%Taylor
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\edef\sum{\displaystyle\xxsum}
\newdimen\picV
\newdimen\picH
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\newtoks\ta\newtoks\tb
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\ta={\\{#1}}\tb=\expandafter{#2}%
\xdef#2{\the\tb\the\ta}}
\def\leftremove#1\to#2{\expandafter\lopoff#1\lopoff#1#2}
\long\def\lopoff\\#1#2\lopoff#3#4{\gdef#4{#1}\gdef#3{#2}}

\def\list{}
\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%
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}

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\vtop to \picV{
      \hrule width \picH height 0pt depth 0pt
      \vfill
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\def\newp{\vfill%%\problemspacing%
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\global\advance\pc by1}

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\newbox\aa\newbox\ab\newbox\ac
\newbox\ad\newbox\ae

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

\hfuzz=20pt

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\def\one{\global\Ax=1 \global\Aspace=\hsize\relax\global\advance\Aspace by-20pt}

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(c)\quad\box\Ac(d)\quad\box\Ad%
(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\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 \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}

\def\xPra#1#2#3#4#5{\switchbox{#1}{\Aa}\switchbox{#2}{\Ab}%
\switchbox{#3}{\Ac}\switchbox{#4}{\Ad}\switchbox{#5}{\Ae}%
\resetboxes\ifcase\Ax \or\praone\or\pratwo\or\prathree\or \or\prafive\fi\vfill}


\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}
 \def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\Lim{\displaystyle\lim}
\def\Int{\displaystyle\int}


\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}}
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\def\+{\tabalign}
\long\gdef\answersheet{
\null\vskip10pt
\centerline{\bf Math 125, Final}
\centerline{December 16, 1998}
\vskip10pt

\itemitem{$\bullet$}
The Honor Code is in effect for this examination.
All work is to be your own.
\itemitem{$\bullet$} Be sure that you have all 15 pages of the test. 
\itemitem{$\bullet$} No calculators are to be used.
\itemitem{$\bullet$} The exam lasts for two hours.
\itemitem{$\bullet$} You are to hand in just the front page.
\vskip5pt
\centerline{Good Luck!}
\vskip10pt
\def\leaderfill{\leaders\hbox to3pt{\hss.\hss}\hfill}
\def\pskip{\vskip -8pt\hbox to\pspacewidth{\leaderfill}\vskip-8pt}
\def\HH{\hskip.42in}
\settabs\+9.\qquad &\HH&\HH&\HH&\HH&\HH&\cr
\centerline{Please mark your answers with an {\bf X}!\qquad Do NOT circle them!}
\par\centerline{\small The dotted lines in the answer box indicate page breaks.}
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\noindent\hbox to0pt{%
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\parindent=0pt
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\ll  2 21345
\ll  3 13425
\ll  4 43512
\ll  5 52413
\ll  6 51423
\ll  7 42315
\ll  8 14325
%\ll  8 14352
\ll  9 51324
\ll 10 45321
\ll 11 24315
\ll 12 15432
\ll 13 53214
\ll 14 14235
\ll 15 14352
\vskip10pt
}\AA=-50pt
\vtop{\obeylines\baselineskip20pt\hsize=2.5in\pspacewidth=2.3in
\parindent=0pt
\ll 16 14253
\ll 17 42351
\ll 18 54312
\ll 19 24315
\ll 20 42531
\ll 21 31245
\ll 22 34215
\ll 23 23514
\ll 24 34125
\ll 25 35412
\vskip10pt
Score: \ul{1in}
}\hss}}{20pt}{5in}%
}

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\leftremove\pagebreaks\to\nextpb
\answersheet}


\long\def\Exam{%
\coversheet{ }%

\vfill\eject\headline{\hfil}
\five
\newp
$$\Lim_{w\to \infty}\ \frac{w^3+3w}{w^5+5w^{3}}\ =$$
\AN{$0$}{$\infty$}{$-\infty$}{$\frac35$}{$\frac53$}
\Pra

