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\centerline{\textbf{\large Exam \#2 solutions $\cdot$ Thursday, March 1, 2007}}
\vspace*{.2in}
\centerline{MATH 124 $\cdot$ Calculus I $\cdot$ Section 8 $\cdot$ Spring 2007}
\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 parts 2c, 2d, and 8b which
specifically ask for verbal responses.}

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

\vspace*{0.1in}

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

\emph{Solution:}
The approximation is $f'(x) \approx \frac{f(x+h)-f(x)}{h}$.  We have $x=2$
throughout, $f(x)=x^x$, and $f(x+h)=(x+h)^{x+h}$.  With $h=0.01$:
$$
\frac{f(x+h)-f(x)}{h} = \frac{2.01^{2.01}-2^2}{0.01} \approx 6.840
$$
With $h=0.001$:
$$
\frac{f(x+h)-f(x)}{h} = \frac{2.001^{2.001}-2^2}{0.001} \approx 6.779
$$
With $h=0.0001$:
$$
\frac{f(x+h)-f(x)}{h} = \frac{2.0001^{2.0001}-2^2}{0.0001} \approx 6.773.
$$
It looks like about 6.77.

% e^(x ln x)
% e^(x ln x) (ln x + 1)
% x^x (ln x + 1)

%% ----------------------------------------------------------------
\textbf{Problem 2.}  On a mountain-climbing expedition, you find that the air
becomes cooler as you climb.  Let $y$ be your altitude above sea level,
measured in feet; let $H$ be the air temperature in degrees Fahrenheit.

\vspace*{0.2in}

\textbf{Part (a).} What are the units of $H'(y)$?

\emph{Solution:}  The notation tells you that $H$ is a function of $y$.
The input units of $H(y)$ are the units of $y$, which are
feet.  The output units of $H(y)$ are the units of $H$, which are degrees
Fahrenheit.  The units of the derivative $H'(y)$ are the original function's
output units over the original functions input units, namely,
degrees Fahrenheit per foot.

\vspace*{0.2in}

\textbf{Part (b).} What is the sign of $H'(y)$?

\emph{Solution:}
The air temperature is getting cooler (decreasing number of degrees) as you
climb (increasing number of feet).  So, the sign must be negative.

\vspace*{0.2in}

\textbf{Part (c).} Give a practical interpretation of $H^{-1}(35)$.

\emph{Solution:}
A function sends inputs (here, feet) to outputs (here, degrees Fahrenheit);
the inverse function sends the original function's outputs (degrees) back to
the original function's inputs (feet).  So, 35 must be degrees Fahrenheit.
Then $H^{-1}(35)$ must be the altitude at which the temperature is $35^\circ
F$.

\vspace*{0.2in}

\textbf{Part (d).} Give a practical interpretation of $H'(7500)$.

\emph{Solution:}
The derivative $H'(y)$ is the rate of change in temperature as a function of
altitude.  At altitude 7500 feet, the temperature is dropping by this many
degrees per additional foot of altitude.

\vspace*{0.2in}

%% ----------------------------------------------------------------
\textbf{Problem 3.}  In a lab experiment, you have microorganisms growing
in a Petri dish.  The number $m$ of microrganisms, in millions, is a
function of time $t$ in days since the start of the experiment.  This
number is given by
$$
	m(t) = 4.1 e^{0.24t}.
$$
Find the rate of change in population on day 5 of the experiment.  In your
answer, please show units.

\emph{Solution:}
The derivative $m'(t)$ is $0.24\cdot 4.1 e^{0.24t}$, using the chain and
exponential rules.  Evaluating at $t=5$ gives
$$
m'(5) = 0.24\cdot 4.1 e^{0.24\cdot 5}
= 0.984 e^{1.2} \approx 3.267 \;\textrm{million microorganisms per day}.
$$

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\newpage
\textbf{Problem 4.}  Let $G(t)=2^{-rt} \sin(at)$.  Find $G'(t)$.

\emph{Solution:}
Using the product and chain rules, we have
\begin{eqnarray*}
	G'(t) &=& -r \ln(2) 2^{-rt} \sin(at) + a 2^{-rt} \cos(at).
\end{eqnarray*}
\vspace*{0.2in}

%% ----------------------------------------------------------------
\textbf{Problem 5.}  Let $f(z)=\tan^{-1}(z)$.  Compute $f''(1)$.

\emph{Solution:}
The first derivative is
$$
	f'(z) = \frac{1}{1+z^2}.
$$
For the second derivative, you can use the quotient rule, or you can use the
exponential and chain rules since we can write $f'(z) = (1+z^2)^{-1}$.
Either way, you should get
$$
	f''(z) = \frac{-2z}{(1+z^2)^2}.
$$
Evaluating at $z=1$ gives
$$
	f''(1) = \frac{-2}{(1+1^2)^2} = \frac{-1}{2}.
$$


\vspace*{0.2in}

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\textbf{Problem 6.}  Let $q(x)=\ln(2+2x+x^2)$.

