\documentclass[11pt,pages,answers]{exam}
\usepackage{amsmath,amsfonts,times, graphicx,multicol,setspace,psfrag,epsfig}

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\newcommand{\ben}{\renewcommand{\theenumi}{\alph{enumi}}\renewcommand{\labelenumi}{(\theenumi)}\begin{enumerate}}
\newcommand{\een}{\end{enumerate}}

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\newcommand{\ea}{\end{longanswer}}


\newcommand{\bp}{\addtocounter{problemnum}{1}\begin{problem}}
\newcommand{\ep}{\end{problem}}

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 \newcommand{\R}{\mathbb{R}}
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\title{Midterm Exam}
\target{Math 158: Spring 2013}

\begin{document}


\begin{exam}{Tuesday, March 17}

\bp Carefully state:
\ben
\item The mean value theorem.
\ba Suppose that $f$ is continuous on $[a,b]$ and differentiable on
$(a,b)$.  Then there exists a number $c\in [a,b]$ with
$f'(c)=\frac{f(b)-f(a)}{b-a}.$ 
\ea
\item L'H\^opital's Rule.
\ba Suppose $\ds \lim_{x\to c }f(x)=\lim_{x\to c}g(x)=0$ and $\ds
\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists.  Then $\ds
\lim_{x\to c}\frac{f(x)}{g(x)}$ exists and $$\ds
\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}.$$
\ea 
\item The fundamental theorem of calculus.
\ba Let $f$ be function that is integrable on the interval $[a,b]$ and
let $c\in [a,b]$.  Define $A:[a,b]\to \R$ by $A(x)=\int_c^x f(t)dt$.  If $f$ is
continuous at $x$ then $A$ is differentiable at $x$ and $A'(x)=f(x)$.
\ea
\een
\ep

% \bp 
% Matching.  The answers are
% $$\frac{1}{3}, \quad \frac{\pi}{8}, \quad \frac{\pi}{4},  
% \quad \frac{1}{2},  \quad \log(2),  \quad
% \frac{\sqrt{\pi}}{2},  \quad 1,  \quad \frac{2}{e},
% \quad \frac{1}{e}, \quad \frac{\pi-2
%   \arccoth(\sqrt{2})}{4\sqrt{2}},\quad \infty
% $$

% \ben 
% \addtolength{\itemsep}{12pt}
% \item $\ds \int_1^\infty \frac{dt}{t^2}$
% \item $\ds \int_1^\infty \frac{dt}{t^2+1}$
% \item $\ds \int_1^\infty \frac{dt}{t^2+t}$
% \item $\ds \int_1^\infty \frac{dt}{e^t}$
% \item $\ds \int_1^\infty \frac{t dt}{e^t}$
% \item $\ds \int_0^\infty \frac{dt}{e^{t^2}}$
% \item $\ds \int_0^\infty \frac{tdt}{e^{t^2}}$
% \item $\ds \int_0^\infty \frac{dt}{t^4}$
% \item $\ds \int_1^\infty \frac{dt}{t^4}$
% \item $\ds \int_1^\infty \frac{dt}{t^4+1}$
% \item $\ds \int_1^\infty \frac{t dt}{t^4+t}$
% \een
% \ep


\bp  Matching
%\ben
%\item Matching

\vspace{0.5in}
\begin{itemize}
\addtolength{\itemsep}{40pt}
\item $\ds \int_1^\infty \frac{1}{t^4}dt=\frac{1}{3}$
\item $\ds \int_1^\infty \frac{1}{1+t^4}dt=\frac{\arccoth(\sqrt{2})}{4\sqrt{2}}$
\item $\ds \int_1^\infty \frac{t}{1+t^4}=\frac{\pi}{8}dt$
\end{itemize}

\bigskip

\hrule
\bigskip

%\item Matching

\vspace{0.5in}
\begin{itemize}
\addtolength{\itemsep}{40pt}
\item $\ds \int_0^\infty \frac{t}{e^t}dt=1$
\item $\ds \int_0^\infty \frac{t }{e^{t^2}}dt=\frac{1}{2}$
\item $\ds \int_0^\infty \frac{1}{e^{t^2}}dt=\frac{\sqrt{\pi}}{2}$
\end{itemize}

