\
&=& \int_0^3 \sqrt{\frac{1}{4} x^4 + x^2 + 1}\ dx \quad\text{(a perfect square under the radical)}\\
&=& \int_0^3 \frac{1}{2}(x^2 + 2)\ dx\\
&=& \left. \frac{x^3}{6} + x\right|_0^3\\
&=& \frac{15}{2}.
\end{eqnarray*}
\end{enumerate}
\item[II.]
\begin{enumerate}
\item[(A)]  The average value is 
$$f_{ave} = \frac{1}{2}\int_0^2 x\sqrt{1 + x^4}\ dx$$
To evaluate the integral, use the substitution $u = x^2$ and \#21 in the table (or
a tangent substitution):
\begin{eqnarray*}
f_{ave} &=& \frac{1}{4} \int_0^4 \sqrt{1 + u^2}\ du\\
&=& \frac{1}{4}\left. \frac{u}{2}\sqrt{1 + u^2} + \frac{1}{2}\ln(u + \sqrt{1 + u^2})\right|_0^4\\
&=& \frac{1}{2}\sqrt{17} + \frac{1}{8}\ln(4 + \sqrt{17}).
\end{eqnarray*}
\item[(B)]  Let the constant density be $\rho$.  The center of mass is at the point 
$(\overline{x},\overline{y}) = (M_y/M, M_x/M)$, where
$$M = \int_0^2 \rho x\sqrt{1 + x^4}\ dx,$$
and 
$$M_x = \int_0^2 \rho \frac{1}{2}(x\sqrt{1 + x^4})^2\ dx,$$
while
$$M_y = \int_0^2 \rho x^2 \sqrt{1 + x^4}\ dx.$$
\item[(C)]  In order for $f(x)$ to be a probability density function, 
we need $\int_{-\infty}^\infty f(x)\ dx = 1$.  Here that means
$$1 = c \int_0^2 x\sqrt{1 + x^4}\ dx.$$
From the result in part (A), we see
$$c = \frac{1}{\sqrt{17} + \frac{1}{4}\ln(4 + \sqrt{17})}.$$

\end{enumerate}
\item[III.]
\begin{enumerate}
\item[(A)]  Show that for any constant $c$, $y = x^2 + \frac{c}{x^2}$
is a solution of the differential equation
\[
y' = 4x - \frac{2}{x}y.
\]

\vspace{.1in}

{\bf Solution:} For $\displaystyle y = x^2 + \frac{c}{x^2}$ we have $\displaystyle y'=2x-\frac{2c}{x^3}$ and $\displaystyle 4x - \frac{2}{x}y= 4x-\frac{2}{x}(x^2+\frac{c}{x^2})=2x-\frac{2c}{x^3}$. Thus $\displaystyle y = x^2 + \frac{c}{x^2}$ is a solution to the differential equation $\displaystyle y' = 4x - \frac{2}{x}y$. \vspace{.1in}  

\item[(B)] All parts of this question refer to the differential equation 
\[
y' = y(4 - y)
\]
\begin{enumerate}
\item[(1)]  Sketch the slope field of this equation, showing 
the slopes at points on the lines $y = 0,1,2,3,4,5$\vspace{.1in}

{\bf Solution:}

\begin{center}
\includegraphics[width=2in]{exam3sol.png}
\end{center}


\item[(2)]  On your slope field, sketch the graph of the 
solution of the equation with $y(0) = 1$.\vspace{.1in}

{\bf Solution:} See figure above. \vspace{.1in}


\item[(3)]  Use Euler's method to approximate the solution of 
this equation with $y(0) = 1$ for $0 \le x \le 1$ using
$n = 4$.  \vspace{.1in}

{\bf Solution:} We have $\Delta x=0.25$.

$$\begin{array}{ll} x_0=0 \ \ \ \ \ \ \ \ & y_0=1\\
x_1=.25 & y_1=y_0+(y_0(4-y_0))\Delta x= 1+ 3(.25)=1.75\\x_2=.5 & y_2=y_1+(y_1(4-y_1))\Delta x= 2.734375\\x_3=.75 & y_3=y_2+(y_2(4-y_2))\Delta x=3.599548340\\ x_4=1 & y_4=y_3+(y_3(4-y_3))\Delta x=3.959909617

