%% Taylor
\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\def\whichexam{Final Exam }
\def\examdate{December 15, 2000}
\def\numberpages{15}
\def\numberofproblems{25}
\def\multiplechoicemax{6}
\def\partialcreditmax{12}

\gdef\pagebreaks{%
\\{1}\\{3}\\{5}\\{7}\\{9}\\{11}\\{12}\\{14}\\{16}\\{18}\\{20}\\{21}\\{23}\\{25}\\{26}%
}
\gdef\answersheetbreaks{\\{1}\\{14}\\{100}}

\def\checkPBN#1@{%
\edef\yy{\number#1}\def\\##1{ \edef\yyy{\number##1}\ifx\yyy\yy \gdef\PBN{1}\fi}%
\xxpagebreaks}
\global\let\xxpagebreaks=\pagebreaks


\input epsf.def

\def\anspic#1#2{\epsfxsize=#1true in%
\vtop to0pt{\vss\hbox{\raise -#1true in\vtop to 0pt{\epsfbox{#2}\vss}}}%
\vrule depth#1 true in height #1true in width0pt}

\def\answersheetbreakcount{}
\def\pagebreakcount{}
\def\leaderfill{\leaders\hbox to3pt{\hss.\hss}\hfill}
\def\pskip{\vskip -8pt\hskip20pt\hbox to 2 true in{\leaderfill}\vskip -8pt}
%%
\newif\ifprelim
\prelimfalse
%\prelimtrue
%%
\def\un#1{\>\hbox{#1}}
\def\fs{\un{ft/sec}}

\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 12345
\ll 2 25341
\ll 3 42153
\ll 4 35214
\ll 5 53412
\ll 6 25314
\ll 7 52134
\ll 8 12345
\ll 9 43125
\ll 10 51324
\ll 11 34215
\ll 12 45123
\ll 13 12345
\ll 14 53142
\ll 15 12345
\ll 16 24315
\ll 17 21534
\ll 18 34512
\ll 19 23145
\ll 20 24351
\ll 21 43251
\ll 22 35124
\ll 23 42513
\ll 24 12345
\ll 25 12345
}

\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}

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\picV=\ht0 %
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}

\def\beginmcbox{\vtop\bgroup\hsize=1 true in}
\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
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\newdimen\pspacewidth
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\edef\sum{\displaystyle\xxsum}

\newdimen\picH\newdimen\picV
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\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\pc<3 \newpage\else\ifprelim\else\newpage\fi\fi\null%
\vskip10pt\centerline{{\bf#1}}\mc=0 %
\ifnum\pc<3\credit=\multiplechoicemax \else\credit=\partialcreditmax \fi}
\newif\ifmultichoice
\def\problemspacing{\ifprelim\vskip10pt\headline{\hfill}\else%
\ifmultichoice%
\ifnum\mc>0 \ifodd\pc\relax\newpage\null\vskip10pt\else\vfill\fi\fi%
\else%
\ifnum\mc>0 \newpage\null\vskip10pt\fi\fi%
\mc=1 \fi}

\long\def\newp#1.#2\endp{
\ifprelim\vskip8pt\else%
\gdef\PBN{0}\checkPBN\pc@%
\def\PBNx{1}%
\ifnum\pc=1 \vskip10pt\else
\ifx\PBN\PBNx\vfill\eject\null\vskip10pt\else\vfill\fi%
\fi\fi%
%\problemspacing%
\noindent\vbox{\noindent{\bf\number\pc.}(\number\credit\ pts.)
\global\advance\pc 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}

\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{%
\gdef\yy{#1}%
\ifx\yy\answersheetbreakcount \ifnum#1>1 \egroup%
\beginmcbox\fi%
\leftremove\Pagebreaks\to\pagebreakcount%
\leftremove\Answersheetbreaks\to\answersheetbreakcount%
\fi%
\ifx\yy\pagebreakcount\ifnum#1>1 \pskip\fi%
\leftremove\Pagebreaks\to\pagebreakcount%
\fi%
\tabalign {#1.}&
\global\ansmark=#2 \relax%
\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 
\rightappenditem{#2#3#4#5#6}\to\list%
}