%\een
\ep

% \bp 
% \vspace{0.5in}
% \begin{multicols}{2}
% \begin{itemize}
% \addtolength{\itemsep}{40pt}
% \item $\ds \int_0^1 \frac{1}{\sqrt{1-t^2}}dt$
% \item $\ds \int_0^1 \frac{ 1}{\sqrt{1+t^2}}dt$
% \item $\ds \int_0^1 \frac{ t}{\sqrt{1-t^2}}dt$
% \end{itemize}

% \bigskip

% \begin{itemize}
% \addtolength{\itemindent}{1.5in}
% \addtolength{\itemsep}{50pt}
% \item $\ds \frac{\pi}{2}$
% \item $\ds 1$
% \item $\ds \log\left(1+\sqrt{2}\right)$
% \end{itemize}
% \end{multicols}

% \vspace{0.5in}
% \ep


\bp The so called \emph{Fresnel integral sine function} is defined to
be $$S(x)=\int_0^x \sin \left(\frac{\pi  t^2}{2}\right) \, dt.$$
% It cannot be evaluated exactly, but it's not difficult to determine
% the intervals on which it is monotonic and to determine the intervals
% on which it is concave up and concave down.

% Your problem:  label the $x$-coordinates of the indicated points on
% the graph of $S$ sketched below.

% \bigskip

% \begin{center}
% \includegraphics[scale=1.1]{graph.pdf}
% \end{center}

Which is the graph of $S$?

\vspace{36pt}

 \begin{center}
 \includegraphics[scale=0.5]{fresA.pdf}\hspace{24pt}\includegraphics[scale=0.5]{fresB.pdf}

\vspace{36pt}

\includegraphics[scale=0.5]{fresC.pdf}\hspace{24pt} \includegraphics[scale=0.5]{fresD.pdf}

 \end{center}


\ba By process of elimination, the answer is the one in the top right.  Note that
$S'(0)=\sin(0)=0$, which eliminates the graph on the top left.
Moreover, $S'(x)=\sin \left(\frac{\pi x^2}{2}\right) \geq 0$ for
$-\sqrt{2}\leq x \leq \sqrt{2}$, so $S$ is increasing on
$[-\sqrt{2},\sqrt{2}]$.  That elimates the two pictures on the bottom.
\ea

\ep


\bp  Here is the unit circle.  The shaded sector $AOC$ has area
$\frac{t}{2}$.  Use the integral formula for arclength to compute the
length of the circular arc from $A$ to $C$.


\begin{center}
\includegraphics[scale=0.5]{sector.pdf}
\end{center}

\ba The arc in question is part of the graph of $f(x)=\sqrt{1-x^2}$.
We compute $\sqrt{1+f'(x)^2}$:
\begin{align*}
\sqrt{1+f'(x)^2}&=\sqrt{1+\left(\frac{-x}{\sqrt{1-x^2}}\right)^2 }\\
&=\sqrt{1+\frac{x^2}{1-x^2}}\\
&=\sqrt{\frac{1}{1-x^2}}\\
&=\frac{1}{\sqrt{1-x^2}}
\end{align*}

The arc is the part of the graph  over the interval
$[\cos(t),1]$ so the length of the arc is given by 
\begin{align*}
\int_{\cos(t)}^1
\frac{1}{\sqrt{1-x^2}} &=\arcsin(t)\biggr]_{\cos(t)}^1 \\
&=\arcsin(1)-\arcsin(\cos(t))\\
&=\frac{\pi}{2}-\arcsin\left((\sin\left(\frac{\pi}{2}-t\right)\right)
\\
&= \frac{\pi}{2}-\frac{\pi}{2}+t\\
&=t
\end{align*}
as expected.
\ea

\ep

\bp 

Consider the region in the first quadrant bound by the unit circle.
\begin{center}
\includegraphics[scale=0.75]{quartercircle.pdf} 
\end{center}
If we rotate this region around the $y$-axis we obtain the upper
half of the unit hemisphere.
\begin{center}
\includegraphics[scale=0.75]{figure.pdf}
\end{center}
Your problem:  Find the volume of the northern hemisphere of the unit
sphere in two different ways:
\ep

