\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
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\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}




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

\def\whichexam{Exam II}
\def\examdate{October 28, 2003}
\def\numberpages{0}
\def\numberofproblems{15}
\def\multiplechoicemax{5}
\def\partialcreditmax{10}
%%
\newif\ifprelim
\prelimfalse
%\prelimtrue
%%
\newif\ifproblemspacing
\problemspacingtrue
\problemspacingfalse
\gdef\f(#1){\smfrac{1}{\cos(#1)}}
\gdef\f(#1){\sec(#1)}
\gdef\smmfrac#1#2{{{#1}\over{#2}}}
%%
%%%%%  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 13245
\ll  2 25431
\ll  3 42135
\ll  4 43251
\ll  5 15324
\ll  6 24315
\ll  7 43512
\ll  8 34512
\ll  9 51432
\ll 10 51243
}

\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}

\long\def\xan#1a.#2b.#3c.#4d.#5e.#6\endp{%
\Ans#1¥
#2¥
#3¥
#4¥
#5¥
#6¥
\endp}
\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
\Newp 1.
A car is traveling along a straight highway.
The position at time $t$ is given by $s(t) =  -t^3 + 10t^2 + 7t$.
At exactly one of the five times below the car is slowing down.
Which time is it?
¥5
a)$4$\\
b)$3$\\
c)$2$\\
d)$1$\\
e)$0$\\
\endp

\Newp 2.
Find $\frac{d^3\hbox to 2pt{\ $\bigl(x^5 + 2x^4-7x^3+3x^2+9x-8\bigr)$\hss}}{dx^3}$
¥3
a)$60x^2 + 48x - 42$\\
b)$60x^2  - 42x +48$\\
c)$60x^2 +24x -6$\\
d)$60x^3 +6$\\
e)$60x^3 - 48x^2 +8x$\\
\endp

\Newp 3.
Find the differential of the function $f(x) = 2x^3 - 4x^2 +7x +10$ when $x=1$ and $dx=0.1$.
¥5
a)$0.5$\\
b)$0.75$\\
c)$0.25$\\
d)$1$\\
e)$0$\\
\endp

\Newp 4.
Consider the function $f(x)=x^3 - 3x +2003$. 
The critical numbers of $f$ are $-1$ and $1$.
The function $f$ has
¥1
a) a local maximum at $x=-1$ and a local minimum at $x=1$.\\
b) an absolute minimum at $x=-1$ and an absolute maximum at $x=1$.\\
c) a local minimum at $x=-1$ and an absolute maximum at $x=1$.\\
d) an absolute maximum at $x=-1$ and a local minimum at $x=1$.\\
e) an absolute minimum at $x=-1$ and a local maximum at $x=1$.\\
\endp

\Newp 5.
On which one of the given intervals is the function $f(x) = \sin(2x)$ concave downwards?
¥5
a)$\bigl(0,\smfrac{\pi}{2}\bigr)$\\
b)$\bigl(\smfrac{\pi}{4},\smfrac{3\pi}{4}\bigr)$\\
c)$\bigl(\smfrac{\pi}{2},\pi\bigr)$\\
d)$\bigl(-\smfrac{\pi}{2},\smfrac{\pi}{2}\bigr)$\\
e)$\bigl(0,\pi\bigr)$\\
\endp

\Newp 6.
Find the following derivative:
$$\frac{d}{dx}\Bigl ( \sin^{2003} \bigl(\sqrt x\ \bigr)\Bigr).$$
¥2
a)$2003 \sin^{2002}\bigl(\sqrt x\ \bigr)\ \cdot\ \cos\bigl(\sqrt x\ \bigr)\ \cdot\ \frac{1}{2\sqrt x}$\\
b)$2003 \sin^{2002}\bigl(\sqrt x\ \bigr)\ \cdot\ \cos\biggl( \frac{1}{2\sqrt x}\biggr)$\\
c)$2003 \sin^{2002}\Biggl(\cos\biggl( \frac{1}{2\sqrt x}\biggr)\Biggr)$\\
d)$2003 \sin^{2002}\biggl(\frac{1}{2\sqrt x}\biggr)$\\
e)$2003 \sin^{2002}\biggl(\frac{1}{\sqrt x}\biggr)$\\
\endp

