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\centerline{\textbf{\large Exam \#2 solutions $\cdot$ Thursday, October 9, 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, except for questions which specifically ask for
verbal responses.}

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

\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 1.}  Let $f(x) = x^{2x}$.  Numerically approximate $f'(1)$
using difference quotients.  Use at least three successively smaller values of
$h$.

\emph{Solution:}
We numerically approximate derivatives by computing
$$
	\frac{f(x+h)-f(x)}{h}
$$
for the specified $x$ and smaller and smaller $h$'s.  Here, this is
\begin{align*}
	\frac{(x+h)^{2(x+h)}-x^{2x}}{h}
	&= \frac{(1+h)^{2(1+h)}-1^{2}}{h}
	= \frac{(1+h)^{2(1+h)}-1}{h} \\
	h &= 0.1:   & \frac{1.1^{2.2}-1}{0.1}       \approx 2.332863 \\
	h &= 0.01:  & \frac{1.01^{2.02}-1}{0.01}    \approx 2.030303 \\
	h &= 0.001: & \frac{1.001^{2.002}-1}{0.001} \approx 2.003003.
\end{align*}
This is beginning to look like $2$.

%% ----------------------------------------------------------------
\textbf{Problem 2.}  The period $T$ (in seconds) of a pendulum depends on the
length $\ell$ (in centimeters) of the pendulum, as follows:
$$
	T = 2\pi \sqrt{\frac{\ell}{g}}
$$
where $g$ is the acceleration due to gravity (a positive constant).

\vspace*{0.1in}

\textbf{Part (a).} What are the units of $dT/d\ell$?

\emph{Solution:}  Units of a derivative are the original function's output
units over the original function's input units.  Here, this is seconds per
centimeter.  Alternatively, the Leibniz-style $dT/d\ell$ notation suggests
putting $T$'s units over $\ell$'s units, i.e. seconds per centimeter.

\textbf{Part (b).} Find the formula for $dT/d\ell$.

\emph{Solution:}
\begin{align*}
	\frac{dT}{d\ell} &= \frac{d}{d\ell} \left(
 	2\pi \sqrt{\frac{\ell}{g}} \right) \\
	&= 2\pi \frac{d}{d\ell} \left( \left(\frac{\ell}{g} \right)^{1/2}\right)
		& & \textrm{(multiplicative constants pull through)} \\
	&= 2\pi \frac{1}{2} \left(\frac{\ell}{g} \right)^{-1/2}
		\frac{d}{d\ell}\left(\frac{\ell}{g}\right)
		& & \textrm{(chain and power rules)} \\
	&= 2\pi \frac{1}{2} \left(\frac{\ell}{g} \right)^{-1/2}
		\frac{1}{g} \\
	&= \frac{\pi}{g} \left(\frac{\ell}{g} \right)^{-1/2}
		& & \textrm{(some simplification to clean up)} \\
	&= \frac{\pi}{g} \sqrt{\frac{g}{\ell}} \\
	&= \frac{\pi}{\sqrt{g \ell}}.
\end{align*}
Alternatively, notice that $\sqrt{\ell/g} = \sqrt\ell/\sqrt g$.  Then
\begin{align*}
	\frac{dT}{d\ell} &= \frac{d}{d\ell} \left(
 		\frac{2\pi}{\sqrt g} \sqrt{\ell} \right) \\
	&= \frac{2\pi}{\sqrt g} \frac{d}{d\ell} \left(
		\sqrt{\ell} \right) \\
	&= \frac{2\pi}{\sqrt g} \frac{d}{d\ell} \left(
		\left( \ell \right)^{1/2} \right) \\
	&= \frac{2\pi}{\sqrt g} \frac{1}{2} \left( \ell \right)^{-1/2} \\
	&= \frac{\pi}{\sqrt g \sqrt \ell} \\
	&= \frac{\pi}{\sqrt {g \ell}}.
\end{align*}

%% ----------------------------------------------------------------
\textbf{Problem 3.}  The horizontal displacement $H$ (in millimeters) of a
weighted spring from its equilibrium position, at time $t$ milliseconds from
the start of an experiment, is
$$
	H(t) = 2.3 + e^{-2 t} \cos(t).
$$

\textbf{Part (a).}  What are the units of $H''(t)$?

\emph{Solution:}  Units of the second derivative are the original function's
output units over the original functions input units squared.  (As before, the
Leibniz notation $d^2H/dt^2$ is a helpful device to remember this fact.) Here,
this is millimeters per millisecond squared:  mm/ms${}^2$.

\textbf{Part (b).}  Find $H''(t)$.

