\magnification1200\footline{\ifnum\pageno=1 \else\number\pageno\fi}
\def\whichexam{Math 126 Exam III}
\def\examdate{April 20, 2004}
\def\numberpages{0}
\def\numberofproblems{14}
\def\multiplechoicemax{5}
\def\partialcreditmax{15}
%%
\newif\ifprelim
\newif\ifproblemspacing

\def\START#1{\ifnum#1=0{\global\prelimtrue\global\problemspacingfalse}%
\else{\ifnum#1=1{\global\prelimfalse\global\problemspacingtrue}%
\else{\global\prelimfalse\global\problemspacingfalse}\fi}%
\fi}

\START 1 %% 0 - initial run for instructors
                  %% 1 - run for the actual exam
                  %% 2 - for the webpage
%%
%%%%%  The macro \correctanswers
%%%%%  sets the number of partial credit problems automatically.
%%%%%  We assume that the first answer listed after the problem is
%%%%%  the correct one.
%%%%%  The permutation indicates in which slots to put the answers
%%%%%  (so the first number in the permutation locates the correct answer!) 
%: Answer permutations
\def\correctanswersA{%
\ll  1 54213
\ll  2 21453
\ll  3 43521
\ll  4 21435
\ll  5 24351
\ll  6 54231
\ll  7 35241
\ll  8 15243
\ll  9 45213
\ll 10 14532
\ll 11 34152
}

%%% MULTIPLE CHOICE ANSWER BOX
%% Two columns
\def\interAskip{\hskip.4in}
\def\interNumberAskip{\hskip.3in}

\def\answercolumnSize{\hsize=2in}
\def\answerBoxSize{4.5in}
\def\answerBoxInset{10pt}

\def\displayanswers{\par\noindent\vtop{\answercolumnSize\correctanswersA}%
\vtop{\answercolumnSize\correctanswersB}}
%% One column
\def\interAskip{\hskip.75in}
\def\interNumberAskip{\hskip.3in}

\def\answercolumnSize{\hsize=3.8in}
\def\answerBoxSize{3.8in}
\def\answerBoxInset{0.5in}

\def\displayanswers{\par\noindent\vtop{\answercolumnSize\correctanswersA}}
%%% END MULTIPLE CHOICE ANSWER BOX

% #1 - path to pdf file
% #2 - scale factor
% #3 #4  - size of box pdf is centered in this box. It can overflow box
\def\LoadPDF#1#2#3#4{
\pdfximage {#1}
\setbox0=\hbox{\pdfrefximage\pdflastximage} % get dimensions of the image
                                            % in order to include the image
                                            % at a specific resolution 
\dimen0=#2\wd0         % calculate the image width at 1200 dpi (0.06 = 72/1200)
\dimen1=#2\ht0
\pdfximage              % include the image at resolution 1200 dpi
    width \dimen0 {#1}  % by setting image width to the calculated value
\divide\dimen0 by 2
\divide\dimen1 by 2
\dimen2=#3 \divide\dimen2 by 2
\dimen3=#4 \divide\dimen3 by 2
\vbox to #4{\vskip\dimen3\hbox to#3{\hskip\dimen2%
\hbox to0pt{\vtop to0pt{\vskip-\dimen1 %
\hbox to 0pt{\hskip -\dimen0\pdfrefximage \pdflastximage\hss}\vss}\hss}%
\hss}\vss}%
}
\def\boxLoadPDF#1#2#3#4{
\pdfximage {#1}
\setbox0=\hbox{\pdfrefximage\pdflastximage} % get dimensions of the image
                                            % in order to include the image
                                            % at a specific resolution 
\dimen0=#2\wd0         % calculate the image width at 1200 dpi (0.06 = 72/1200)
\dimen1=#2\ht0
\pdfximage              % include the image at resolution 1200 dpi
    width \dimen0 {#1}  % by setting image width to the calculated value
\divide\dimen0 by 2
\divide\dimen1 by 2
\dimen2=#3 \divide\dimen2 by 2
\dimen3=#4 \divide\dimen3 by 2
\boxit{\vbox to #4{\vskip\dimen3\hbox to#3{\hskip\dimen2%
\hbox to0pt{\vtop to0pt{\vskip-\dimen1 %
\hbox to 0pt{\hskip -\dimen0\pdfrefximage \pdflastximage\hss}\vss}}%
\hss}\vss}}{1pt}{1pt}%
}


