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\centerline{\textbf{\large Exam \#3 solutions $\cdot$ Tuesday, April 3, 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, unless a problem specifically requests a verbal
response.}

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

\vspace*{0.1in}

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\textbf{Problem 1.}
Consider the following graph of $f'(x)$:

\vspace*{0.1in}

\begin{center}
\includegraphics{e3_1.eps}
\end{center}

\vspace*{0.1in}

I repeat for emphasis:  I am showing you the graph of $f'(x)$, but
I will ask you questions about $f(x)$.

\vspace*{0.1in}

\textbf{Part (a).}
Which of the labeled points are critical points of $f(x)$?

\vspace*{0.1in}
\emph{Solution:}  Critical points of $f$ occur where $f'$ is zero or undefined.
There are none of the latter; $f'$ is zero at points $A$, $D$, and $F$.
\vspace*{0.1in}

\textbf{Part (b).}
Which of those critical points are local maxima (not minima) of $f(x)$?
(In this problem, I am not interested in local extrema at boundary points.)

\vspace*{0.1in}
\emph{Solution:}  Local maxima must occur at the critical points ($A$, $D$, and
$F$) or boundary points; the latter are not of interest here.  We can use
the first-derivative test on each of these three points.
\begin{itemize}
\item At point $A$, $f'(x)$ is negative on both sides of $A$.  This means $f$
is decreasing on either side of $A$, so $A$ is a false alarm of $f$.
\item At point $D$, $f'(x)$ is negative to the left and positive to the
right.  Thus $D$ is a local minimum of $f$.
\item At point $F$, $f'(x)$ is positive to the left and negative to the
right.  Thus $F$ is a local maximum of $f$.
\end{itemize}

Alternatively, we can try the second-derivative test.
\begin{itemize}
\item At point $A$, $f'(x)$ has zero slope, i.e. $f''(A)$ is 0.
The second-derivative test is inconclusive here.
\item At point $D$, $f'(x)$ has positive slope, i.e. $f''(D)>0$.
Thus $D$ is a local minimum of $f$.
\item At point $F$, $f'(x)$ has negative slope, i.e. $f''(F)<0$.
Thus $F$ is a local maximum of $f$.
\end{itemize}
\vspace*{0.1in}

In conclusion, $f$ has a local maximum at $F$.

\textbf{Part (c).}
Which labeled points are inflection points of $f(x)$?

\vspace*{0.1in}
\emph{Solution:}  Inflection points of $f$ occur where $f''$ is zero,
i.e. where $f'$ has zero slope.  These are points $A$, $C$, and $E$.
\vspace*{0.1in}

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\newpage
\textbf{Problem 2.}
Find $dy/dx$ if
$\ln(y)\sin(y) = \cos(x)$.

\vspace*{0.1in}
\emph{Solution:}  Using implicit differentiation and the product
rule, we have
\begin{eqnarray*}
	\frac{1}{y}\sin(y)\frac{dy}{dx} + \ln(y)\cos(y)\frac{dy}{dx}
		&=& -\sin(x) \\
	\left(\frac{1}{y}\sin(y) + \ln(y)\cos(y)\right)\frac{dy}{dx} &=& -\sin(x) \\
	\frac{dy}{dx}
	&=& \frac{-\sin(x)}{\left(\frac{1}{y}\sin(y) + \ln(y)\cos(y)\right)}
	= \frac{-y\sin(x)}{\sin(y) + y\ln(y)\cos(y)}.
\end{eqnarray*}
\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 3.}
Let $f(z) = z^4+a/z^4$.
Find $a$ such that $f(z)$ has a minimum at $z=2$.

\vspace*{0.1in}
\emph{Solution:}  If $f(z)$ is to have a minimum at $z=2$, we must
have $f'(2) = 0$.  Differentiating, we have
$$
	f'(z) = 4z^3 - 4a/z^5;
$$
evaluating at $z=2$ we have
$$
	f'(2) = 32 - 4a/32.
$$
Setting this to zero and solving for $a$ gives us
\begin{eqnarray*}
	32 - 4a/32 &=& 0 \\
	32 &=& 4a/32 \\
	4a &=& 1024 \\
	a &=& 256.
\end{eqnarray*}
\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 4.}
Let $g(x)=\sin^2(x)-\cos(x)$.

\vspace*{0.1in}

\textbf{Part (a).}
Find an equation for the tangent line to $g(x)$ at $x=\pi/2$.

\vspace*{0.1in}
\emph{Solution:}  As usual, we use point-slope form: $y = y_0 + m(x-x_0)$.
Here, $x_0 = \pi/2$; $y_0 = g(x_0) = 1$; $m=g'(x_0)$.  Using the chain rule we
get $g'(x) = 2\sin(x)\cos(x) + \sin(x)$; evaluating at $x_0$ we have $g'(\pi/2)
= 1$.  Putting these together we obtain
$$
	y = 1 + (x - \pi/2) = x + 1 - \pi/2.
$$
\vspace*{0.1in}

\textbf{Part (b).}
Write down an equation for the error function involving $g(x)$ and the linear
approximation.