\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}



%%%%%%%
\def\eat#1{}
\def\exam#1{\ifprelim\Exam{Bullwinkle}\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{
\global\let\Answersheetbreaks=\answersheetbreaks
\global\let\Pagebreaks=\pagebreaks
\leftremove\Answersheetbreaks\to\answersheetbreakcount
\leftremove\Pagebreaks\to\pagebreakcount
\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 two hours.
\itemitem{$\bullet$} You will only hand in this page, so be sure
you have marked the answer sheet below correctly. Dotted lines and new columns indicate page breaks
in the test.
\itemitem{$\bullet$} Be sure that you have all \numberpages\ pages of the test.
\vskip5pt
\centerline{Good Luck!}
\vskip4pt
\def\HH{\hskip.45in}
\settabs\+9.\qquad &\HH&\HH&\HH&\HH&\HH&\cr
%\hskip.5 true in%
\boxit{
\vbox{\hfuzz=10in\baselineskip20pt
\noindent
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
\vskip2pt\noindent
\hbox to 6.5true in{{\beginmcbox\correctanswers\egroup}
}}}{10 true pt}{6.5 true in}
%%\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!
\vskip4pt
\tabalign {\bf Final Exam Total:}&\quad\ul{1in}\cr

\tabalign\hfil {\bf Course Total:}&\quad\ul{1in}\cr
}}{10pt}{1in}\vss}}}}
\answersheet
\vfill\eject\headline{\hfil}
\multichoicetrue
\newchoice{Multiple Choice}

\newp 1.
Find the limit
$\Lim_{x\to 2}\>\frac{x^3+4}{x^2-1}$.
\Ans 5
$4$€
$2$€
$6$€
$+\infty$€
Does not exist.€
\endp

\newp 2.
A body moves along a straight line with a constant acceleration of $5\un{m/sec}^2$.
Initially it is moving at a velocity of $11\un{m/sec}$ at a distance of $20\un{m}$ from the zero
position on the line.
Which formula below is a formula for the body's position at time $t$?
\Ans 4
$\frac52 t^2+11t+20$€
$\frac{11}2t^2+5t+20$€
$10t^2+11t+5$€
$\frac52 t^2+11t$€
$\frac{11}2 t^2+20$€
\endp

\newp 3.
A $13\un{ft}$ ladder is leaning against the side of a building when its base begins to slide 
away from the building.
By the time the base is $5\un{ft}$ from the building, the base is moving at a rate of $4\fs$.
How fast is the top of the ladder sliding down the wall at this moment?
\Ans 5
$\frac53\fs$€
$4\fs$€
$\frac35\fs$€
$\frac14\fs$€
$\frac5{12}\fs$€
\endp

\newp 4.
Let $f(x)=\Int^{1+x^2}_2\frac{1}{t^2-1}\> dt$. Evaluate $f^\prime(3)$.
\Ans 5
$\frac2{33}$€
$\frac19$€
$\frac{10}9$€
$\frac9{10}$€
$\frac1{99}$€
\endp

\newp 5.
Find an equation for the tangent line to the curve $y=x^4-15x^2+30$ at the point $(2,-14)$.
\Ans 2
$y=-28x+42$€
$y=x-16$€
$y=(4x^3-30x)(x-2)-14$€
$y=\frac1{28}x-14$€
$y=32x-78$€
\endp

\newp 6.
$\Lim_{x\to\infty}\>\frac{3x^4-5x^3+1x^2-19x+11}{5x^4-6x^3+7x^2-78x+199}=\ ?$
\Ans 5
$\frac35$€
$\frac{11}{199}$€
$\frac{11}{5}$€
$\frac{3}{199}$€
$\frac17$€
\endp

