\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\def\whichexam{Exam III}
\def\examdate{December 2, 2003}
\def\numberpages{0}
\def\numberofproblems{15}
\def\multiplechoicemax{5}
\def\partialcreditmax{10}
%%
\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!) 
\def\correctanswers{%
\ll  1 25314
\ll  2 51324
\ll  3 32415
\ll  4 13254
\ll  5 15324
\ll  6 43512
\ll  7 32415
\ll  8 52431
\ll  9 21354
\ll 10 43125
}

\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)
\pdfximage              % include the image at resolution 1200 dpi
    width \dimen0 {#1}  % by setting image width to the calculated value
\vtop to#4{\vss\hbox to#3{\hbox to \dimen0{\pdfrefximage \pdflastximage}\hss}}
}


\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�#2�#3�#4�#5�#6�#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}
\let\Correctanswers=\correctanswers
\def\correctanswers{\global\numbermultiplechoice=0 \gdef\list{}\Correctanswers}
\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}

\newdimen\picH\newdimen\picV
\def\picture #1 by #2 (#3 scaled #4){
\picV=#2 \divide\picV by1000 \multiply\picV by#4 
\picH=#1 \divide\picH by1000 \multiply\picH by#4
\vtop to \picV{
      \hrule width \picH height 0pt depth 0pt
      \vfill
      \special{picture #3 scaled #4}}}

\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}

\long\def\Newp#1.#2�#3a)#4\endp{
\newp#1.#2
\Xan#3
a)#4\endp}


\long\def\newp#1.#2\endp{\problemspacing%
\noindent\vbox{\noindent{\bf\number\pc.}(\number\credit\ pts.)
\global\advance\pc by1 %
\global\advance\breakpc by1 %
#2\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\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=20pt
\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
\long\def\Xan#1a)#2b)#3c)#4d)#5e)#6\endp{%
\Ans#1�
#2�
#3�
#4�
#5�
#6�
\endp}

%%%%%%%
\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.
\vskip5pt
\centerline{Good Luck!}
\vskip4pt
\def\HH{\hskip.75in}
\settabs\+9.\qquad &\HH&\HH&\HH&\HH&\HH&\cr
\hskip.5 in\boxit{\hbox to 3.8in{\vbox{\obeylines\baselineskip20pt
\noindent
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
\correctanswers
}\hss}}{10pt}{5in}
%%\vskip10pt

\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
\Newp 1.
Evaluate
$\displaystyle\mathop{\lim}_{x \rightarrow - \infty} \frac{\sqrt{4x^6 + 3}}{x^3 + 2} $
�5
a)$-2$\\
b)$ 4$\\
c) $3/2$\\
d)$6$\\
e) $2$\\
\endp

\Newp 2. 
 If $f'(x)= \sqrt{x} + \frac {1}{\sqrt{x}}$ and  $f(1) = \frac{8}{3}$, find
the value of $f(4)$.
�5
a)$ 28/3$\\
b)$ 27/3$\\
c)$ 26/3$\\
d)$25/3$\\
e)$24/3$\\
\endp

\Newp 3. 
Estimate the area under the graph of $f(x)=\frac{1}{x}$ from
$x=1$ to $x=3$ using a subdivision of the interval into $4$ equal
subintervals and using values of $f(x)$ at the {\bf right-hand end-points}
of the  subintervals.
�5
a)$19/20$\\
b) $9/10$\\
c)$17/20$\\
d)$1$\\
e)$4/5$\\
\endp

\Newp 4. 
Evaluate the following indefinite integral
$\displaystyle\int \ \frac{\sin x}{\ \sqrt{2+\cos x\>}\ }\ dx$
�2
a) $-2\sqrt{2+\cos x\>}+ C$\\
b) $\sqrt{(2+\cos x)^3\>}+ C$\\
c) $ \frac{1}{\sqrt{(2+\cos x)^3\>}} + C$\\
d) $2\sqrt{2+\sin x\>} + C$\\
e) $\frac{2\cos x}{\sqrt{2+\sin x\>}} + C$\\
\endp

\Newp 5.
Evaluate the following definite integral
$\displaystyle\int_{-8}^0 \sqrt{64 - x^2\>}\>dx $
�5
a) $16 \pi$\\
b) $64\pi$\\
c) $32 \pi$\\
d) $8 \sqrt{\pi}$\\
e) $0$\\
\endp