\long\def\hide #1\endhide{}

\def\rhb#1{\noindent\hbox to\hsize{\hfil#1}}

\def\longans#1{\vtop{\noindent\hsize=6.25 true in #1}}
\def\arctan{{\rm arctan\>}}

\long\def\Ans#1\ui#2\ui#3\ui#4\ui#5\ui#6\ui#7\endp{
\ifcase#1\relax\or\one\or\two\or\three\or\four\or\five\fi
\AN{#2}{#3}{#4}{#5}{#6}\Pra}

\def\dxd{\displaystyle}

\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\newcount\cnumberofproblems
\cnumberofproblems=\numberofproblems
\global\advance\cnumberofproblems by 1 

\font\ld=cmbx12
\def\endp{\relax}
\newcount\totalproblems\totalproblems=\numberofproblems
\newtoks\ta\newtoks\tb
\long\def\rightappenditem#1\to#2{%
\ta={\\{#1}}\tb=\expandafter{#2}%
\xdef#2{\the\tb\the\ta}}
\def\leftremove#1\to#2{\expandafter\lopoff#1\lopoff#1#2}
\long\gdef\lopoff\\#1#2\lopoff#3#4{\gdef#4{#1}\gdef#3{#2}}

\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\def\loop#1\repeat{\def\body{#1}\iterate}
\def\iterate{\body\let\next=\iterate\else\let\next=\relax\fi\next}
\font\small=cmr7
\def\newpage{\vfill\eject}
\newdimen\pspacewidth
\newcount\credit
\newcount\pspace\pspace=0
\newcount\pc\pc=1
\newcount\breakpc\breakpc=1
\hfuzz=5pt
\let\xxsum=\sum
\edef\sum{\displaystyle\xxsum}

\newcount\mc
\def\newchoice#1{\ifnum\breakpc<3 \newpage\else\ifprelim\else%
\ifproblemspacing\newpage\else\bigskip\fi\fi\fi\null%
\vskip10pt\centerline{{\bf#1}}\mc=0 %
\ifnum\breakpc<3\credit=\multiplechoicemax \else\credit=\partialcreditmax \fi}
\newif\ifmultichoice
\def\problemspacing{\ifprelim\vskip10pt\headline{\hfill}\else%
\ifproblemspacing
\ifmultichoice%
\ifnum\mc>0 \ifodd\breakpc\relax\newpage\null\vskip10pt\else\vfill\fi\fi%
\else%
\ifnum\mc>0 \newpage\null\vskip10pt\fi\fi%
\else \vskip10pt\headline{\hfill}\fi%
\mc=1 \fi}
%: Multiple choice problem macro
\long\def\Newp #1\ui#2#3\\#4\\#5\\#6\\#7\\#8\endp{%
\newp#1
\Ans#2\ui
#3\ui
#4\ui
#5\ui
#6\ui
#7\ui
\endp}


\long\def\newp#1\endp{\problemspacing%
\noindent\vbox{\noindent{\bf\number\pc.}(\number\credit\ pts.)
\global\advance\pc by1 %
\global\advance\breakpc by1 %
#1\endp}\vfil}


\newbox\Aa\newbox\Ab\newbox\Ac
\newbox\Ad\newbox\Ae
\newbox\aa\newbox\ab\newbox\ac
\newbox\ad\newbox\ae

\newdimen\Aspace
\newcount\Ax
\newcount\tmpPC
\def\rememberPC{\global\tmpPC=\pc\global\advance\tmpPC by-1 }

\hfuzz=20pt

\def\five{\global\Ax=5 \global\Aspace=.2\hsize\relax\global\advance\Aspace by-20pt}
\def\four{\global\Ax=4 \global\Aspace=.25\hsize\relax\global\advance\Aspace by-20pt}
\def\three{\global\Ax=3 \global\Aspace=.3\hsize\relax\global\advance\Aspace by-20pt}
\def\two{\global\Ax=2 \global\Aspace=.5\hsize\relax\global\advance\Aspace by-20pt}
\def\one{\global\Ax=1 \global\Aspace=\hsize\relax\global\advance\Aspace by-20pt}

