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\centerline{Mathematics 136, section 2 -- Advanced Placement Calculus}
\centerline{Final Examination Solutions -- December 18, 2009}
\vskip 10pt
\noindent
I. Consider the functions:
$$y = f_1(x) = \frac{2 \sin(x) + 1}{1+x^2}, \qquad y = f_2(x) = x + x^2 + 1, 
\qquad y = f_3(x) = x e^{3x} + e^{x^3}$$  
\begin{itemize}
\item[A)] (15) Find the derivatives of $f_1,f_2,f_3$.
\sol
By the quotient rule,
$$f_1'(x) = \frac{2\cos(x) + 2x^2\cos(x) - 4x \sin(x) - 2x}{(1 + x^2)^2}.$$
By the derivative power rule,
$$f_2'(x) = 1 + 2x$$
By the Chain and Product rules,
$$f_3'(x) = e^{3x} + 3xe^{3x} + 3x^2 e^{x^3}.$$
\item[B)] (5) Find the equation of the tangent line to 
the graph $y = f_1(x) + f_3(x)$  at $(0,2)$.
\sol  $f_1(0) + f_3(0) = 1 + 1 = 2$ and $f_1'(0) + f_3'(0) = 2 + 1 = 3$.
So the tangent line is $y - 2 = 3(x - 0)$, or $y = 3x + 2$.  
\end{itemize}
\vskip 10pt
\noindent
II.  Figure 1 (at the top of the next page)
shows the graph $y = f(x)$ for some function $f(x)$.
\begin{itemize}
\item[A)] (10) Sketch the graph $y = -f(x + 1)$.  
\sol See Figure 2 on a later page.  The graph is shifted one unit left and reflected across the $x$-axis.
\item[B)] (10) Sketch the graph $y = f'(x)$, 
and explain the features of your graph, including its $x$-axis intercepts, 
its local maxima/minima, points where $f'(x)$ is undefined, and so forth.
\sol See Figure 3 on a later page.  
The graph $y = f(x)$ has a local maximum at about $x = -.6$ this
gives the $x$-intercept on the graph $y = f'(x)$.  $y = f(x)$ has a 
cusp at $x = 0$, which produces the vertical asymptote on $y = f'(x)$. 
$y = f(x)$ is concave down starting at $x = -1$ to about $x = 0.5$, 
then concave up after that.  So $y = f'(x)$ changes from decreasing 
to increasing at about $0.5$.  This gives the local minimum.
\end{itemize}
\begin{figure}
\begin{center}
\includegraphics[width=3in]{FI.ps}
\caption{$y = f(x)$ in Problem II}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=3in]{FIIA.ps}
\caption{$y = -f(x+1)$ in Problem II}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=3in]{FIIB.ps}
\caption{$y = f'(x)$ in Problem II}
\end{center}
\end{figure}
\vskip 10pt
\noindent
III.  
\begin{itemize}
\item[A)]  (10) State the definition of the definite
integral of a function $f(x)$ from $a$ to $b$.
\sol The definite integral of $f(x)$ over $[a,b]$
is the limit of the Riemann sums 
$$\int_a^b f(x)\ dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta x,$$
provided that the limit exists (independently of the choice of 
the $x_i^*$).
\item[B)]  (10) State both parts of the Fundamental
Theorem of Calculus.
\sol  Let $f(x)$ be continuous on $[a,b]$.  Then:
\begin{enumerate}
\item  The function 
$$G(x) = \int_a^x f(t)\ dt$$
is an antiderivative of $f(x)$ on $(a,b)$ (that is $G'(x) = f(x)$
for all $x$ in $(a,b)$).
\item  If $F(x)$ is any antiderivative of $f(x)$,
then 
$$\int_a^b f(x)\ dx = F(b) - F(a).$$
\end{enumerate}
\end{itemize}  
\vskip 10pt
\noindent
IV.  The region $R$ in the 
$xy$-plane is bounded by $y = x^3 - 1$, $x = 1$, $x = 2$, and the 
$x$-axis. 
\begin{itemize}
\item[A)]  (10)  Find the volume of the solid obtained by rotating $R$
about the $x$-axis.