\Newp 6.
Let $F(x) = \Int_0^{x^3} \sec t \>dt $ for $0 \leq x \leq 1$.
What is $F'(x)$?
�3
a) $3x^2 \sec (x^3)$\\
b) $\sec x$\\
c) $\sec (3x^2) + C$\\
d) $\sec x^3$\\
e) $\frac{x^4}{4} \sec x^3 $\\
\endp

\Newp 7. 
Evaluate the definite integral
$\displaystyle \int_0^1 x^2(\sqrt{x} + 3)\ dx = ? $
�5
a)$9/7$\\
b)$9/2$\\
c)$7/5$\\
d)$5/3$\\
e)$5/2$\\
\endp

\Newp 8. 
Evaluate the definite integral
$\displaystyle\int_0^\pi (2x+1) \sin(x^2 + x)\ dx $.
�2
a) $-\cos(\pi^2 + \pi) + 1$\\
b) $\sin(\pi^2 + \pi) - 1$\\
c) $-\cos(\pi^2) + \cos( \pi + 1)$\\
d) $2 \cos(\pi^2 + \pi) - 2$\\
e) 0\\
\endp

\Newp 9. 
Find the area of the region bounded by the curves $y = x^3 -
x$ and $ y = 3x$.
�5
a) $8$\\
b) $4$\\
c) $-4$\\
d) $0$\\
e) $16$\\
\endp

\Newp 10.
The volume of the solid obtained by rotating the region
bounded by $y = \sec x$, $y = 1$, $x = -1$ and $x = 1$ about the $x$-axis
is given by which of the following integrals:
�2
a) $\pi \Int_{-1}^1 (\sec^2 x -1) dx $\\
b) $\pi \Int_{-1}^1 (1 - \sec^2 x) dx$\\
c) $2\pi \Int_0^1 x(1 - \sec x)^2 dx$\\
d) $\pi \Int_{-1}^1 x(\sec^2 x -1) x dx$\\
e) $\pi \Int_{-1}^1 (\sec x - 1)^2 dx$\\
\endp

\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}



\newp 11.
\dimen0=3in
 Consider the function
$\displaystyle y=\frac{x^2}{x^2 + 3} $

\bigskip
\noindent
(a) On which intervals is the function decreasing?
\bigskip
Answer: $\underline{\hskip\dimen0}$

\vskip 4true cm
\noindent
(b) On which intervals is the function concave up?
\bigskip
Answer: $\underline{\hskip\dimen0}$

\vskip6true cm
\noindent
(c) Does the function have any horizontal asymptotes and if so what are
they?
\bigskip
Answer: $\underline{\hskip\dimen0}$

\vskip 3true cm
\noindent
(d) Sketch the function on the next page.
The top graph is for you to do your scratch work.
Use the bottom graph for your final answer.
\endp\vfill\eject
\null\vskip20pt
\centerline{\sl Scratch work}\vskip-80pt
\noindent\null\hskip-20pt\LoadPDF{BlankGraph.pdf}{0.8}{3 true in}{3 true in}
\vskip4true in
\centerline{\bf Final answer}\vskip-80pt
\noindent\null\hskip-20pt\LoadPDF{BlankGraph.pdf}{0.8}{3 true in}{3 true in}
\vfill\eject\endp


\newp 12. 
A cylindrical can without a top is made to contain 1000
$\rm{cm}^3$ of liquid.  Find the dimensions of the can that will
minimize the cost of the metal to make the can.
Be sure to show that your answer is actually an absolute minimum.
\endp

\newp  13. 
A particle is moving in a straight line with acceleration
$$a(t)=1+\cos t$$
and initial velocity $v(0)=0$ and initial position $s(0)=0$.
Find the position of the particle at time $t$.
Show your work.
\endp

\newp 14. 
In attempting to solve the equation $2 - \sec x  = \tan x$ on
$[0,\frac{\pi}{2})$ by Newton's method we begin with
$x_1=\frac{\pi}{4}$. Find the value of $x_2$ in this process.
Show your work.
\endp

\newp 15.
 Use the washer method to find the volume of the solid
obtained by rotating the region bounded by $y = x+2$ and $y = 4x - x^2$
about the $y$-axis. XXXX line $y=3$.
Show your work.
\endp

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


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


