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                                       \hfill Math 217\hfill}}
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\begin{document}

\bigskip

Show all work. {\bf Circle your answer.}

No books, no notes, no
electronic devices at all.

Solutions will be posted shortly after the exam:
www.math.wisc.edu/$\sim$miller/m217

\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\begin{center}
{\bf Hand in to your TA -- Circle your DIScussion section (column one):} \\
\end{center}
\begin{center}
\begin{tabular}{l|l|l|l}
DIS 321 & 7:45   TR  & 901  Van Vleck & Forray, Susan Jane     \\
DIS 322 & 8:50   TR  & 6224 Soc Sci   & Bisgard, James William \\
DIS 324 & 9:55   TR  & 123  Ingraham  & Bisgard, James William \\
DIS 326 & 11:00  TR  & B235 Van Vleck & Leete, Jennifer A.     \\
DIS 327 & 11:00  TR  & 113  Ingraham  & Lu, Tianshi            \\
DIS 328 & 12:05p TR  & B235 Van Vleck & Leete, Jennifer A.     \\
DIS 329 & 12:05p TR  & 6228 Soc Sci   & Lu, Tianshi            \\
DIS 330 & 1:20p  TR  & B321 Van Vleck & Forray, Susan Jane     \\
DIS 332 & 2:25p  TR  & B135 Van Vleck & Hur, Youngmi           \\
DIS 334 & 3:30p  TR  & B329 Van Vleck & Hur, Youngmi           \\
\end{tabular}
\end{center}





% end file with \finishtbl
%\bigskip

\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
  \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}

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\prob{10} % P 9-2
Sketch the ellipse with equation:
 $$ {x^2\over 25}+{y^2\over 9}=1 $$
Label all four  points where the ellipse intersects the axis.
\par\noindent Find, draw, and label the focal points.

\answerx The focal points are at $(\pm 4,0)$.  The
ellipse intersects the axises at $(\pm 5,0)$ and $(0,\pm 3)$.

\prob{10} % C 5-10
Find the center of mass of a thin homogeneous plate covering
the first quadrant of the disk bounded by the circle $x^2+y^2=a^2$.

\answer $({4\over 3\pi}a, {4\over 3\pi}a)$

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\prob{10} % P4-2
The amount of \$4000 is invested at the interest rate
of 6\% compounded monthly.  How much will you have after
3 years and 4 months?

\answer $4000(1+{.06\over 12})^{40}=4000(1.005)^{40}=4883.1769$ dollars

\prob{10} % P4-6
A roasted Turkey is taken from an oven when its temperature has reached 170
degrees and placed on a table in a room where the temperature is 70 degrees.
 If the temperature of the Turkey is 150 degrees after 10 minutes, when will
its temperature be 140 degrees?

\answer $10{\ln(.7)\over\ln(.8)}=15.984103$ minutes


\prob{10} % Exam 3 problem 6
Prove Pythagorus's Theorem:  In a right triangle with
hypotenus $c$ and sides $a$ and $b$, we have that

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\answerx See Exam 3 problem 6.

\prob{10}  % 4.8
(A) (Do either problem 6A or problem 6B.  {\bf Do not do both.})

Prove the following part of the fundamental theorem of calculus:
\par \bigskip
Suppose $f(x)$ is continuous on the interval $[a,b]$,
$F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with
the $F^\prime(x)=f(x)$ for all $x$ in $(a,b)$.  Let
$n$ be a large integer.
\par Define $\triangle x={{b-a}\over n}$.
\par Define $x_k=a+k\triangle x$ for $k=0,1,\ldots,n$.
\par\bigskip
  Show there exists $c_k$ with $x_{k-1}\leq c_k\leq x_k$
for $k=1,2,\ldots,n$ and such that
$$F(b)-F(a)=\sum_{k=1}^n \; f(c_k)\triangle x$$

\newpage
\noindent
6. (10 pts) (B) (Do either problem 6A or problem 6B.  {\bf Do not do both.})

Prove the following version of the fundamental theorem of
calculus:

\bigskip
Suppose $f(x)$ is a continuous function on the interval
$[a,b]$ and $f(x)\geq 0$ for all $x$ in $[a,b]$.  For
any $x\in [a,b]$ let $A(x)$ be the area under $f$ and above the
interval $[a,x]$ on the $x$-axis.

Prove
$${dA\over dx}=f(x)$$

\def\delx{\triangle x}

\answerx See Exam 1 problem 6 for 6A.  For 6B see pages 196-198
of your calculus book:
We must prove that
$$\lim_{\delx\to 0}{{A(x+\delx)-A(x)}\over \delx}=f(x)$$
We consider the limit for $\delx\to 0^+$
the proof for $\delx\to 0^-$ is similar.
Let $\delx>0$ be small.
Then $A(x+\delx)-A(x)$ is the area below $f$ and above the line
segment $[x,x+\delx]$ on the $x$-axis.  Let $m$ be the minimum
value of $f$ on $[x,x+\delx]$ and $M$ the maximum value.  Then
it is clear that
$$m\delx \leq A(x+\delx)-A(x) \leq M\delx$$
Hence
$$m \leq {{A(x+\delx)-A(x)}\over \delx} \leq M$$
By the extreme value theorem and intermediate value theorem for
continuous functions there exists $x^*$ with $x\leq x^*\leq x+\delx$
such that
$$f(x^*)= {{A(x+\delx)-A(x)}\over \delx}$$
Since $f$ is continuous
$$\lim_{\delx\to 0^+}f(x^*)=f(x)$$
Hence
$$\lim_{\delx\to 0^+}{{A(x+\delx)-A(x)}\over \delx}= \lim_{\delx\to 0^+}f(x^*)
=f(x)$$

\finishtbl
\printanswers

\end{document}