\sol  The cross-sections by planes perpendicular to the $x$-axis are
disks with radius $x^3 - 1$.  So the volume is
$$V = \int_1^2 \pi (x^3 - 1)\ dx = \pi \int_1^2 x^6 - 2x^3 + 1\ dx = \frac{163\pi}{14}.$$
\item[B)]  (5) Set up the integral to compute the volume of 
the solid obtained by rotating $R$ about the line $x = 1$.  You
{\it do not} need to evaluate.  
\sol  The cross-sections by planes perpendicular to the line $x = 1$
are washers with inner radius $(y+1)^{1/3} - 1$ and outer radius $2 - 1 = 1$.  
So the integral to compute the volume is
$$V = \int_0^7 \pi(1)^2 - \pi((y+1)^{1/3} - 1)^2\ dy.$$
\item[C)]  (10)  Find the coordinates of the centroid of $R$.
\sol  The centroid is at the point $(\overline{x},\overline{y})$
with 
$$\overline{x} = \frac{\int_1^2 x(x^3 - 1)\ dx}{\int_1^2 x^3 - 1\ dx} = \frac{94}{55}$$
and
$$\overline{y} = \frac{\int_1^2 \frac{1}{2}(x^3 - 1)^2\ dx}{\int_1^2 x^3 - 1\ dx} = \frac{163}{77}$$
\end{itemize}
\vskip 10pt
\noindent
V.  Compute each of the following integrals.  If you use an entry 
from the table, identify it by number.
\begin{itemize}
\item[A)] (5) $\displaystyle{\int x^2 \sin(3x^3) \cos(3x^3)\ dx}$
\vskip 10pt
\sol  Let $u = \sin(3x^3)$.  Then $du = 9x^2 \cos(3x^3)$ by the Chain
Rule.  So the form is 
$$\frac{1}{9}\int u\ du = \frac{u^2}{18} + C = \frac{\sin^2(3x^3)}{18} + C.$$ 
\item[B)] (5) $\displaystyle{\int_0^\infty \frac{dx}{x^2 + 9}}$
\vskip 10pt
\sol  Because of the infinite limit of integration, this is improper.  We
evaluate as usual:
\begin{align*}
\int_0^\infty \frac{dx}{x^2 + 9} &= \lim_{b\to\infty} \int_0^b \frac{dx}{x^2+9}\\
&= \lim_{b\to\infty} \left.\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)\right|_0^b \quad\text{using \# 17 in table}\\
&= \lim_{b\to\infty} \frac{1}{3}\tan^{-1}\left(\frac{b}{3}\right)\\
&= \frac{\pi}{6}.
\end{align*}
\item[C)] (10) $\displaystyle{\int x^2 \ln(2x)\ dx}$  (use parts)
\vskip 10pt
\sol  Let $u = \ln(2x)$ and $dv = x^2\ dx$.  Then $du = \frac{2}{2x} = \frac{1}{x}$
and $v = \frac{x^3}{3}$.  So
\begin{align*}
\int x^2\ln(2x)\ dx &= \frac{x^3\ln(2x)}{3} - \int \frac{x^3}{3x}\ dx\\
&= \frac{x^3\ln(2x)}{3} - \int \frac{x^2}{3}\ dx\\
&= \frac{x^3\ln(2x)}{3} - \frac{x^3}{9} + C.\\
\end{align*}
(This can be checked with \# 101 from the table.)
\item[D)] (15) $\displaystyle{\int \frac{x^2}{\sqrt{25 - x^2}}\ dx}$ (use a trigonometric substitution)
\sol  From the form of the integrand we use the substitution $x = 5\sin\theta$
and $dx = 5\cos\theta\ d\theta$.  So the integral is:
\begin{align*}
\int \frac{x^2}{\sqrt{25 - x^2}}\ dx &= \int \frac{25\sin^2\theta\ d\theta}{\sqrt{25(1 - \sin^2\theta)}}\\
&= \int \frac{125\sin^2\theta\cos\theta\ d\theta}{5\cos\theta}\\
&= 25\int \sin^2\theta\ d\theta\\
&= \frac{25}{2}(\theta - \sin\theta\cos\theta) + C\quad \text{using \# 63 or \# 73 in the table}\\
&= \frac{25}{2}(\sin^{-1}\left(\frac{x}{5}\right) - \frac{x}{5}\cdot\frac{\sqrt{25 - x^2}}{5} + C\\
&= \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) - \frac{x}{2} \sqrt{25-x^2} + C.
