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\centerline{\textbf{\large Exam \#4 solutions $\cdot$ Thursday, December 4, 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}

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\textbf{Problem 1.}
Let
\begin{align*}
	x(t) &= 2t^3 - 15t^2 + 24t + 7 \\
	y(t) &= t^2 + t + 1.
\end{align*}
Find all values of $t$ such that this curve has a vertical tangent line.

\emph{Solution:}  The tangent line is vertical when $dx/dt=0$:
\begin{align*}
	x(t) &= 2t^3 - 15t^2 + 24t + 7 \\
	\frac{dx}{dt} &= 6t^2 - 30t + 24 \\
	6t^2 - 30t + 24 &= 0 \\
	t^2 - 5t + 4 &= 0 \\
	(t-1)(t-4) &= 0 \\
	t &= 1, 4.
\end{align*}

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\vspace*{0.1in}
\textbf{Problem 2.}

\textbf{Part (a).}
Find the average value of
$$
	G(y) = \frac{A}{y} + B y
$$
over the interval $-4 \le y \le -1$, where $A$ and $B$ are positive constants.

\emph{Solution:}  The average value is
$$
	\frac{1}{b-a} \int_{y=a}^{y=b} G(y)\,dy
$$
with $a=-4$ and $b=-1$.  This is
\begin{align*}
	\frac{1}{-1 - -4} \int_{y=-4}^{y=-1} \left(\frac{A}{y} + B y\right) \,dy
	&= \frac{A}{3} \int_{y=-4}^{y=-1} \frac{1}{y} \,dy
		+ \frac{B}{3} \int_{y=-4}^{y=-1} y \,dy \\
	&= \frac{A}{3} \left[ \ln(|y|) \right]_{y=-4}^{y=-1}
		+ \frac{B}{3} \left[ \frac{y^2}{2} \right]_{y=-4}^{y=-1} \\
	&= \frac{A}{3} \left( \ln(1) - \ln(4) \right)
		+ \frac{B}{3} \left( \frac{1}{2} - \frac{16}{2} \right) \\
	&= -\frac{A\ln(4)}{3} - \frac{7.5 B}{3} \\
	&= -\frac{A\ln(4)}{3} - 2.5B.
\end{align*}

\textbf{Part (b).}
If the units of $y$ and $G(y)$ are inches and gallons, respectively,
what are the units of the average value found in part (a)?

\emph{Solution:}
Intuitively, the average value of a gallons function should be gallons.
Mechanically, the units of the integral are gallons times inches; the
denominator $(-1 - -4)$ has units of inches.  Thus,
$$
	\frac{1}{\textrm{inches}} \cdot
	(\textrm{gallons} \cdot \textrm{inches})
	= \textrm{gallons}.
$$

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\vspace*{0.1in}
\textbf{Problem 3.}
Suppose the rate of change of the price of a stock is $R(t)$ where $t$ is
measured in days since the start of the year and $R(t)$ is dollars per day.

\textbf{Part (a).}  What are the units of $\int_{31}^{59} R(t) dt$?

\emph{Solution:}  The units of the integral are $R$ units times $t$ units.
This is dollars per day times days, i.e. dollars.

\textbf{Part (b).}  Give a practical interpretation of $\int_{31}^{59} R(t) dt$.

\emph{Solution:}  This is the change in the stock price between the 31st and
59th days of the year.  (Note that Jan. 1 would be the 0th day.) You may also
notice that days 31-59 are the month of February (in a non-leap year).

\textbf{Part (c).} If $R(t) < 0$ for $30 \le t \le 60$, can you tell what the
sign of $\int_{31}^{59} R(t) dt$ is?  If so, what is the sign?  If not, why
not?

\emph{Solution:}
If the integrand is negative throughout the interval of integration, then the
integral's value will be negative:  all the area will be below the horizontal
axis.

\textbf{Part (d).} If $R(t) < 0$ for $30 \le t < 40$ and $R(t) > 0$ for $40 < t
\le 60$, can you tell what the sign of $\int_{31}^{59} R(t) dt$ is?  If so,
what is the sign?  If not, why not?

\emph{Solution:} The falling-price period (9 days) is shorter than the
rising-price period (19 days), so one might be tempted to say there was a net
gain.  However, we don't know anything about the magnitude of $R$ --- the area
below the horizontal axis for days 31-40 might be more or less than the area
above the horizontal axis for days 40-59.  So, we can't determine the sign of
the integral based on the information we have.

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\vspace*{0.1in}
\textbf{Problem 4.}
Find the exact area of the region between $f(x) = x^{2/3}$ and $g(x)=\sin(x)$
over the interval $1/8 \le x \le \pi$.

\emph{Solution:}  Graphing, we see that $x^{2/3}$ is on top of $\sin(x)$.
So, we compute
\begin{align*}
	\int_{x=1/8}^{x=\pi} (f(x) - g(x)) \,dx
	&= \int_{x=1/8}^{x=\pi} (x^{2/3} - \sin(x)) \,dx \\
	&= \left[ \frac{3}{5} x^{5/3} + \cos(x) \right]_{x=1/8}^{x=\pi} \\
	&= \left( \frac{3}{5} \pi^{5/3} + \cos(\pi) \right)
	 - \left( \frac{3}{5} \left(\frac{1}{8}\right)^{5/3} + \cos(1/8) \right) \\
	&= \frac{3}{5} \pi^{5/3} - 1 - \frac{3}{5} \frac{1}{32} - \cos(1/8) \\
	&= \frac{3}{5} \pi^{5/3} - \cos(1/8) - 1 - \frac{3}{160} \\
	&= \frac{3}{5} \pi^{5/3} - \cos(1/8) - \frac{163}{160}.
\end{align*}

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\vspace*{0.1in}
\textbf{Problem 5.}
Label the following statements as correct (C) or incorrect (I):

\textbf{Part (a).}
If $f'(x) < 0$ on an interval then $f(x)$ is decreasing on that interval.