\newp 7.
Which equation below is the solution to the initial value problem
$$\frac{dy}{dx}=\frac{1}{3+\sin^2 x}\quad;\qquad y(1)=2$$
\Ans 2
$2+\Int_1^x \frac{1}{3+\sin^2 w}\>dw$€
$3+\Int_1^x \frac{1}{2+\sin^2 w}\>dw$€
$2+\Int_1^x \frac{1}{3+\sin^2 w\cos w}\>dw$€
$1+\Int_2^x \frac{1}{3+\sin^2 w}\>dw$€
$1+\Int_2^x \frac{1}{3+2\sin w\cos w}\>dw$€
\endp

\newp 8.
Evaluate $\displaystyle\int_1^9\frac{1}{\sqrt x\bigl(1+2\sqrt x\bigr)^2}\> dx$.
\Ans 4
$\frac4{21}$€
$\frac14$€
$\frac17$€
$\frac89$€
The integral does not exist.\hss€
\endp

\newp 9.
The equation $x^5+x-1=0$ has one solution between $0$ and $1$. 
Find the result of one iteration
of Newton's Method applied to this equation with $1$ as the starting point.
\Ans 5
$\frac56$€
$1$€
$\frac34$€
$\frac57$€
$\frac12$€
\endp

\newp 10.
Consider $\displaystyle\int_{1}^3 x^3\>dx$.
Divide the interval of integration into $5$ equal pieces. Which summation below
is the Riemann sum for this partition where the point in each interval is a point at
which $f(x)$ obtains its maximum in that interval?
\Ans 3
$\frac25\sum_{i=1}^5\biggl(1+\frac{2i}5\biggr)^3$€
$\frac25\sum_{i=0}^4\biggl(1+\frac{2i}5\biggr)^3$€
$\frac25\sum_{i=1}^5\biggl(\frac{2i}5\biggr)^3$€
$\frac25\sum_{i=0}^4\biggl(\frac{2i}5\biggr)^3$€
$\frac15\sum_{i=1}^5\biggl(1+\frac{2i}5\biggr)^3$€
\endp

\newp 11.\par\noindent\vskip -16pt \hbox{\raise 170pt\vtop to 0pt{\hsize=4 true in\noindent%
Which sum below is the result of applying the Trapezoid rule to the integral
$\Int_{-2}^{2}\sqrt{x^4+1\ }\> dx$ where we divide the interval into $8$ pieces?
\vss}\hskip-40pt%
\vtop to0pt{\hbox to0pt{\vbox{\epsfbox{graph11.eps}}\hss}\vss}}
\Ans 1
$\frac{1}4\bigl(1\cdot \sqrt{17}+2\cdot\frac{\sqrt{97}}{4} + 2\cdot\sqrt{2}+
2\cdot\frac{17}{4}+2\cdot 1+2\cdot\frac{17}{4}+ 2\cdot\sqrt{2}+2\cdot\frac{\sqrt{97}}{4}
+1\cdot \sqrt{17}\bigr)$€
$\frac{1}2\bigl(1\cdot \sqrt{17}+2\cdot\frac{\sqrt{97}}{4} + 2\cdot\sqrt{2}+
2\cdot\frac{17}{4}+2\cdot 1+2\cdot\frac{17}{4}+ 2\cdot\sqrt{2}+2\cdot\frac{\sqrt{97}}{4}
+1\cdot \sqrt{17}\bigr)$€
$\frac{1}3\bigl(1\cdot \sqrt{17}+2\cdot\frac{\sqrt{97}}{4} + 4\cdot\sqrt{2}+
2\cdot\frac{17}{4}+4\cdot 1+2\cdot\frac{17}{4}+ 2\cdot\sqrt{2}+4\cdot\frac{\sqrt{97}}{4}
+1\cdot \sqrt{17}\bigr)$€
$\frac{1}4\bigl(1\cdot \sqrt{17}+2\cdot\frac{\sqrt{97}}{4} - 2\cdot\sqrt{2}+
2\cdot\frac{17}{4}+2\cdot 1-2\cdot\frac{17}{4}+ 2\cdot\sqrt{2}-2\cdot\frac{\sqrt{97}}{4}
+1\cdot \sqrt{17}\bigr)$€
$\frac{1}4\bigl(1\cdot \sqrt{17}+2\cdot\frac{\sqrt{97}}{4} - 4\cdot\sqrt{2}+
2\cdot\frac{17}{4}+2\cdot 1-4\cdot\frac{17}{4}+ 2\cdot\sqrt{2}-4\cdot\frac{\sqrt{97}}{4}
+1\cdot \sqrt{17}\bigr)$€
\endp