\long\def\AN#1#2#3#4#5{%
\setbox\Aa=\hbox to\Aspace{#1\hfil}%
\setbox\Ab=\hbox to\Aspace{#2\hfil}\setbox\Ac=\hbox to\Aspace{#3\hfil}%
\setbox\Ad=\hbox to\Aspace{#4\hfil}\setbox\Ae=\hbox to\Aspace{#5\hfil}}

\newdimen\Ls\Ls=10pt

\def\prafive{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad%
(e)\quad\box\Ae}

\def\prafour{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad\vskip\Ls\noindent%
(e)\quad\box\Ae}

\def\prathree{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad%
(e)\quad\box\Ae}

\def\pratwo{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\praone{\vskip\Ls\noindent(a)\quad\box\Aa\vskip\Ls\noindent(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\resetboxes{\setbox\Aa=\box\aa\setbox\Ab=\box\ab\setbox\Ac=\box\ac%
\setbox\Ad=\box\ad\setbox\Ae=\box\ae}
\def\switchbox#1#2{\ifcase#1 \relax\or\setbox\aa=\box#2\or\setbox\ab=\box#2\or%
\setbox\ac=\box#2\or\setbox\ad=\box#2\or\setbox\ae=\box#2 \fi}

\newcount\pcc
\def\mooseAntlers#1#2#3#4#5@{\gdef\pA{#1}\gdef\pB{#2}\gdef\pC{#3}\gdef\pD{#4}\gdef\pE{#5}}
\def\Pra{\leftremove\list\to\moose%
\expandafter\mooseAntlers\moose@%
\switchbox{\pA}{\Aa}\switchbox{\pB}{\Ab}%
\switchbox{\pC}{\Ac}\switchbox{\pD}{\Ad}\switchbox{\pE}{\Ae}%
\resetboxes%
\ifcase\Ax \or\praone\or\pratwo\or\prathree\or\prafour\or\prafive\fi%
\pcc=\pc\advance\pcc by -1 %
\immediate\write16{Problem \number\pcc - Correct Answer \number\pA}%
}


\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\def\smsmfrac#1#2{{{\scriptscriptstyle#1}\over{\scriptscriptstyle#2}}}
\def\Lim{\displaystyle\lim}
\def\Int{\displaystyle\int}
\def\arcsec{\rm arcsec}

\def\frac#1#2{{\displaystyle{{\textstyle#1}\over{\textstyle#2}}}}

\long\def\boxit#1#2#3{\vbox{\hrule\hbox{\vrule\kern#2
\vbox{\hsize=#3\kern#2#1\kern#2}\kern#2\vrule}\hrule}}
\newcount\pc
\def\pr{\vskip10pt\noindent\number\pc.\global\advance\pc by 1}
\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}

\def\rhs#1#2{\par\noindent\hbox to\hsize{\hfill\hbox to#1{#2\hfill}}}
\def\ul#1{\vrule width#1 height -3pt depth3.1pt}
\def\UL#1#2{\hbox to0pt{\qquad #2\hss}\ul{#1}}
\newif\ifnotmarkanswers\notmarkanswerstrue
\newcount\ansmark\newcount\numbermultiplechoice\newcount\numberpartialcredit
\newdimen\AA\AA=10pt
\def\ll#1 #2#3#4#5#6{\tabalign {#1.}&
\ifprelim
\global\ansmark=1 \relax%
\else
\global\ansmark=#2 \relax%
\fi%
\ifnotmarkanswers \global\ansmark=0 \fi%
\ifnum\ansmark=1 \relax($\bullet$)\else(a)\fi&
\ifnum\ansmark=2 \relax($\bullet$)\else(b)\fi&
\ifnum\ansmark=3 \relax($\bullet$)\else(c)\fi&
\ifnum\ansmark=4 \relax($\bullet$)\else(d)\fi&
\ifnum\ansmark=5 \relax($\bullet$)\else(e)\fi&
\hskip\AA\cr\global\advance\numbermultiplechoice by1 
\ifprelim
\rightappenditem{12345}\to\list
\else
\rightappenditem{#2#3#4#5#6}\to\list%
\fi%
}

