\documentclass[11pt]{amsart}
\parindent=0pt

% The next four lines override the settings in the amsart style file.
\voffset=-0.8in    % Moves text up
\hoffset=-1.2in    % Moves text left
\textwidth=6.8in   % Sets new page text width
\textheight=9.3in  % Sets new page text height

\usepackage{amssymb}
\usepackage{graphicx} % For including EPS files

\input{isomath}
\input{mathenv}

%% ================================================================
\begin{document}

\thispagestyle{empty} % Removes page numbering for this page only
\centerline{\textbf{\large Exam \#3 solutions $\cdot$ Thursday, November 6, 2008}}
\vspace*{.2in}
\centerline{MATH 124 $\cdot$ Calculus I $\cdot$ Section 26 $\cdot$ Fall 2008}
\vspace*{0.2in}

\emph{Note:  Some of my solutions are wordy, for the sake of explanation.  All
I expect of you is computation, unless a problem specifically requests a verbal
response.}

John Kerl (\texttt{kerl.john.r@gmail.com}).

\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 1.} 
Consider the following table of values of a function $f$ and its first
two derivatives.
$$
\begin{tabular}{|r||r|r|r|r|r|}
\hline $x$      & $-3$   & $-1$  & $1$   & $3$   & $5$    \\
\hline
\hline $f(x)$   & $3.0$  & $4.0$ & $5.0$ & $8.0$ & $9.0$  \\
\hline $f'(x)$  & $1.0$  & $0.0$ & $0.5$ & $1.0$ & $0.3$  \\
\hline $f''(x)$ & $-2.0$ & $0.0$ & $1.4$ & $0.0$ & $-0.4$ \\
\hline
\end{tabular}
$$
Furthermore, $f'$ and $f''$ have no zeroes other than the ones shown.

\textbf{Part (a).}
Does $f$ have a critical point at $x=-1$?  Why or why not?

\emph{Solution:} True:  definition of critical point.

\textbf{Part (b).}
Does $f$ have a local maximum at $x=-1$?  Why or why not?

\emph{Solution:}
False: $f'$ is positive on either side, so $f$ fails
the first derivative test at $x=-1$.
The second derivative test is inconclusive since $f''(-1)$ is 0.

\textbf{Part (c).}
Does $f$ have an inflection point at $x=3$?  Why or why not?

\emph{Solution:}
True:  $f''(3)=0$ and $f''$ changes sign at $x=3$.

%% ----------------------------------------------------------------
\textbf{Problem 2.}
Consider the curve
$$
	\ln(xy) = 2x.
$$

\textbf{Part (a).}  Find $dy/dx$.

\emph{Solution:}  Differentiating implicitly, we have
\begin{align*}
	\frac{d}{dx}\left( \ln(xy) \right) &= \frac{d}{dx}\left( 2x \right) \\
	\frac{1}{xy}\left( y + x \frac{dy}{dx} \right) &= 2 \\
	y + x \frac{dy}{dx} &= 2xy \\
	x \frac{dy}{dx} &= 2xy - y \\
	\frac{dy}{dx} &= \frac{y(2x - 1)}{x}.
\end{align*}

\textbf{Part (b).} Find an equation for the tangent line the curve at the point
$(x,y)=(1,e^2)$.

\emph{Solution:}
Plugging $x=1$ and $y=e^2$ into the above, we have
\begin{align*}
	\frac{dy}{dx}\Bigg|_{x=1,y=e^2} &= \frac{e^2(2\cdot 1 - 1)}{1} = e^2.
\end{align*}
This is the $m$.  The $a$ and $b$ are $1$ and $e^2$.  So we have
\begin{align*}
	y &= m(x-a) + b \\
	&= e^2(x-1) + e^2 \\
	&= e^2 x.
\end{align*}

Note:  We can also differentiate this explicitly if we like.  First solve
for $y$:
\begin{align*}
	\ln(xy) &= 2x \\
	xy &= e^{2x} \\
	y &= \frac{e^{2x}}{x} \\
	\frac{dy}{dx} &= \frac{2xe^{2x} - e^{2x}}{x^2}
		= \frac{e^{2x}(2x - 1)}{x^2} \\
	\frac{dy}{dx}\Bigg|_{x=1,y=e^2} &= \frac{e^{2}(2 - 1)}{1^2}
		= e^2.
\end{align*}
The tangent-line calculation is the same.

%% ----------------------------------------------------------------
\textbf{Problem 3.}

\textbf{Part (a).}
Find $B$ so that
$$
	G(x) = B 2^x + 2^{-x}
$$
has a critical point at $x=-1$.

\emph{Solution:}
A critical point is where $G'(x)=0$ or is undefined.  This is an everywhere
differentiable function, so we need to solve $G'(x)=0$.
\begin{align*}
	G'(x) &= B \ln(2) 2^x - \ln(2) 2^{-x} \\
		&= \ln(2) \left( B 2^x - 2^{-x}\right).
\end{align*}
Solving for $B$ in $G'(-1)=0$, we have
\begin{align*}
	\ln(2) \left( B 2^{-1} - 2^{1}\right) &= 0 \\
	B 2^{-1} = 2^{1} \\
	\frac{1}{2} B  = 2 \\
	B &= 4.
\end{align*}

\textbf{Part (b).}  Is this critical point a minimum, maximum, or neither?
Explain your reasoning.

\emph{Solution:}  This is a minimum, as can be read off the graph.
Alternatively, note that the second derivative is $G''(x) = \ln(2)^2 (B2^x +
2^{-x})$.  This is positive for all $x$, and in particular for $x=-1$.  So, by
the second-derivative test, this critical point is a local (in fact, global)
minimum.