\newp 12.
The slope of the tangent line to the curve
$y^2=x^3-3x^2+2x$ at the point $(3,-\sqrt6)$ is
\Ans 5
$-\frac{11}{2\sqrt6}$€
$\frac{11}{2\sqrt6}$€
$\frac{3}{2\sqrt6}$€
$\frac{\sqrt6}{2}$€
$\frac{-\sqrt6}{2}$€
\endp

\newp 13.
What is $\frac{d^2y}{dx^2}$ for the parameterized curve $x(t)=1+\sin t$, $y(t)=t+\cos t$ 
when $t=0$?
\Ans 4
$-1$€
$1$€
$0$€
$\tan(1)$€
The curve is not differentiable at $t=0$.\hss€
\endp

\newp 14.
Consider a solid in space which is sliced by planes perpendicular to the $x$ axis.
The base of the solid is in the $yz$ plane. 
At distance $x>0$ from the $yz$ plane, the slice is a {\bf square} with the length
of one side being $\sqrt{1-x^3\ }$.
Which integral below computes the volume?
\Ans 3
$\Int_0^1(1-x^3)\>dx$€
$2\pi\Int_0^1x\sqrt{1-x^3\ }\>dx$€
$\pi\Int_0^1(1-x^3)\>dx$€
$\Int_0^1x\sqrt{1-x^3\ }\>dx$€
$2\pi\Int_0^1\sqrt{1-x^3\ }\>dx$€
\endp

\newp 15.
Which statement below holds for the autonomous differential
equation
$$\frac{dy}{dx}=\frac{y}{1+y^2}\ ?$$
\Ans 2
If $y(0)>0$ then $y(2)>y(0)$.€
If $y(0)>0$ then $y(2)<y(0)$.€
If $y(0)<0$ then $y(2)>y(0)$.€
$y=\frac25$ is a solution.€
The equation has no solution for which $y$ is a constant.\hss€
\endp

\newp 16.
Consider the region in the first quadrant bounded by the lines
$y=2x+1$ and $x=3$.
Rotate this region around the $y$ axis.
Which integral below computes the volume of the resulting solid of revolution?
\Ans2
$2\pi\Int_0^3 x(2x+1)\> dx$€
$\pi\Int_0^3 (2x+1)^2-x^2\> dx$€
$2\pi\Int_0^3 3(2x+1)\> dx$€
$\pi\Int_0^3 y(1-y)\> dy$€
$\pi\Int_0^7 \bigl(3^2-(2x+1)^2\bigr)\> dx$€
\endp

\newp 17.
Where does the graph of the linearization of the function $f(x)=3x^3-12$ at $x=2$ cross the $y$ axis?
\Ans5
At $y=-60$.€
At $y=-12$.€
At $y=24$.€
At $x=2$.€
No where.€
\endp

\newp 18.
On which interval below is the function $2x^3-15x^2+24x$ decreasing?
\Ans 5
$[1,4]$€
$[3,5]$€
$[0,2]$€
$[0,4]$€
$[2,8]$€
\endp

\newp 19.
How many inflection points does the curve $y=4x^5-5x^4-9$ have?
 \Ans 5
{$1$}€
{$2$}€
{None}€
{$3$}€
{$4$}€
\endp

\newp 20.
\par\noindent
\vtop to2 true in{\noindent\hbox{\raise160pt\vtop to0pt{\hsize=3true in\noindent%
If the following is a graph of the function $f(x)$
which graph among the answers
is the graph of $\Int_0^x f(t)\>dt\>$?}\hskip40pt
\epsfxsize=1.7true in\epsfbox{graph21.eps}}}%