\emph{Solution:}
\begin{align*}
	H(t) &= 2.3 + e^{-2 t} \cos(t). \\
	H'(t) &= \frac{d}{dt}( e^{-2 t} \cos(t) ) \\
	&= \frac{d}{dt}(e^{-2 t}) \cos(t)
		+ e^{-2 t} \frac{d}{dt}(\cos(t))
		&& \textrm{(product rule)} \\
	&= -2 e^{-2 t} \cos(t)
		- e^{-2 t} \sin(t)
		&& \textrm{(chain and trig rules)} \\
	H''(t) &=
		-2 \frac{d}{dt}(e^{-2 t} \cos(t))
		- \frac{d}{dt}(e^{-2 t} \sin(t))
		\\
	&=
		-2 \frac{d}{dt}(e^{-2 t}) \cos(t)
		-2 e^{-2 t} \frac{d}{dt}\cos(t)
		- \frac{d}{dt}(e^{-2 t}) \sin(t)
		- e^{-2 t} \frac{d}{dt}(\sin(t))
		\\
	&=
		4 e^{-2 t} \cos(t)
		+2 e^{-2 t} \sin(t)
		+ 2 e^{-2 t} \sin(t)
		- e^{-2 t} \cos(t) \\
	&= 3 e^{-2 t} \cos(t) +4 e^{-2 t} \sin(t) \\
	&= e^{-2 t} (3 \cos(t) + 4 \sin(t).
\end{align*}

Several of you found $H'(t)$ as above, then factored out the $e^{-2t}$
to obtain
\begin{align*}
	H'(t) &= e^{-2t} (-2 \cos(t) - \sin(t)).
\end{align*}
Then we can use the product rule to get
\begin{align*}
	H''(t) &=
		-2 e^{-2t} (-2 \cos(t) - \sin(t)) +
		e^{-2t} (2 \sin(t) - \cos(t)) \\
	&= e^{-2t} (4 \cos(t) + 2\sin(t) + 2 \sin(t) - \cos(t)) \\
	&= e^{-2t} (3 \cos(t) + 4\sin(t)).
\end{align*}
You get the same answer either way, but as it happens, the latter works out a
little nicer.

%% ----------------------------------------------------------------
\textbf{Problem 4.}  The reaction time $T$ (in minutes) of a chemical
experiment is a function of the amount $m$ (in grams) of catalyst added to a
solution.  That is, $T = f(m)$.

\vspace*{0.1in}

\textbf{Part (a).}  Describe the significance of $f(200)$, including units
in your answer.

\emph{Solution:}  $f$ takes grams to minutes.  So, this is number of minutes
the reaction takes to complete when $200$ grams of catalyst are used.

\textbf{Part (b).}  Describe the significance of $f^{-1}(6)$, including units
in your answer.

\emph{Solution:}  $f^{-1}$ takes minutes to grams.  So, this is the number
of grams of catalyst needed to make the reaction complete in 6 minutes.

\textbf{Part (c).}  Suppose you discover the following:  As you repeat the
experiment, with one more gram of catalyst each time, you find that the
reaction time decreases.  However, each additional gram has less and less
effect.  What is sign of $f''(m)$?  Please justify your answer.  (Sketch a
graph if you like.  If you do so, please label your axes and explain verbally
what you are doing.)

\emph{Solution:}  Here's a sketch which is compatible with the information
presented:  the reaction time $T$ decreases as $m$ increases, but decreases
by less and less each time:

\vspace*{0.2in}
\begin{center}
\includegraphics{figures/e24csoln.eps}
\end{center}
\vspace*{0.2in}

From the sketch it is clear that the $T$ vs. $m$ curve is concave up, i.e.
$f''(m)$ is positive.

A non-graphical way to think of this is that $f'(m)$ is negative, but less and
less so as $m$ increases.  That is, $f''(m)$ is positive.

%% ----------------------------------------------------------------
\textbf{Problem 5.}  Let $f(x)=\frac{1}{x^2+1}$.

\vspace*{0.1in}

\textbf{Part (a).}  Find $f'(x)$.

\emph{Solution:}
\begin{align*}
	\frac{d}{dx} \left(\frac{1}{x^2+1} \right)
	&= \frac{(x^2+1)(0) - (1)(2x)}{(x^2+1)^2} \\
	&= \frac{-2x}{(x^2+1)^2}.
\end{align*}

\textbf{Part (b).} Find an equation for the tangent line to $f(x)$ at $x=2$.