\end{array} $$ \vspace{.1in}

\item[(4)]  This is a separable equation, find the general solution
and determine the constant of integration from the initial condition $y(0) = 1$. \vspace{.1in}

{\bf Solution:} After separating the variables we have $\displaystyle \int \frac{1}{y(4-y)}\, dy= \int \, dx$. For the integral in $y$ we use partial fractions: $\displaystyle \frac{1}{y(4-y)}=\frac{A}{y}+\frac{B}{4-y}$. We find that $A=B=1/4$ and thus $\displaystyle \int \frac{1}{y(4-y)}\, dy= \frac{1}{4}\ln|y| - \frac{1}{4}\ln|4-y|$. Therefore,  $\displaystyle \frac{1}{4}\ln\left| \frac{y}{4-y}\right|=x+C$. Then $ \displaystyle \left|  \frac{y}{4-y}\right| = e^{4x}\cdot e^{4C}$ and thus $ \displaystyle  \frac{y}{4-y}= A \cdot e^{4x}$. Solving for $y$, we obtain $ \displaystyle y= \frac{4Ae^{4x}}{1+Ae^{4x}}$.

The initial condition $y(0)=1$ gives $ \displaystyle 1=\frac{4A}{1+A}$ and thus $A=1/3$.\vspace{.1in}


\end{enumerate}
\item[(C)]  Find the general solutions of the following differential equations
\begin{enumerate}
\item[(1)]  ${\displaystyle y' = \frac{y}{x(x+1)} }$\vspace{.1in}

{\bf Solution:} This is a separable differential equation. \vspace{.1in}

We have $\displaystyle \int \frac{dy}{y}= \int \frac{dx}{x(x+1)}$. For the integral on the right we use partial fractions. $\displaystyle \frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$. 

Thus $\displaystyle \int \frac{1}{x(x+1)} \, dx = \ln|x|-\ln|x+1| + C=\ln\left|\frac{x}{x+1}\right| +C$. 

We have $\displaystyle \ln|y|=\ln\left|\frac{x}{x+1}\right| +C$ and thus $\displaystyle |y|= e^{\ln\left|\frac{x}{x+1}\right| +C}= \left|\frac{x}{x+1}\right| \cdot e^C$. \vspace{.1in}

Therefore  $\displaystyle y=A \frac{x}{x+1}$ is the general solution of the given differential equation.\vspace{.1in}


\item[(2)]  ${\displaystyle y' = \frac{\sqrt{1 - x^2}}{e^{2y}}.}$

\vspace{.1in}

{\bf Solution:} This is a separable differential equation. \vspace{.1in}

We have $\displaystyle \int e^{2y}\, dy= \int \sqrt{1-x^2}\, dx$. For the integral on the right we use the trigonometric substitution $x=\sin \theta$, $dx=\cos \theta \, d\theta$. Thus $\displaystyle  \int \sqrt{1-x^2}\, dx= \int \sqrt{1-\sin \theta} \cos \theta \, d \theta= \int \cos^2 \theta \, d \theta = \int \frac{1+ \cos 2 \theta}{2} \, d \theta= \frac{1}{2}\theta + \frac{1}{4}\sin 2 \theta + C= $  \vspace{.1in}


$\displaystyle \frac{1}{2}\theta + \frac{1}{4}2 \sin  \theta \cos \theta+ C = \frac{1}{2} \arcsin x + \frac{1}{2} x \sqrt{1-x^2}+C$\vspace{.1in} 

Therefore $\displaystyle \frac{1}{2} e^{2y}=\frac{1}{2} \arcsin x + \frac{1}{2} x \sqrt{1-x^2}+C$ or $\displaystyle  e^{2y}=\arcsin x + x \sqrt{1-x^2}+D$ and we have that  $\displaystyle  y=\frac{1}{2} \ln(\arcsin x + x \sqrt{1-x^2}+D)$ is the general solution to the given differential equation. \vspace{.1in}


\end{enumerate}
\item[(D)]  Newton's Law of Cooling states that the 
rate at which the temperature of an object changes is 
proportional to the difference between the object's temperature
and the surrounding temperature.  A hot cup of tea with temperature
$100^\circ$C is placed on a counter in a room maintained at constant 
temperature $20^\circ$C.  Ten minutes later the tea has cooled to 
$76^\circ$C.  How long will it take to cool off to $45^\circ$C?  
(Express Newton's Law as a differential equation, solve it for the 
temperature function, then use that to answer the question.)