\two
\newp
Which choice below completes the next phrase to a true sentence?
The function $f(x)=\vert\sin(x)\vert$ is
\AN{differentiable except at integer multiples of $\pi$.}
{differentiable everywhere.}{differentiable nowhere.}
{continuous except at integer multiples of $\pi$}
{continuous only at integer multiples of $\pi$.}
\Pra

\newpage%% page 3 top
\three
\newp
If $y=\Int_0^x \sec w\ dw$, for $-{\pi \over 2} < x < {\pi \over 2}$,
 then $\frac{d^2y}{dx^2}=$
\AN{$\sec x\tan x$}
{$\sec x$}
{$\tan x$}
{$\cos x$}
{\hbox to 0pt{The second derivative does not exist.\hss}}
\Pra

\two
\newp
If
$$ x^4 + y^4 = 8xy+y\ ,$$
find $\frac{dy}{dx}$ at the point $(2,1)$.
\AN{$\frac{24}{13}$}{$0$}{$\frac{17}{19}$}{It does not exist.}{\hbox to0pt{It may exist but it can not be determined.\hss}}
\Pra

\newpage%% page 4 top
\two
\newp %% tied to #4
Suppose $f$ and $h$ are two functions which are differentiable at $8$.
If $f(8)=2$, $f^\prime(8)=5$, $h(8)=3$ and $h^\prime(8)=7$.
Let $F(x)=f\bigl(h(x)\bigr)$. 
Then $F^\prime(8)\ =\ ?$
\AN{Can not be determined from the above data.}{$14$}{$15$}{$35$}{$42$}
\Pra

\five\newp
The value of $\Int^2_1\frac{dx}{x}$ is $0.6931$ to two decimal places.
Which number below best approximates $\Int^1_0\frac{x}{x^2+1}\> dx$?
\AN{$0.3465$}{$0.6931$}{$1.3862$}{$2.7724$}{$0.1732$}
\Pra

\newpage
\two
\newp
$$\dd{(1+x^2-x^3)^{37}}{x}=\ ?$$
\AN{$37(1+x^2-x^3))^{36}(2x-3x^2)$}
{$37(1+x^2-x^3))^{36}$}
{$\frac{(1+x^2-x^3))^{38}}{38(2x-3x^2)}$}
{$\frac{(1+x^2-x^3))^{38}}{38}$}
{$\frac{(1+x^2-x^3))^{38}(2x-3x^2)}{38}$}
\Pra

\three\newp
The absolute maximum of the function $x+\frac 1{2x^2}$ on the interval $[{{1}\over{4}},4]$ is
\AN{$8\frac{1}4$}{$4\frac{1}{32}$}{$2\frac{1}{8}$}{$\frac32$}{\hbox to0pt{The function has no absolute maximum.\hss}}
\Pra

\newpage%% page 6 top
\two
\vbox to0pt{%
\newp
Suppose that the graph below
is the graph of the function $f(x)$.

\vskip-.25in\picH=3.5in\px{function.ps}

\vskip-.4in\noindent Which graph below best represents the graph of the function $\Int_0^x f(t)\> dt$.
%\picH=1.85in
\picH=2.3in
\AN{\px{ans.ps}}
{\px{b.ps}}
{\px{c.ps}}
{\px{d.ps}}
{\px{e.ps}}
\Pra
\vss}

\newpage%% page 7 top
\two
\newp
What are the asymptotes (both vertical and horizontal) for the graph of
$$y=\frac{1-4x^2}{(x-3)(x+2)}\quad ?$$
\AN{$y=-4$, $x=3$ and $x=-2$}{$y=\pm\frac12$, $x=3$ and $x=-2$}{$x=\frac12$, $y=3$ and $y=-2$}
{$x=3$ and $y=\pm\frac12$}{There are no asymptotes}
\Pra

\three
\newp
Where does the linearization of the function $f(x)=3x^3-12$ at $x=2$ cross the $x$ axis?
\AN{$x=\frac53$}{$x=\frac74$}{$x=2$}{$x=\root3\of4$}{Never.}
\Pra

\newpage%% page 8 top
\five
\newp
\rememberPC
How many inflection points does the curve $y=3x^4-4x^3-3$ have?
 \AN{$2$}{None}{$1$}{$3$}{$4$}
\Pra

\newp
How many solutions in real numbers does the equation $3x^4-4x^3-3=0$ have?