\emph{Solution:}  This is correct.

\textbf{Part (b).}
If $f'(x) = 0$ at a point then $f(x)$ has an inflection point at that point.

\emph{Solution:}  This is incorrect.  Change ``inflection point'' to ``critical
point'' and it would be correct.  A critical point of $f(x)$ is, by defintion,
a point where $f'(x)$ is zero or undefined.

\textbf{Part (c).}
If $f'(x)$ has a maximum or minimum at a point then $f(x)$ has a critical
point there.

\emph{Solution:}  This is incorrect.  Change ``critical point'' to ``inflection
point'' and it would be correct.

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\vspace*{0.1in}
\textbf{Problem 6.}
Let
$$
	\frac{dP}{dt} = t(1-t).
$$

\textbf{Part (a).}  Find a general solution for $P$.

\emph{Solution:}
Antidifferentiating both sides, we get
\begin{align*}
	P &= \int t(1-t)\,dt \\
	&= \int (t-t^2)\,dt \\
	&= \frac{t^2}{2} - \frac{t^3}{3} + C.
\end{align*}

\textbf{Part (b).}  Find a specific solution for $P$ if $P(3)=2$.

\emph{Solution:}
\begin{align*}
	P &= \frac{t^2}{2} - \frac{t^3}{3} + C \\
	2 &= \frac{3^2}{2} - \frac{3^3}{3} + C \\
	2 &= \frac{9}{2} - 9 + C \\
	2 &= -\frac{9}{2} + C \\
	C &= 2 + \frac{9}{2} = \frac{13}{2} \\
	P &= \frac{t^2}{2} - \frac{t^3}{3} + \frac{13}{2}.
\end{align*}

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\vspace*{0.1in}
\textbf{Problem 7.}
Let $F(x)= \int_{10}^x f(t) dt$, with the following known values for $f(t)$:
$$
\begin{tabular}{|c||c|c|c|c|c|}
\hline $t$    & $0$     & $10$   & $20$   & $30$   & $40$   \\
\hline $f(t)$ & $-1.25$ & $-1.2$ & $-1.1$ & $-0.9$ & $-0.7$ \\
\hline
\end{tabular}
$$
Estimate $F(20)$ and $F(30)$.

\emph{Solution:}
We have
$$
	F(20) = \int_{10}^{20} f(t)\,dt
	\quad\textrm{and}\quad
	F(30) = \int_{10}^{30} f(t)\,dt.
$$
There are various ways to estimate this.
\begin{itemize}

\item Using left-hand sums, we have
$$
	\int_{10}^{20} f(t)\,dt
	\approx 10\cdot (-1.2) = -12
$$
and
$$
	\int_{10}^{30} f(t)\,dt
	\approx 10\cdot(-1.2) + 10\cdot(-1.1) = -12-11=-23.
$$

\item Using right-hand sums, we have
$$
	\int_{10}^{20} f(t)\,dt
	\approx 10\cdot (-1.1) = -11
$$
and
$$
	\int_{10}^{30} f(t)\,dt
	\approx 10\cdot(-1.1) + 10\cdot(-0.9) = -11-9=-20.
$$

\end{itemize}

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\vspace*{0.1in}
\textbf{Problem 8.}

\textbf{Part (a).}
Find
$$
	\frac{d}{d\theta} \;
	\int_{\pi/2}^\theta \frac{\sin(x)}{x} dx.
$$

\emph{Solution:}
When we see the derivative of an integral function, we should think of the
Second Fundamental Theorem of Calculus:
\begin{align*}
	\frac{d}{d\theta} \;
	\int_{\pi/2}^\theta \frac{\sin(x)}{x} \,dx
	&= \frac{\sin(\theta)}{\theta}.
\end{align*}

\textbf{Part (b).}
Find
$$
	\frac{d}{d\theta} \;
	\int_{\pi/2}^{\pi-e^{-\theta}} \frac{\sin(x)}{x} \,dx.
$$

\emph{Solution:}
To use the Second Fundamental Theorem of Calculus and the chain rule, we first
do ourselves the favor of naming the integrand:
\begin{align*}
	f(x) &= \frac{\sin(x)}{x}.
	%F(\theta) &= \int_{\pi/2}^{\theta} \frac{\sin(x)}{x} \,dx.
\end{align*}
Then the upper and lower limits are $u(\theta) = \pi - e^{-\theta}$ and
$\ell(\theta) = \pi/2$, respectively.  We drop the upper and lower limits
into the integrand, factoring in the speeds of the limits:
\begin{align*}
	\frac{d}{d\theta} \;
		\int_{\pi/2}^{\pi-e^{-\theta}} \frac{\sin(x)}{x} \,dx
	%&= \frac{d}{d\theta} \; F(u(\theta)) \\
	&= f(u(\theta)) \cdot u'(\theta) - f(\ell(\theta)) \cdot \ell'(\theta) \\
	&= \frac{\sin(\pi - e^{-\theta})}{\pi - e^{-\theta}} \cdot e^{-\theta}
	- \frac{\sin(\pi/2)}{\pi/2} \cdot 0 \\
	&= \frac{\sin(\pi - e^{-\theta})}{\pi - e^{-\theta}} \cdot e^{-\theta}.
\end{align*}

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