\def\nomark{\pageno=1 %
\global\notmarkanswerstrue
\headline{\vbox{\rhs{3in}{Name:\ \UL{3in}{\hfil}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{\hfill}\hss}%
}}
\answersheet}

\def\Nomark#1{\pageno=1 %
\global\notmarkanswerstrue
\headline{\vbox{\rhs{3in}{Name:\ \UL{3in}{\hfil}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{\quad #1\hfil}\hss}%
}}
\answersheet}


\def\marked{\pageno=1 %
\global\notmarkanswersfalse
\headline{\vbox{\rhs{3in}{Name:\ \UL{3in}{ANSWERS}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{ANSWERS}\hss}%
}}
\answersheet}

%%\let\\=\relax
%%%%%%%
\font\sfm=cmr8

\def\eat#1{}


\def\exam#1{\ifprelim\Exam{ }
\else\expandafter\EExam#1,@\fi%
}

\def\EExam#1,#2@{\Exam{#1}\def\xx{#2}%
\ifx\empty\xx\let\next=\eat\else\let\next=\EExam\fi\next#2@}

\def\+{\tabalign}

\long\def\Exam#1{%
\pc=1 \pageno=1 \pspace=0 %
\headline{\vbox{\rhs{3in}{Name:\ \ul{3in}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{#1}\hss}%
}}


\long\gdef\answersheet{
\null\vskip10pt
\centerline{\whichexam}
\centerline{\examdate}
\vskip10pt

\itemitem{$\bullet$}
The Honor Code is in effect for this examination.
All work is to be your own.
\itemitem{$\bullet$} No calculators.
\itemitem{$\bullet$} The exam lasts for one hour.
\itemitem{$\bullet$} Be sure that your name is on every page in case
pages become detached. 
\itemitem{$\bullet$} Be sure that you have all \numberpages\ pages of the test. 
\itemitem{$\bullet$} The backs of pages may be used if you need additional room
to work on a problem.
\vskip5pt
\centerline{Good Luck!}
\vskip4pt
\settabs\+\interNumberAskip&\interAskip&\interAskip&\interAskip&\interAskip&\interAskip&\cr
\hskip\answerBoxInset\boxit{\hbox to \answerBoxSize{\vbox{\obeylines\baselineskip20pt
\noindent
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
\global\numbermultiplechoice=0 \gdef\list{}%

\displayanswers}\hss}}{10pt}{10in}

\global\numberpartialcredit=\numbermultiplechoice
\global\advance\numberpartialcredit by 1
\settabs\+Total multiple choice: &\hskip 2in\cr
\hfill{{\vbox to0pt{\boxit{\vbox{\obeylines\hsize=3in\baselineskip26pt
DO NOT WRITE IN THIS BOX!
\tabalign Total multiple choice: &\quad\ul{1in}\cr
\loop\ifnum\numberpartialcredit< \cnumberofproblems
\tabalign\hfill\number\numberpartialcredit.&\quad\ul{1in}\cr\global\advance\numberpartialcredit by 1
\repeat