\end{align*}
\end{itemize}
\vskip 10pt
\noindent
VI. (15)  Find the arclength of the catenary curve 
$\displaystyle{y = \frac{e^x + e^{-x}}{2}}$
from $x = 0$ to $x = 2$.   
\begin{itemize}
\sol The arclength is computed by the following integration.  Note
that this is one of the examples where the stuff $1 + (y')^2$ under the radical
is again a perfect square, after some algebra:
\begin{align*}
L &= \int_0^2 \sqrt{1 + (y')^2}\ dx\\
&= \int_0^2 \sqrt{1 + \left(\frac{e^x - e^{-x}}{2}\right)^2}\ dx\\
&= \int_0^2 \sqrt{1 + \frac{e^{2x}}{4} - \frac{1}{2} + \frac{e^{-2x}}{4}}\ dx\\
&= \int_0^2 \sqrt{\frac{e^{2x}}{4} + \frac{1}{2} + \frac{e^{-2x}}{4}}\ dx\\
&= \int_0^2 \frac{e^x + e^{-x}}{2}\ dx\\
&= \left.\frac{e^x - e^{-x}}{2}\right|_0^2\\
&= \frac{e^2 - e^{-2} - 1 + 1}{2}\\
&=  \frac{e^2 - e^{-2}}{2} \doteq 3.63.
\end{align*}
\end{itemize}
\vskip 10pt 
\noindent
VII.  All parts of this question refer to the differential equation
$$\frac{dy}{dx} = y^2 - y.$$
\begin{itemize}
\item[A)] (10) Sketch the slope field 
showing the slopes at points on the lines $y = -1/2, 0, 1/2, 1, 3/2$.
\sol See Figure 4.
\begin{figure}
\begin{center}
\includegraphics[width=3in]{FVII.ps}
\caption{The slope field in VII.}
\end{center}
\end{figure}
\item[B)] (10)  Find the solution with $y(0) = 1/2$.  
\sol  We separate variables and integrate, using partial
fractions on the $y$-integral:
\begin{align*}
\int \frac{dy}{y(y-1)} &= \int\ dx\\
\int \frac{-1}{y} + \frac{1}{y-1}\ dy &= x + c\\
-\ln|y| + \ln|y-1| &= x + c\\
\ln\left|\frac{y-1}{y}\right| &= x + c\\
\frac{y-1}{y} &= k e^x \quad \text{where $k = \pm e^c$}\\
y &= \frac{1}{1 - ke^x}.
\end{align*}
Then from the initial condition $1/2 = \frac{1}{1 - k}$.  So $k = -1$, and
our particular solution is $y = \frac{1}{1 + e^x}$.
\end{itemize}
\vskip 10pt
\noindent
VIII.  A hive of honey bees is attacked by a fungal infection
at $t = 0$, and the population declines at 
a rate proportional to the {\it square root\/} of the population.
\begin{itemize}
\item[A)] (5) Express this statement about the rate of growth of the 
population $P$ as a differential equation.
\sol  If the population is $P$, then the differential equation 
can be written as 
$$\frac{dP}{dt} = k\sqrt{P}.$$
(Since the population is declining, $k < 0$.)
\item[B)]  (10) Use separation of variables to find the general solution
of your differential equation.
\sol  Separating variables and integrating,
\begin{align*}
\int\frac{dP}{\sqrt{P}} &= \int k\ dt\\
2 P^{1/2} &= kt + c\\
P &= \left(\frac{kt + c}{2}\right)^2.
\end{align*}
\item[C)]  (5) At $t = 0$ there were 900 bees; 441 were
left after 6 weeks.  When did the bee population disappear entirely?