\vspace{.1in}

{\bf Solution:} Let $T(t)$ denote the temperature of the cup at time $t$ measured in minutes from the time it was placed on the counter. The differential equation modeling this scenario is $\displaystyle \frac{dT}{dt}=k(T-20)$. In fact, this is an initial value problem: $T(0)=100$ and we have the additional information $T(10)=76$. This will help us find the constant of proportionality $k$. The differential equation is separable and we have $\displaystyle \int \frac{dT}{T-20}=\int k \, dt$. Integrating both sides we obtain $\displaystyle   \ln|T-20|=kt +C$ and thus $T-20=Ae^{kt}$. Therefore $T(t)=20+Ae^{kt}$. Since $T(0)=100$, we have $A=80$. Since $T(10)=76$, we have $76=20+80e^{10k}$. Thus $\displaystyle k=\frac{1}{10} \ln \frac{56}{80}$ and $\displaystyle T(t)=20+80e^{1/10 \ln(7/10)t}$. To find the time when the tea has cooled to $45^{\circ}C$, we sove $\displaystyle 20+80e^{1/10 \ln(7/10)t}=45$. Thus $\displaystyle e^{1/10 \ln(7/10)t}= 25/80=5/16$ and the tea will be at $45^{\circ}C$ after $\displaystyle t=10 \frac{\ln(5/16)}{\ln(7/10)}\approx 32.6$ minutes. 
\vspace{.1in}


\end{enumerate}
\item[IV.]
\begin{enumerate}
\item[(A)]  Does the infinite series ${\displaystyle \sum_{n=1}^\infty n \ln(1 + n)}$ converge?
Why or why not?
\vspace{.1in}

{\bf Solution:} Since $\displaystyle \lim_{n \rightarrow \infty} n \ln(1 + n) \neq 0$, the series ${\displaystyle \sum_{n=1}^\infty n \ln(1 + n)}$ diverges (by the $n$th Term Divergence Test).\vspace{.1in}

\item[(B)]  Use the Integral Test to determine whether or not
  \[
  \sum_{k=1}^\infty\frac{k}{e^k}
  \]
  converges.
  \vspace{.1in}

{\bf Solution:} The function $\displaystyle f(x)=\frac{x}{e^x}$ is continuous and positive. Since $\displaystyle f'(x)=\frac{e^x-xe^x}{e^{2x}}=\frac{e^x(1-x)}{e^{2x}}<0$ for $x>1$, f(x) is also decreasing for $x>1$. 

Consider 
$$\int_1^{\infty} \frac{x}{e^x}\, dx= \lim_{b \rightarrow \infty}\int_1^b x e^{-x}\, dx.$$ 
Using integration by parts, $u=x$, $du=dx$, $dv=e^{-x}dx$, $v=-e^{-x}$, the improper integral equals $\displaystyle \lim_{b \rightarrow \infty} \left( -be^{-b}+e^{-1}+\int_1^b e^{-x}\, dx\right) = \lim_{b \rightarrow  \infty} \left( -be^{-b} +e^{-1}-e^{-b}+e^{-1}\right)$. 
Since 
$$\displaystyle \lim_{b \rightarrow \infty} e^{-b}=0$$ and 
$$\displaystyle \lim_{b \rightarrow \infty} be^{-b}= \displaystyle \lim_{b \rightarrow \infty} \frac{b}{e^b} \stackrel{l'H}{=} \displaystyle \lim_{b \rightarrow \infty} \frac{1}{e^b}=0,$$ the improper integral converges to $2e^{-1}$. By the Integral Test, the series $\displaystyle  \sum_{k=1}^\infty\frac{k}{e^k}$ converges. \vspace{.1in}


\item[(C)]  Use the Ratio Test to determine whether or not
  \[
  \sum_{k=0}^\infty \frac{3^n}{n!}
  \]
  converges. \vspace{.1in}

{\bf Solution:} $$\lim_{n \rightarrow \infty}\frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}}= \lim_{n \rightarrow \infty}\frac{3}{n+1}=0<1.$$By the Ratio Test, the series converges. \vspace{.1in}
  