\tabalign\hfil {\bf Total:}&\quad\ul{1in}\cr
}}{10pt}{1in}\vss}}}}


\setbox0=\vbox{\answersheet}\pageno=2
\multichoicetrue
\newchoice{Multiple Choice}
\def\kg{{\rm kg}}
\def\log{{\rm log}}
\credit=5
\null\vskip6pt
%material covered: 
%10.6 (linear eqs), 
%11.1(parametric curves), 
%11.2(parametric area, arc length, surface area), 
%11.3(polar coordinates), 
%11.4(polar length and area), 
%12.1(sequences), 
%12.2(series), 
%12.3(integral test), 
%12.4(comparison tests), 
%12.5(alternating series)
%end material covered
%: Question 1
\Newp
Find the solution to the differential equation with initial value:
$$x^3 y^\prime + x^2 y = 3x^4 + x\qquad y(1)=0$$
\ui3
$y = x^2 + \frac{\ln x}{x} - \frac{1}{x}$\\
$y = \frac{3x^4}{4} + x -\frac{7}{4}$\\
$ y = e^x -e$\\
$y = e^x(x^2-3x+2)$\\
$y = \ln(x+e^x-1) - 1$.\\
\endp
%: Question 2
\Newp
Which integral below computes the area above the $x${--}axis and 
below the parameterized curve 
$x(t) = t - \sin(t) + e^t$ and $y(t)=\cos(t)$ for $0\leq t \leq \pi/2$.
\ui2
$\dxd\int_0^{\pi/2} \cos(t) (1 - \cos(t) +e^t)\ dt$\\
$\dxd\int_0^{\pi/2} \sin(t) (1 - \cos(t) +e^t)\ dt$\\
$\dxd\int_0^{\pi/2} \sin(t) (1 - \sin(t) +e^t)\ dt$\\
$\dxd\int_0^{\pi/2} \cos(t) (1 - \sin(t) +e^t)\ dt$\\
$\dxd\int_0^{\pi/2} \sin(t) (1 + 2\cos(t) +e^t)\ dt$\\
\endp
%: Question 3
\Newp
Which graph below is the graph of the polar equation $r = 3 \cos(3\theta)$.
\ui2
\lower 1in\vbox{\LoadPDF{./polarGraphs/problem2b.pdf}{.35}{2in}{2in}}\\
\lower 1in\vbox{\LoadPDF{./polarGraphs/problem2a.pdf}{.35}{2in}{2in}}\\
\lower 1in\vbox{\LoadPDF{./polarGraphs/problem2c.pdf}{.35}{2in}{2in}}\\
\lower 1in\vbox{\LoadPDF{./polarGraphs/problem2d.pdf}{.35}{2in}{2in}}\\
\lower 1in\vbox{\LoadPDF{./polarGraphs/problem2e.pdf}{.35}{2in}{2in}}\\
\endp
\global\advance\breakpc by 1
%: Question 4
\Newp
Which integral below computes the length of the parameterized curve \par\noindent
$x(t) = t  - e^t$ and $y(t)= t^2 + e^t$ for $0\leq t \leq \pi$.
\ui2
$\dxd\int_0^\pi \sqrt{1- 2e^t + 2e^{2t} +4t^2 + 4t e^t \ }\ dt$\\
$\dxd\int_0^\pi \sqrt{t^2 - 2te^t + 2e^{2t} +t^4 + 2t^2 e^t \ }\ dt$\\
$\dxd\int_0^\pi \sqrt{t^2+ 6te^t - 2e^{2t} +4t^2 \ }\ dt$\\
$\dxd\int_0^\pi \sqrt{1- 2e^t  + 2e^{2t} +t^4 + 2t^2 e^t \ }\ dt$\\
$\dxd\int_0^\pi \sqrt{1+ 2e^t - 4t^2 - 4t e^t \ }\ dt$\\
\endp

%: Question 5
\Newp
Which integral below is the area of the region inside the circle $r = 2\sin\theta$ 
and outside the circle $r =  \sin\theta$.

\hskip100pt \vbox{\LoadPDF{./areaMUProblem.pdf}{.5}{2in}{2in}}
\ui3
$\frac{1}{2}\dxd\int_0^\pi 3\sin^2\theta\ d\theta$\\
$\frac{1}{2}\dxd\int_0^\pi 3\cos^2\theta\ d\theta$\\
$\frac{1}{2}\dxd\int_0^\pi 4\cos^2\theta\ d\theta$\\
$\frac{1}{2}\dxd\int_0^\pi 4\sin^2\theta\ d\theta$\\
$\frac{1}{2}\dxd\int_0^\pi 5\cos^2\theta\ d\theta$\\
\endp