\Newp 7.
Find an equation of the tangent line to the curve $x^2-4y^2=0$ at the point $(2,1)$.
¥3
a)$y=\frac{1}{2} x$\\
b)$y=\frac{1}{2} x + 1$\\
c)$y= x - 1$\\
d)$y=\frac{1}{4} x + \frac{1}{2}$\\
e)$y=\frac{1}{3} x + \frac{2}{3}$\\
\endp

\Newp 8.
Find $y^\prime(x)$ for $x^2y^2 - 2x=4-4y$.
¥3
a)$y^\prime(x)= \frac{2-2xy^2}{2x^2y+4}$\\
b)$y^\prime(x)= \frac{2-xy^2}{2x^2y+1}$\\
c)$y^\prime(x)= \frac{1+2xy^2}{x^2y+4}$\\
d)$y^\prime(x)= \frac{2-4xy^2}{2x^2y-1}$\\
e)$y^\prime(x)= \frac{2+xy^2}{2x^2y-4}$\\
\endp

\Newp 9.
 Find $y^{\prime\prime}(x)$ for the following function:
$$y=(3x-1)^5 + (2x+1)^2\ .$$
¥2
a)$y^{\prime\prime} = 180(3x-1)^3 + 8$\\
b)$y^{\prime\prime} = 160(3x-1)^4 + 6$\\
c)$y^{\prime\prime} = 120(3x-1)^3 + 4$\\
d)$y^{\prime\prime} = 100(3x-1)^4 + 2$\\
e)$y^{\prime\prime} = 160(3x-1)^3 + 8$\\
\endp

\Newp 10.
 Find all critical numbers of
$$f(x)= \frac{2x^2}{x+2}.$$
¥2
a)$x=0$ and $x=-4$\\
b)$x=0$, $x=-2$ and $x=-4$\\
c)$x=0$, $x=-2$ and $x=-1$\\
d)$x=1$ and $x=-4$\\
e)$x=0$, $x=-2$ and $x=4$\\
\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.
On the interval $[0, 1]$ the polynomial $p(x)=x^5 + 2x^3 + 2x - 3$ has at least
one root by the Intermediate Value Theorem since a polynomial is continuous
on any interval and $p(0) = -3 < 0$
and $p(1) = 1 + 5 + 2 - 3 = 5 > 0$.
Show that there is only one root of this polynomial on this interval.
\endp

\newp 12. 
Consider the function $f(x) = 3x^4 - 4x^3 + 2003$ defined on the real axis.
\item{a)} find the absolute minimum and absolute maximum of $f$ (if they exist);
\item{b)} find the intervals of increase and decrease;
\item{c)} find the intervals of concavity and inflection points.
\endp

\newp 13.
Suppose an oil spill has taken the form of a circular region
and its area is increasing at the rate of $100$ square meters per
hour. At what rate the radius of the region increasing when the
radius is $200$ meters? Justify your answer.
\endp

\newp 14.
After $t$ hours, the number of bacteria in a laboratory culture
is given by $$n = 6t^2+200\ .$$ Use differentials to approximate $\Delta n$, the
change in the number of bacteria when t changes from $5$ hours to $5.2$
hours. Justify your answer.
\endp

\newp 15.
Find the absolute maximum and minimum values of the following
function
$$f(x)=2-x^{\smfrac{2}{3}}$$
on the interval $[-2,2]$. Justify your answer.
\endp


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


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

\Nomark{Taylor}
\vfill\eject
\Nomark{Alber}
\vfill\eject
\Nomark{Derwent}
\vfill\eject
\end