\addtocounter{problemnum}{-1}
\bp \textbf{Continued.}
\ben
\item By discs.
\ba Rotating a horizontal slice of the quarter circle at $y$ yields a disc
of radius $r=x$, which has area $\pi r^2=\pi (x^2)=\pi (1-y^2).$  So we have
\begin{align*}
Volume &= \int_0^1 \pi (1-y^2) dy \\
&= \pi y-\frac{y^3}{3}\biggr]_0^1 \\
&=\frac{2}{3}\pi.
\end{align*}
\ea
\item By shells.
\ba Rotating a vertical slice of the quarter circle at $x$ yields a
cylindrical shell of height $y=\sqrt{1-x^2}$ and radius $x$, which has
area $2 \pi r h=2 \pi x \sqrt{1-x^2}.$
So we have 
\begin{align*}
Volume&=\int_0^1 2 \pi x \sqrt{1-x^2}dx \\
&=-\frac{2}{3}\pi \left(1-x^2\right)^{3/2} \biggr]_0^1\\
&=\frac{2}{3}\pi.
\end{align*}
\ea
\een
\ep

% \bp True or False.
% \ben
% \addtolength{\itemsep}{2in}
% \item $\ds \lim_{x \to 0}\frac{\cos(x
%     \sqrt{x})-1}{\sinh(x)\log(1+x^2)}=-\frac{1}{2}.$
% \item $\ds \lim_{x \to \infty}\frac{x+\sin(x)}{x}=1.$
% \item If $\ds \lim_{x\to c} f(x)$ exists and $\ds \lim_{x\to
%     c}g(x)$ does not exist, then $\ds \lim_{x\to c} f(x)g(x)$ does not exist.
% \een
% \ep

% \bp True or False.
% \ben
% \addtolength{\itemsep}{2in}
% \item $\ds \int_0^\infty \left(\frac{\sin(x)}{x}\right)^2dx $ converges.
% \item If  $\ds \int_1^\infty f(x) dx$ converges then $\ds \lim_{x\to
%     \infty} f(x)=0.$
% \item If  $\ds \lim_{x\to
%     \infty} f(x)=0$ then $\ds \int_1^\infty f(x) dx$ converges.
% \een
% \ep

% \bp Find the range of $f:[-3,3]\to \R$ defined by 
% $$f(x)=3 x^4-4 x^3-24 x^2+48 x+12$$
% % then $f([-3,3])=[-100,75].$
% \een
% \ep



% \bp True or False.  Right answer +1, wrong answer -1, no answer 0.
% \ben
% \addtolength{\itemsep}{0.5in}
% \item For every number $p$, the integral $\ds \int_0^\infty \frac{dt}{t^p}$ diverges.
% \item For any number $p>0$, $\ds \lim_{x \to
%     \infty}\frac{\log(x)}{x^p}=0.$ 
% \item For any number $p>0$, $\ds \lim_{x\to \infty}\frac{x^p}{e^x}=0$.
% \item If $\ds \int_1^\infty f(x)dx$ converges, then $\ds \lim_{x\to \infty}f(x)=0.$
% \item If $\ds \lim_{x\to c}f(x)=L$ and $\ds \lim_{x \to c }g(x)$ DNE, then
% $\ds \lim_{x \to c} f(x) g(x)$ DNE.
% % \item If $\ds \lim_{x\to c}f(x)=L$, $\ds \lim_{x \to c }g(x) DNE$, and
% % $\ds \lim_{x \to c} f(x) g(x) $ exists, then $L=0$.
% \item If $\ds \lim_{x\to c}f(x)=0$, $\ds \lim_{x \to c }g(x)=0$
%   and $\ds \lim_{x \to c}\frac{f'(x)}{g'(x)}$ DNE then 
%   $\ds \lim_{x \to c}\frac{f(x)}{g(x)}$ DNE.
% \item If $\ds \lim_{x \to c}\frac{f'(x)}{g'(x)}=L$ exists, then
%   $\ds \lim_{x \to c}\frac{f(x)}{g(x)}=L$ also.
% \item $\ds \lim_{x \to 0}\frac{\cos(x \sqrt{x})-1}{\sinh(x)\log(1+x^2)}=-\frac{1}{2}.$
% \een
% \ep



\end{exam}
\end{document}