%: Question 6
\Newp
Find the length of the polar spiral $r = e^{2\theta}$, $0\leq \theta\leq 3$.
\ui3
$\frac{\sqrt{5 }}{2}\>(e^6-1)$\\
$\frac{\sqrt{5 }}{2}\>(e^6+1)$\\
$\frac{1}{2}\>(e^6-1)$\\
$\frac{1}{2}\>(e^6+1)$\\
$\frac{\sqrt{5 }}{2}\>e^6 - 1$\\
\endp
%: Question 7
\Newp
Which statement below is true about the series
$\sum_{n=1}^\infty\ \frac{2^n}{n^2 + 2^n}$
\ui1
$\dxd\lim_{n\to\infty}\ \frac{2^n}{n^2 + 2^n} = 1$ so the series diverges.\\
$\dxd\lim_{n\to\infty}\ \frac{2^n}{n^2 + 2^n} = 1$ so the series converges.\\
$\dxd\lim_{n\to\infty}\ \frac{2^n}{n^2 + 2^n} = 0$ so the series diverges.\\
$\dxd\lim_{n\to\infty}\ \frac{2^n}{n^2 + 2^n} = 0$ so the series converges.\\
$\dxd\lim_{n\to\infty}\ \frac{2^n}{n^2 + 2^n}$ does not exist so the series converges.\\
\endp

%: Question 8
\Newp
Sum the series $\sum_{n=1}^\infty\ \frac{2^n}{5^{2n}}$.
\ui5
$\frac{2}{23}$\\
$\frac{100}{23}$\\
$\frac{92}{25}$\\
$\frac{25}{23}$\\
$\frac{5}{3}$\\
\endp

%: Question 10
\Newp 
Consider the following two series:
\bigskip
\centerline{I) $\sum_{n=2}^\infty \frac{1}{n^3-n^2}$\qquad and \qquad
II) $\sum_{n=2}^\infty \frac{1}{n^2}$\ .}
\bigskip
Which statement below is true?
\ui1
Both series converge.\\
Both series diverge.\\
Series I) converges but series II) diverges.\\
Series II) converges but series I) diverges.\\
None of the other statements are true.\\
\endp

%: Question 9
\Newp
Which of the statements below is true about the three series\par
I) $\sum_{n=1}^\infty\ \frac{(-1)^{n-1}}{n}$\hfil
II) $\sum_{n=1}^\infty\ \frac{(-1)^{n-1}}{n^2}$\hfil
III) $\sum_{n=1}^\infty\ \frac{(-1)^{n-1}}{n!}$
\ui1
I) conditionally converges; II) and III) absolutely converge\\
They all absolutely converge.\\
They all conditionally converge.\\
I) diverges, II) conditionally converges and III) absolutely converges.\\
I)  absolutely converges, II) conditionally converges and III) diverges.\\
\\
\endp

%: Question 11
\Newp
Which series below conditionally converges?
\ui4
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\root 7\of {n^3\ }}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}\sqrt{n}}{n^2}$\\
$\sum_{n=1}^\infty \frac{(-1)^n n!}{n^2}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\root 3\of {n^4\ }}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}e^n}{\root 5\of {n^2\ }}$\\
\endp

%: Start partial credit
\multichoicefalse
\def\vvbv{\vbox{\hbox to 6in{\hfill Partial Credit\hfill}%
\hbox to 6in{\rm%
\hfill You must show your work on the partial credit problems 
to receive credit!\hfill}%
\vskip10pt}}
\newchoice{\vvbv}

%: Question 12
\newp
Use the Limit Comparison Test to show that the series\quad
$$\sum_{n=3}^\infty \frac{n^3-6n^2+4n-2}{2n^5-n^4+3n^2}$$
converges.
\endp
%: Question 13
\newp
Set up an integral which computes the shaded area. 
The polar equation of the region is $r = 2\sin\theta - 1$.
\par\bigskip\noindent{\bf Remarks:} $\sin\left(\frac{\pi}{6}\right) = \frac 12$.
The answer is the difference of two areas.
\vskip 20pt
\vbox{\LoadPDF{./areaMUPolar.pdf}{.7}{5in}{4in}}
\endp

%: Question 14
\newp
Set up an integral which computes the surface area of the surface obtained 
by rotating the parameterized  ellipse $x(t)=5+4\cos(t)$, $y(t)=3\sin(t)$, $0\leq t\leq 2\pi$ about the
line $x=9$ .

\vbox{\LoadPDF{./surfaceAreaProblem.pdf}{.7}{5in}{4in}}

\endp

\xdef\numberpages{\number\pageno}\vfill\eject
}


\exam{}
\marked{}
\vfill\eject
%\nomark
\Nomark{}
\end