\Ans 3
\anspic{1.25}{graph21a.eps}€
\anspic{1.25}{graph21b.eps}€
\anspic{1.25}{graph21c.eps}€
\anspic{1.25}{graph21d.eps}€
\anspic{1.25}{graph21e.eps}€\endp

\newp 21.%
Which answer below identifies {\bf all} of the asymptotes of the curve
$$y=\frac{x^2+2x+1}{x-1}\ ?$$
\Ans 1
$x=1$ is a vertical asymptote; $y=x+3$ is an oblique one.€
$x=1$ is a vertical asymptote; $y=1$ is a horizontal one.€
$x=1$ is a vertical asymptote.€
$y=3$ is a horizontal asymptote.€
$x=1$ is a vertical one; $y=-1$ is a horizontal asymptote.€
\endp

\newp 22.
\par\noindent\hbox{\raise50pt\vbox{\hsize=4 true in\noindent%
The curves $y=x^4-3$ and $y=-x^4+5$ enclose an area.
Set up a definite integral which calculates the area of this region.}\hskip40pt
\epsfxsize=1.25 true in\epsfbox{graph22.eps}}
\Ans 2
$\Int_{-\sqrt2}^{\sqrt2}(8-2x^4)\>dx$€
$\Int_{-\sqrt2}^{\sqrt2} 2\>dx$€
$\Int_{-1}^{1}(8-2x^4)\>dx$€
$\Int_{-1}^{1} 2\>dx$€
$\Int_{0}^{\root 4\of3}(8-2x^4)\>dx$€
\endp

\newp 23.
Evaluate $\Int (1-\sin^2 x)\cos x\> dx$.
\Ans 2
$\sin x-\frac{\sin^3 x}{3}+C$€
$\cos x-\frac{\cos^3 x}{3}+C$€
$\frac13\cos(3x)-\frac23\cos^2x+C$€
$\frac13\sin(3x)-\frac23\cos^2x+C$€
$\frac13\cos(3x)-\frac23\sin^2x+C$€
\endp

\newp 24.
Which integral below gives the length of the curve
$x(t)=2\cos t$, $y(t)=5\sin t$ from $t=0$ to $t=\frac{\pi}2$ ?
\Ans 2
$\Int_0^{\smfrac{\pi}2}\sqrt{4\sin^2t+25\cos^2t}\>dt$€
$\Int_0^{\smfrac{\pi}2}\sqrt{4\cos^2t+25\sin^2t}\>dt$€
$\Int_0^{\smfrac{\pi}2}\sqrt{4\sin^2t+4\cos^2t}\>dt$€
$\Int_0^{\smfrac{\pi}2}\sqrt{1+4\sin^2t+25\cos^2t}\>dt$€
$\Int_0^{\smfrac{\pi}2}\sqrt{1+4\cos^2t+25\sin^2t}\>dt$€
\endp

\newp 25.
\def\fp{\hbox{ft$\cdot$lbs}}
\vskip-12 true pt
\noindent\hbox{\raise20pt\vbox{\hbox to2.5 true in{\hss\epsfbox{tank.eps}}}
\hskip-10pt\raise50pt\vbox{\hsize=4 true in\noindent%
Find the work done in pumping a liquid over the rim of a tank.
The tank is $50\un{ft}$ long and has a semi{--}circular end of radius
$10\un{ft}$. Suppose that the tank is filled to a depth of $7\un{ft}$ and
that the liquid has a density of $100\un{\fp/ft}^3$.}}
%%\par\noindent\vrule width\hsize depth0pt height .1pt\par
\vskip-40true pt\hskip120pt\epsfbox{tankend.eps}
\Ans 2
$-10^4\Int_{-10}^{-3}y\sqrt{100-y^2\ }\>dy\ \fp$€
$-10^4\Int_{-10}^{-7}y\sqrt{100-y^2\ }\>dy\ \fp$€
$-10^4\Int_{-10}^{-3}\sqrt{100-y^2\ }\>dy\ \fp$€
$-10^4\Int_{-10}^{-7}\sqrt{100-y^2\ }\>dy\ \fp$€
$0\ \fp$€
\endp

\vfill\eject}


\exam{}\marked{}
\end