\noindent Hints: The function is the same as the function in \#\number\tmpPC.
You might also want to consider where the function is increasing/decreasing.
\AN{$2$}{None}{$1$}{$3$}{$4$}
\Pra


\newpage%% page 9 top
\two
\newp
Which function below is a solution to the initial value problem
$$y^\prime=\sin(\sqrt x\>)\quad\hbox{and}\quad y(2)=4\ ?$$
\AN{$4+\Int_2^x\sin(\sqrt t\>)\>dt$}{$\cos(\sqrt x\>)+4-\cos(\sqrt2\>)$}
{$4+\Int_2^{\sqrt x}\sin t\>dt$}
{$4+\Int_0^x\sin(\sqrt t\>)\>dt$}
{$\cos(\sqrt x\>)+4$}
\Pra

\five
\newp
The value of $\Int^1_0\frac{3x^2+3}{(x^3+3x+4)^2}\>dx$ is
\AN{$\frac18$}{$-\frac14$}{$8$}{$4$}{$\frac14$}
\Pra

\newpage%% page 10 top
\three
\newp
Find the value of $c$ for which $f(c)$ is the average value of $f$ for the function
$f(x)=x^2$ on the interval $[0,6]$.
\AN{$2\sqrt{3}$}{$\sqrt6$}{$3$}{$\sqrt3$}{\hbox to 0pt{There is no $c$ in the interval $[0,3]$.\hss}}
\Pra

\five
\newp
About how accurately should you measure a variable $t$ to insure that
your calculation of the area of the rectangle with sides $a=t$, and $b=3t$ 
is within $5\%$ of its true value?
\AN{$2.5\%$}{$3.5\%$}{$.05\%$}{$200\%$}{$5\%$}
\Pra

\newpage
\three
\newp
You are to plan a road between two towns, $A$ and $B$, separated by a canyon.
You decide to build a bridge from town $A$ to some point on the
opposite edge of the canyon, and then run the road along the edge
of the canyon to town $B$.
Naturally, it costs more to build the bridge than the road along the rim.
Specifically, it costs $\$5$ per unit distance to build the bridge verses
$\$3$ per unit distance to build along the rim.
The canyon is $12$ units wide and it is $20$ units from one town to the
point on the edge opposite the other.
What value of ${x}$ in the graph below will minimize the cost?

\noindent\setbox0=\hbox{\noindent\epsfxsize=4in \epsfbox{:problem18:Bridge.ps}}%
\picV=\ht0 %
\vbox to 0pt{\box0 \vss}\vrule width 0pt height 0pt depth\picV%
\AN{$9$}{$8$}{$12$}{$10$}{$20$}
\Pra

\newpage%% page 11 top
\five
\newp
A lighthouse has a searchlight that rotates clockwise at a constant speed of $0.6$ radians per second:
${{d\theta}\over{dt}}=0.6$.
The light is a mile offshore.
How fast does the light appear to be moving across the beach at the point closest to the light in miles per second?
\noindent\setbox0=\hbox{\noindent\epsfxsize=4in \epsfbox{:problem19:lighthouse.ps}}%
\picV=\ht0 %
\vbox to 0pt{\box0 \vss}\vrule width 0pt height 0pt depth\picV%