\item[(D)] Determine (with justification!) whether or not the following
  series converge:
  \[
  \sum_{k=1}^\infty \frac{1}{\sqrt{k}},\qquad
  \sum_{n=0}^\infty (-1)^n\frac{3^n}{\pi^{2n}},\qquad
  \sum_{n=1}^\infty \frac{1}{n^{1.01}}.
  \] \vspace{.1in}

{\bf Solution:} The series $\displaystyle  \sum_{k=1}^\infty \frac{1}{\sqrt{k}}$ is the $p$-series with $p=1/2$ and thus it diverges. \vspace{.1in}

The series $\displaystyle \sum_{n=0}^\infty (-1)^n\frac{3^n}{\pi^{2n}}$ is the geometric series with ratio $\displaystyle \frac{-3}{\pi^2}$. Since the ratio is less than $1$ in absolute value, the series converges. (The sum of the series is $\displaystyle \frac{1}{1+\frac{3}{\pi^2}}$.)\vspace{.1in}

The series $\displaystyle  \sum_{n=1}^\infty \frac{1}{n^{1.01}}$ is the $p$-series with $p=1.01$. Since $p>1$, the $p$-series converges. \vspace{.1in}
  


\item[(G)] For each of the given power series, find the interval of
  convergence.
  \[
  f(x)= \sum_{k=0}^\infty x^k, \qquad
  f(x)=\sum_{n=1}^\infty\frac{(2x)^n}{\sqrt{n}},\qquad
  g(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-5)^n}{n\cdot3^n}.
  \]
  (In particular, give the radius of convergence, 
  and investigate convergence at the endpoints.)
  
  \vspace{.1in}

{\bf Solution:} 
For $f(x) = \displaystyle{f(x) = \sum_{k=0}^\infty x^k}$, we see that this is a
geometric series with ratio $x$.  The series converges if and only if $|x| < 1$.  
The radius of convergence is $1$, the series does not converge for either $x = \pm 1$.
The interval of convergence is $(-1,1)$.  (Note:  all of this could also be derived
through use of the Ratio Test.)

For $\displaystyle  g(x)=\sum_{n=1}^\infty\frac{(2x)^n}{\sqrt{n}}$, consider the Ratio Test. $$\lim_{n\rightarrow \infty} \left|\frac{\frac{(2x)^{n+1}}{\sqrt{n+1}}}{\frac{(2x)^n}{\sqrt{n}}}\right|=\lim_{n\rightarrow \infty} 2|x| \frac{\sqrt{n}}{\sqrt{n+1}}=2|x|.$$ The series converges if $|x|<1/2$ and it diverges if $|x|>1/2$. Since the series is centered at $0$ the radius of convergence is $1/2$.

If $x=1/2$,   the series equals $\displaystyle  \sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ which is the $p$-series with $p=1/2$. Since $p<1$, the series diverges. 

If $x=-1/2$, the series equals $\displaystyle  \sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}$ . Since the sequence $\displaystyle \frac{1}{\sqrt{n}}$ is decreasing and it converges to $0$ as $b \rightarrow \infty$, the series converges by the Alternating Series Test. 

The interval of convergence for the first series is $[1/2,1/2)$.\vspace{.1in}


We consider the Ratio Test for $\displaystyle h(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-5)^n}{n\cdot3^n}$. $$\lim_{n\rightarrow \infty} \frac{\frac{|x-5|^{n+1}}{(n+1)3^{n+1}}}{\frac{|x-5|^n}{n \cdot 3^n}}=\lim_{n\rightarrow \infty} \frac{|x-5|\cdot n}{3(n+1)}=\frac{|x-5|}{3}.$$ The series converges if $|x-5|<3$ and it diverges if $|x-5|>3$. Thus the radius of convergence is $3$. 

If $x-5=3$, \emph{i.e.}, $x=8$, the series becomes the alternating harmonic series and it converges. 

If $x-5=-3$, \emph{i.e.}, $x=2$, the series equals $\displaystyle g(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(-1)^n}{n}= - \sum_{n=1}^{\infty} \frac{1}{n}$ which is the negative of the harmonic series and thus it diverges. 

The interval of convergence for the third series is $(2,8]$. 





\end{enumerate}
\end{enumerate}

\end{document}  