%\vspace*{2.5in}

%%% ----------------------------------------------------------------
%\textbf{Problem 4.}
%Let $f(t) = te^{-at}$.
%Find $a$ such that $f(t)$ has an inflection point at $t=7$.

%\vspace*{0.2in}

%% ----------------------------------------------------------------
\textbf{Problem 4.}
For each of the following, does the limit exist?
If so, what is it, and why?  If not, why not?
\vspace*{0.1in}

\textbf{Part (a).}
$
	\lim_{t\to 0} \frac{1-\cosh t}{t^2}
$

\emph{Solution:}  Since $\cosh(0)=1$, this is a 0/0 situation and
l'H\^opital's rule applies.  Differentiating top and bottom, we have
\begin{align*}
	\lim_{t\to 0} \frac{1-\cosh t}{t^2}
	&=
	\lim_{t\to 0} \frac{-\sinh t}{2t}.
\end{align*}
This is still a 0/0 situation, so we need to use l'H\^opital's rule again:
\begin{align*}
	\lim_{t\to 0} \frac{-\sinh t}{2t}
	&=
	\lim_{t\to 0} \frac{-\cosh t}{2}.
\end{align*}
Now we can set $t=0$ to obtain $-1/2$.  (This also matches what you can see if
you graph $(1-\cosh(t))/t^2$ on your calculator.)

\textbf{Part (b).}
$
	\lim_{z\to 0} \frac{3^z}{z^3}
$

\emph{Solution:}
The numerator goes to 1 as $z \to 0$; the denominator goes to zero as $z \to
0$.  Thus, the limit does not exist.  (This is neither a $0/0$ case nor an
$\infty/\infty$ case, so l'H\^opital's rule does not apply.)

%\textbf{Part (c).}
%$
%	\lim_{z\to 0} \frac{z^3}{3^z}
%$
%
%\textbf{Part (d).}
%$
%	\lim_{x\to -\infty} \frac{1 + 4x + 2x^2}{-7 - 3x^2}
%$

%\vspace*{0.2in}

%% ----------------------------------------------------------------
\textbf{Problem 5.}
The east and west sides of a rectangular enclosure cost \$50 per meter; the
north and south sides cost \$80 per meter.  Find the dimensions of the
enclosure with least cost enclosing an area of 1000 square meters.

\emph{Solution:}
Let $x$ be the length of the north and south sides;
Let $y$ be the length of the east and west sides.
Writing down the given information, we have
\begin{align*}
	xy &= 1000 \\
	C &= 160x + 100y.
\end{align*}
We need to optimize the cost $C$, which is currently in terms of two variables.
Using the area expression to eliminate a variable we get
\begin{align*}
	y &= \frac{1000}{x} \\
	C &= 160x + \frac{100000}{x}.
\end{align*}
Extrema occur at boundary points and critical points.  When $x=0$, there
is no area; when $x$ gets big (so $y$ goes to 0) there is no area.
So, it suffices to look at critical points of $C$.  We have
\begin{align*}
	\frac{dC}{dx} &= 160 - \frac{100000}{x^2}. \\
	160 - \frac{100000}{x^2} &= 0 \\
	\frac{100000}{x^2} &= 160 \\
	x^2 &= \frac{100000}{160} = \frac{10000}{16} \\
	x &= \sqrt{\frac{10000}{16}} = \frac{100}{4} = 25.
\end{align*}
Then
\begin{align*}
	y &= \frac{1000}{25} = 40.
\end{align*}

%% ----------------------------------------------------------------
\textbf{Problem 6.}
The trajectory of an orbiting object is described by
$$
	r(1 + 0.2\cos(\theta)) = 10.
$$
(The units of $r$ are thousands of kilometers, or megameters if you like, and
the units of $\theta$ are radians.) Find $d\theta/dt$ when $\theta=\pi/3$,
given that $dr/dt = -20$ megameters per hour when $\theta=\pi/3$.  Compute your
answer to three decimal places.

\emph{Solution:}
First note, as I announced in class, that you don't need to know polar
coordinates to do this problem.

Here the quantities $r$ and $\theta$ both vary in time; these quantities are
related by the above equation.  Differentiating both sides with respect
to time, we get
\begin{align*}
	\frac{dr}{dt}(1 + 0.2\cos(\theta)) - 0.2 r \sin(\theta) \frac{d\theta}{dt}
		&= 0 \\
	\frac{dr}{dt}(1 + 0.2\cos(\theta)) &= 0.2 r \sin(\theta) \frac{d\theta}{dt}
		\\
	\frac{d\theta}{dt} &=
		\frac{\frac{dr}{dt}(1 + 0.2\cos(\theta))}{0.2 r \sin(\theta)}.
\end{align*}
To evaluate this at $\theta=\pi/3$, we need $dr/dt$ there, which is $-20$;
$\sin(\theta)$ which is $\sqrt{3}/2 \approx 0.866$; $\cos(\theta)$ which is
$0.5$; and $r$.  The latter is
\begin{align*}
	r &= \frac{10}{1 + 0.2\cos(\theta)}
	= \frac{10}{1 + 0.2\cdot 0.5}
	\approx 9.091.
\end{align*}
Then
\begin{align*}
	\frac{d\theta}{dt}\Bigg|_{\theta=\pi/3} &=
		\frac{-20(1 + 0.2\cdot 0.5)}{0.2 \cdot 9.091 \cdot 0.866} \\
	&\approx -13.972 \;\textrm{radians/meter}.
\end{align*}

\end{document}