\AN{$0.6$}{$0.36$}{$0.3\pi$}{$\frac\pi6$}{$1.6$}
\Pra


\two
\newp
The two curves $y=16-x^4$ and $y=4-x^2$ enclose a bounded region.
Which integral computes its area?
\AN{$\Int_{-2}^{2}(12+x^2-x^4)\>dx$}{$\Int_{-2}^{2}(x^4-x^2-12)\>dx$}
{$\Int_{-2}^{2}20-x^2-x^4\>dx$}
{$\Int_{-2}^{2}20+x^2+x^4\>dx$}
{$\Int_{-2}^{2}\bigl((16-x^4)^2-(4-x^2)^2\bigr)\>dx$}
\Pra

\newpage%% page 12 top
\two
\newp
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 disk of radius $\sqrt{1-x^4\ }$.
Which integral below computes the volume?
\AN{$\pi\Int_0^1(1-x^4)\>dx$}
{$2\pi\Int_0^1\sqrt{1-x^4\ }\>dx$}
{$\Int_0^1(1+x^4)\>dx$}
{$\pi\Int_0^1x\sqrt{1-x^4\ }\>dx$}
{$\pi\Int_0^1x(1-x^4)\>dx$}
\Pra

\newp
Consider the region in the plane bounded by the two curves 
$y=4+\sqrt{4-x^2\ }$ and $y=x^2$.
Rotate this region about the $x$ axis.
Which integral below computes the resulting volume?
\AN{$\pi\Int^2_{-2}\bigl((4+\sqrt{4-x^2\ })^2-(x^2)^2\bigr)\> dx$}
{$\pi\Int^2_{-2}\bigl((x^2)^2-(4+\sqrt{4-x^2\ })^2\bigr)\> dx$}
{$\pi\Int^1_{-1}\bigl((4+\sqrt{4-x^2\ })^2-(x^2)^2\bigr)\> dx$}
{$\pi\Int^1_{-1}\bigl((x^2)^2-(4+\sqrt{4-x^2\ })^2\bigr)\> dx$}
{$2\pi\Int^2_{0}x\bigl(4+\sqrt{4-x^2\ }-x^2\bigr)\> dx$}
\Pra

\newpage%% page 13 top

\newp
Consider the region in the first quadrant bounded by the lines
$y=x+2$ and $x=4$.
Rotate this region around the $y$ axis.
Which integral below computes the volume of the resulting solid of revolution?
\AN{$2\pi\Int_0^4 x(x+2)\> dx$}
{$2\pi\Int_0^4 x(2-x)\> dx$}
{$2\pi\Int_0^4 4(x+2)\> dx$}
{$2\pi\Int_0^6 y(4-y)\> dy$}
{$\pi\Int_0^4 \bigl(4^2-(x+2)^2\bigr)\> dx$}
\Pra

\two
\newp
Which integral below gives the arc length of the graph of the
curve $x=\sqrt{8-2y^2\ }$ from $y=0$ to $y=2$?
\AN{$\Int_0^2\sqrt{\frac{8+2y^2}{8-2y^2}}\>dy$}
{$\Int_0^2{\frac{2y}{\sqrt{8-2y^2}\>}}\>dy$}
{$\Int_0^2\sqrt{9-2y^2\ }\>dy$}
{$\Int_0^2\sqrt{1+\frac{1}{8-2y^2}}\>dy$}
{$\Int_0^2\sqrt{\frac{8+2y^2}{\sqrt{8-2y^2}}}\>dy$}
\Pra

\newpage%% page 14 top
\two\newp
Consider the graph of the function $y=x^3$ from $x=0$ to $x=2$.
Rotate this curve around the $x$ axis.
Which integral below gives the surface area of the resulting solid of revolution?
\AN{$2\pi\Int_0^2(x^3)\sqrt{1+9x^4\ }\>dx$}
{$\pi\Int_0^2(x^3)^2\>dx$}
{$2\pi\Int_0^2x\> x^3\>dx$}
{$2\pi\Int_0^2(x^3)(3x^2)\>dx$}
{$2\pi\Int_0^8y\sqrt{1+y^2\ }\>dy$}
\Pra

\newpage

}

\Exam %%Dwyer, Cholak, Taylor, Wong, Cao, Jarre, {Nollet}
\eject
\marked{}
\end