\vspace*{0.2in}

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

\emph{Solution}:
Using the log and chain rules, we have
$$
	q'(x) = \frac{2+2x}{2+2x+x^2}.
$$

\vspace*{0.2in}

\textbf{Part (b).} Find an equation for the tangent line to $q(x)$ at $x=3$.

\emph{Solution}:
As usual, use point-slope form at the specified point.  The horizontal
coordinate of the point is $x_0=3$; the vertical coordinate of the
point is
$$y_0 = q(3) = \ln(2+6+9) = \ln(17).$$
The slope is
$$
	m = q'(3) = \frac{2+6}{2+6+9} = \frac{8}{17}.
$$
Putting these together we have
%$$
%	y = y_0 + m(x-x_0) = \ln(17) + \frac{8}{17}(x-3).
%$$
\begin{eqnarray*}
	y &=& y_0 + m(x-x_0) \\
	y &=& \ln(17) + \frac{8}{17}(x-3).
\end{eqnarray*}
As a sanity check, you can graph the original function and the tangent line.
You should see

\includegraphics{e2_6.eps}

\vspace*{0.2in}

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\newpage
\textbf{Problem 7.}  Let $F(t)$ and $G(t)$ be given by the following
graphs.

\includegraphics{e2F.eps}
\qquad
\includegraphics{e2G.eps}

\textbf{Part (a).} Find $G'(40)$.

\emph{Solution}:
This is undefined: $G$ has a corner at $t=40$.

\vspace*{0.2in}

\textbf{Part (b).} Let $H(t) = \frac{F(t)}{G(t)}$.  Find $H'(50)$.

\emph{Solution}:
Using the quotient rule, we have
$$
	H'(t) = \frac{F'(t)G(t)-F(t)G'(t)}{G(t)^2}.
$$
At $t=50$, this is
$$
	H'(50) = \frac{F'(50)G(50)-F(50)G'(50)}{G(50)^2}.
$$
We can read all these values off the graph:
$$
	F (50) = 40, \qquad
	F'(50) =  2, \qquad
	G (50) = 40, \qquad
	G'(50) =  0.
$$
Then
$$
	H'(50) = \frac{2 \cdot 40 - 40 \cdot 0}{40^2}
	= \frac{80}{1600} = 0.05.
$$

\vspace*{0.2in}

\textbf{Part (c).} Let $H(t) = F(G(t))$.  Find $H'(30)$.

\emph{Solution}:
Using the chain rule, we have
$$
	H'(t) = F'(G(t))G'(t).
$$
We can read all the necessary values off the graph:
$$
	G (30) = 50, \qquad
	F'(50) =  2, \qquad
	G'(30) = -1.
$$
Then
$$
	H'(30) = F'(G(30))G'(30) = 2\cdot (-1) = -2.
$$

\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 8.}  Let
$$
	f(x) = \begin{cases}
		\sin(x), & x \ge 0 \\
		x-x^3, & x \le 0. \\
	\end{cases}
$$
\vspace*{0.2in}

\textbf{Part (a).}  Find $f'(x)$.  Write it as a piecewise function.

\emph{Solution}:
Differentiating the pieces, we have
$$
	f'(x) = \begin{cases}
		\cos(x), & x \ge 0 \\
		1-3x^2, & x \le 0. \\
	\end{cases}
$$
\vspace*{0.2in}

\vspace*{0.2in}

\textbf{Part (b).}  Is the original function $f(x)$ differentiable at $x=0$?
Why or why not?
(Hint:  graph it.)

\emph{Solution}:
Graphing the function, you should see

\includegraphics{e2_8.eps}

The slopes match up at $x=0$ so $f'(x)$ is defined there.

\vspace*{0.2in}

%% ----------------------------------------------------------------
\textbf{Problem 9.}  When is $g(x)=x^3+bx^2+cx+d$ concave up?  Assume $b,c,d$
are constants.  (You will need to solve an inequality.)

\emph{Solution}:
A function is concave up when $g''(x)>0$.  Computing this, we have
\begin{eqnarray*}
	g  (x) &=&  x^3+bx^2+cx+d \\
	g' (x) &=& 3x^2+2bx+c \\
	g''(x) &=& 6x+2b \\
	g''(x) &>& 0 \\
	6x+2b &>& 0 \\
	6x &>& -2b \\
	x &>& \frac{-b}{3}.
\end{eqnarray*}

\vspace*{0.2in}

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