\vspace*{0.1in}
\emph{Solution:}  The error function is the original function minus the tangent-line
approximation:
$$
	E(x) = (\sin^2(x)-\cos(x)) - (x + 1 - \pi/2)
	= \sin^2(x)-\cos(x) - x - 1 + \pi/2.
$$
\vspace*{0.1in}

%\textbf{Part (c).}
%Using your calculator, graph the error function on the interval $0 \le x \le
%3$.  Using the space provided, sketch the original function, the linear
%approximation, and the error function.  From the graph, estimate the worst
%error on that interval.
%
%\vspace*{1.5in}

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\newpage
\textbf{Problem 5.}
Find $b$ and $d$ such that $f(x) = x^3+bx^2+d$ has an inflection point at
$x=3$ and $y$-intercept $-5$.

\vspace*{0.1in}
\emph{Solution:}
The $y$-intercept of a function $f(x)$ is $f(0)$.  Thus $d=-5$.
If $f$ is to have an inflection point at $x=3$, we must have $f''(3)=0$.
Differentiating, we have
\begin{eqnarray*}
	f' (x) &=& 3x^2+2bx \\
	f''(x) &=& 6x+2b.
\end{eqnarray*}
Evaluating at $x=3$ and setting equal to zero, we get
\begin{eqnarray*}
	6\cdot 3 + 2b &=& 0 \\
	2b &=& -18 \\
	b &=& -9.
\end{eqnarray*}
\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 6.}
Does the limit
$$
	\lim_{x\to \infty} \frac{x^2}{e^{0.0024x}}
$$
exist?  If so, what is it, and why?  If not, why not?

\vspace*{0.1in}
\emph{Solution:}
This limit is of the form $\infty/\infty$, so l'H\^opital's rule applies.
Differentiating top and bottom we get
$$
	\lim_{x\to \infty} \frac{2x}{0.0024 e^{0.0024x}}
$$
which is still of the form $\infty/\infty$.  Using l'H\^opital's rule
again, we have
$$
	\lim_{x\to \infty} \frac{2}{0.0024^2 e^{0.0024x}}.
$$
This has a constant on top and a function increasing without bound
on the bottom, so the limit is zero.
\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 7.}
A rectangular region is to be painted inside a semicircular region
as follows:

\vspace*{0.1in}

\begin{center}
\includegraphics{e3_hoop.eps}
\end{center}

\vspace*{0.1in}

The semicircle has radius 7 feet.  Find $x$ that maximizes the area of the
rectangle.  (You must show all your work and use calculus.)

\vspace*{0.1in}
\emph{Solution:}
The area of the rectangle is its length times its width, i.e. $A=2xy$.  To be
able to do calculus on $A$ we need to write it in terms of a single variable.
The equation of a circle of radius 7 centered at the origin is $x^2+y^2=49$;
the equation for the upper semicircle is $y=\sqrt{49-x^2}$.  Substituting this
into $A$ and differentiating (using the product and chain rules), we have
\begin{eqnarray*}
	A &=& 2x\sqrt{49-x^2} \\
	\frac{dA}{dx} &=& 2\sqrt{49-x^2} + 2x \frac{-2x}{2\sqrt{49-x^2}} \\
	&=& 2\sqrt{49-x^2} - \frac{2x^2}{\sqrt{49-x^2}}.
\end{eqnarray*}
Setting this equal to zero and solving for $x$ we have
\begin{eqnarray*}
	2\sqrt{49-x^2} - \frac{2x^2}{\sqrt{49-x^2}} &=& 0 \\
	\sqrt{49-x^2} &=& \frac{x^2}{\sqrt{49-x^2}} \\
	49-x^2 &=& x^2 \\
	2x^2 &=& 49 \\
	x &=& 7/\sqrt{2}.
\end{eqnarray*}
The maximum of $A$ happens at a critical point (which we just found) or a
boundary point.  The boundary values of $x$ are 0 and 7; areas there are both
zero.  Thus the maximum area occurs at $x=7/\sqrt{2} \approx 4.950$.
\vspace*{0.1in}

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\textbf{Problem 8.}
An amusement-park ride is upright and circular with 50-foot radius.  Passengers
are seated around the perimeter while the wheel turns counterclockwise at some
(as yet unknown) constant rate:

\vspace*{0.2in}

\begin{center}
\includegraphics{e3_ferris.eps}
\end{center}

\vspace*{0.1in}

The height of the labeled passenger above the ground is given by
$$
	y = 50 + 50 \sin(\theta).
$$
If the passenger's height above ground is increasing by 3.2 feet per second at
angle $\theta=\pi/3$, how fast is the wheel turning?  Include units in your
answer.

\vspace*{0.1in}
\emph{Solution:}
The quantities $y$ and $\theta$ both vary with time and are related by the
equation $y = 50 + 50 \sin(\theta)$.  Differentiating both sides with respect
to time we find
$$
	\frac{dy}{dt} = 50\cos(\theta)\frac{d\theta}{dt}.
$$
Since we are given the rate of change of $y$ and we are asked to find the
rate of change of $\theta$, we need to solve for $d\theta/dt$.  This is
$$
	\frac{d\theta}{dt} = \frac{dy/dt}{50\cos(\theta)}.
$$
When $\theta=\pi/3$, this is
$$
	\frac{d\theta}{dt}\bigg{|}_{\theta=\pi/3} = \frac{3.2}{50\cos(\pi/3)}.
	= \frac{3.2}{50\cdot 0.5} = 0.128.
$$
The units of $d\theta/dt$ are the units of $\theta$ over the units of $t$,
i.e. radians per second.
\vspace*{0.1in}

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