\emph{Solution:} Using point-slope form, the tangent line has equation
$$
	y=m(x-a)+b.
$$
We have $a=2$, $b=f(2)$, and $m=f'(2)$.  The latter two are
\begin{align*}
	f(2)  &= \frac{1}{2^2+1} = \frac{1}{5} \\
	f'(2) &= \frac{-2(2)}{(2^2+1)^2} = \frac{-4}{25}.
\end{align*}
Then
\begin{align*}
	y &= \frac{-4}{25} (x-2) + \frac{1}{5} \\
	&= \frac{-4x}{25} + \frac{8}{25} + \frac{1}{5} \\
	&= \frac{-4x}{25} + \frac{13}{25} \\
	&= -0.16x + 0.52
		& & \textrm{(if you prefer)}.
\end{align*}

%% ----------------------------------------------------------------
\textbf{Problem 6.}  Some values of $V(t)$ and $H(t)$ and their first
derivatives are given by the following table.
$$
\begin{tabular}{|c||c|c|c|c|}
\hline $t$ & $V(t)$ & $H(t)$ & $V'(t)$ & $H'(t)$ \\
\hline $6$ & $8$    & $-7$   & $0.5$   & $-0.2$  \\
\hline
\end{tabular}
$$

\vspace*{0.1in}

\textbf{Part (a).} Let $E(t) = V(t) + H(t)$.  Find $E'(6)$.

\emph{Solution:}  Using the sum rule, we have
$E'(t) = V'(t) + H'(t)$ and thus
$E'(6) = V'(6) + H'(6)$.
Reading the values off the table, this is $0.5-0.2=0.3$.

\textbf{Part (b).} Let $E(t) = V(t)H(t)$.  Find $E'(6)$.

\emph{Solution:}  Using the product rule, we have
$E'(t) = V'(t)H(t) + V(t) H'(t)$ and thus
$E'(6) = V'(6)H(6) + V(6) H'(6)$.
Reading the values off the table, this is $0.5(-7) + 6(-0.2) =-3.5-1.2=-4.7$.

\textbf{Part (c).} Let $E(t) = \frac{V(t)}{H(t)}$.  Find $E'(6)$.

\emph{Solution:}  Using the product rule, we have
$$
	E'(t) = \frac{V'(t)H(t) - V(t) H'(t)}{H(t)^2}
$$
and thus
$$
	E'(6) = \frac{V'(6)H(6) - V(6) H'(6)}{H(6)^2}.
$$
Reading the values off the table, this is
$$
	\frac{(0.5)(-7) - (8)(-0.2)}{(-7)^2}
	= \frac{-3.5 + 1.6}{49}
	= \frac{-1.9}{49}.
$$

%% ----------------------------------------------------------------
\textbf{Problem 7.}
Let $g(x)=x e^{-ax}$.  Assume $a$ is a positive constant.  Over what
interval(s) is $g(x)$ increasing?  (You will need to solve an inequality.)

\emph{Solution:}
The function $g'(x)$ is increasing when its derivative $g'(x)$ is positive.
We need to solve the inequality $g'(x)>0$.
First, we need to find $g'(x)$:
\begin{align*}
	g'(x) &= \frac{d}{dx}(x) e^{-ax} + x \frac{d}{dx}(e^{-ax})
		& & \textrm{(product rule)} \\
	&= e^{-ax} - ax e^{-ax}
		& & \textrm{(chain rule)} \\
	&= (1-ax) e^{-ax}.
\end{align*}
Then the inequality:
\begin{align*}
	(1-ax) e^{-ax} &> 0 \\
	(1-ax) &> 0
		& & \textrm{($e^{\textrm{anything}}$ is positive)} \\
	1 &> ax
		\\
	x &< 1/a
		& & \textrm{($a$ is positive)}.
\end{align*}

%% ----------------------------------------------------------------
\textbf{Problem 8.}  Let $C(\theta) = \frac{\cos(\theta) - 1}{\theta}$.

\vspace*{0.1in}

\textbf{Part (a).}  Does $C'(1)$ exist?  If so, find its value exactly.
If not, explain why not.

\emph{Solution:}  A function is differentiable at a point (``innocent
until proven guilty'') unless one of the following happens:
\begin{itemize}
\item \emph{The original function isn't defined at that point}:
	No, since we can take cosine minus 1 of anything, and we can divide by
	$\theta$ as long as it's non-zero --- which is the case since $\theta=1$.

\item \emph{Discontinuity at that point}:
	No, as may be checked from the graph.

\item \emph{A corner at that point}:
	No, as may be checked from the graph.

\item \emph{A vertical tangent line at that point}:
	No, as may be checked from the graph.

\end{itemize}

We can use the quotient rule to find the derivative:
\begin{align*}
	C'(\theta) &= \frac{\theta(-\sin(\theta)) - (\cos\theta-1)(1)}\theta^2{} \\
	&= \frac{-\theta\sin(\theta) - \cos\theta + 1}\theta^2{} \\
	C'(1) &= -\sin(1) - \cos(1) + 1.
\end{align*}

\textbf{Part (b).}  Does $C'(0)$ exist?  If so, find its value exactly.
If not, explain why not.

\emph{Solution:}  $C(0)$ is not defined (division by zero) so the
derivative cannot be defined there either.

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