\sol  Substituting $t =0$ into the equation for $P$, we see
$900 = (c/2)^2$, so $c = 60$.  Then at $t = 6$, 
$441 = (3k + 30)^2$, so $k = -3$.  The formula for the population is then
$P = \left(\frac{-3t}{2} + 30\right)^2$.  This equals zero when $3t/2 = 30$, so 
$t = 20$.  After 20 weeks, the bee population has gone to zero.   
\end{itemize}
\vskip 10pt
\noindent
IX.  Series.
\begin{itemize}
\item[A)] (5)  Find the sum of the infinite series 
$$4 - \frac{8}{3} + \frac{16}{9} - \frac{32}{27} + \cdots.$$
\sol  This is a geometric series with $a = 4$ and $r = \frac{-2}{3}$.
So the sum is
$$\frac{a}{1 - r} = \frac{4}{5/3} = \frac{12}{5}.$$
\item[B)] (10) Find the Taylor series of $g(x) = e^{-x}$
at $a = 0$.
\sol We have $g^{(n)}(x) = (-1)^n e^{-x}$, so $g^{(n)}(0) = (-1)^n$. Therefore
the Taylor series is
$$\sum_{n=0}^\infty \frac{(-1)^n x^n}{n!} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots .$$
\item[C)] (10) Compute the Taylor polynomial of degree 2 
for $f(x) = \sqrt{x}$ at $a = 1$ and use it to approximate
$\sqrt{1.3}$.  How close is the approximation?
\sol  We have 
$f(1) = 1$, then $f'(x) = \frac{1}{2} x^{-1/2}$, so $f'(1) = \frac{1}{2}$, and
finally $f''(x) = \frac{-1}{4} x^{-3/2}$, so $f''(1) = \frac{-1}{4}$.  
The Taylor polynomial is
$$p_2(x) = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2.$$
The approximation is 
$$p_2(1.3) = 1 + \frac{1}{2}(.3) - \frac{1}{8}(.3)^2 \doteq 1.13875.$$
The calculator value is something like $1.140175$.  So the approximation
is correct to 2 decimal places (after rounding).  
\end{itemize}
\vskip 10pt
\noindent
{\it Extra Credit.}  For a water tank with cross-section area $A(y)$ at
height $y$ above the bottom and an opening
of area $1$ at $y = 0$, Torricelli's Law implies that as 
the water drains out through the opening,
the height $y$ of the water in the tank changes according to the equation
$$A(y)\frac{dy}{dt} = -\sqrt{2gy}$$
(where $g$ is the acceleration of gravity at the surface of the 
Earth).  
\begin{itemize}
\item[A)] (5) Suppose that the tank has the shape of the 
solid of revolution formed by rotating the area between a curve
$x = f(y)$ and the $y$-axis, $0 \le y \le h$, about the $y$-axis.
How should $f(y)$ be chosen in order for the 
water level to decrease at a {\it constant rate\/}?
\sol In order for the water level to be decreasing at a constant
rate, we need $\frac{dy}{dt}$ constant.  This will happen when
$A(y)$ is proportional to $\sqrt{y}$  (note that the $\sqrt{2g}$ is
constant.  Now $A(y)$ is the cross-section area of the 
solid of revolution, so $A(y) = \pi f(y)^2$.  Hence we want
$f(y)^2$ proportional to $\sqrt{y}$.  This says that 
$f(y) = c y^{1/4}$ for some constant $c$, or equivalently
$y = k x^4$ (a surprisingly simple equation!)
\item[B)]  (5)  Suppose you had such a tank. Explain
how you could use it as a {\it time-measuring device\/}.  (Note:  
The ancient Egyptians actually constructed tanks of this
shape and used them to measure time 
as early as the 16th century BC (i.e. long before calculus!).  
The Greeks took over the idea and their
name for this sort of device was a {\it clepsydra}.)  
\sol  To use such a tank as a time-measurement device, the
simplest approach would be to put some type of measuring stick
vertically in the tank.  If the stick was marked in
{\it intervals of equal length}, then as the water dropped, the 
same amount of time would elapse between all pairs of
consecutive marks.  
\end{itemize}
\vskip 10pt
\centerline{\it Have a joyous and peaceful holiday season!}

\end{document}
