\documentclass[11pt]{article}
\def\Z{\mathbb{Z}}
\def\Q{\mathbb{Q}}
\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
\usepackage{latexsym,amsmath,amssymb,subfig,epsfig}
\usepackage[all]{xy}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{9.0in}
\setlength{\topmargin}{-0.5in}
\begin{document}
\centerline{Mathematics 136, section 2 -- Advanced Placement Calculus}
\centerline{Solutions for Exam 2 -- November 4, 2009}
\begin{enumerate}
\item[I.]  Consider the function $f(x) = x^3 + 6x^2 - 15x$.  
\begin{enumerate}
\item[A)] (10) Find all the critical numbers of $f$ and classify
each as a local maximum, local minimum, or neither.
\item[] {\it Solution:}  We have $f'(x) = 3x^2 + 12x - 15 = 3(x + 5)(x - 1)$.
This is defined for all $x$ and zero only at $x = -5,1$ so those are 
the critical numbers.  We have $f''(x) = 6x + 12$.  Hence $f''(-5) = -18$
and $f''(1) = 18$.  By the Second Derivative Test, $x = -5$ is a local
maximum, and $x = 1$ is a local minimum.  (Note: this can also be determined
by the First Derivative Test, finding the intervals where $f'$ is positive
($x < -5$ and $x > 1$)
and the interval where $f'$ is negative ($-5 < x < 1$).   
\vskip 10pt
\item[B)] (5)  What are the absolute (global) maximum and minimum
values of $f(x)$ on the interval $[0,3]$? 
\item[] {\it Solution:}  The critical number $x = 1$ is in this interval.
$f(0) = 0$, $f(1) = -8$ and $f(3) = 36$.  So the absolute maximum is $f(3) = 36$
and the absolute minimum is $f(1) = -8$.  
\end{enumerate}
\vskip 10pt
\item[II.]  (10) Evaluate the following limit:
\[
\lim_{\theta\to 0} \frac{\cos(5 \theta) - 1}{\theta^2}.
\]
\vskip 10pt
\noindent
{\it Solution:}  This is indeterminate of the form $0/0$, so we can apply 
L'Hopital's Rule (twice):
\begin{align*}
\lim_{\theta\to 0} \frac{\cos(5 \theta) - 1}{\theta^2} 
&= \lim_{\theta\to 0} \frac{-5\sin(5\theta)}{2\theta}\quad \text{(still $0/0$)}\\
&= \lim_{\theta\to 0} \frac{-25 \cos(5 \theta)}{2}\\
&= \frac{-25}{2}.
\end{align*}

\item[III.]  The velocity of a accelerating car is measured at each 
of the following times, yielding a table of values:  
\[
\begin{array}{lccccccc}
       t = {\rm time (sec)}& \vline &1&3&5&7&9&11\\ \hline
       v(t) = {\rm velocity (ft/sec)}& \vline &1&3.5&5.1&7.8&9.0&11.2
\end{array}
\]
\begin{enumerate}
\item[A)] (5) Give an estimate for the total distance traveled by the car between $t = 1$ and $t = 11$ using a {\it left-hand} Riemann
sum for the velocity function and $\Delta t = 2$.  
\item[] {\it Solution:}  The left-hand Riemann sum estimate is
\[
v(1) \Delta t + v(3) \Delta t + v(5) \Delta t + v(7) \Delta t + v(9) \Delta t = 
2 + 7 + 10.2 + 15.6 + 18 = 52.8 {\rm ft}.
\] 
\item[B)] (5)  Is your result from A) less than or greater than
the actual total distance traveled, assuming $v$ is always increasing
between $t = 1$ and $t = 11$?  Explain.
\item[] {\it Solution:}  If $v$ is always increasing on this interval,
then each term in the left-hand Riemann sum \emph{underestimates} the
actual distance traveled on that interval.  Hence the answer in A is 
\emph{less} than the actual distance traveled.
\end{enumerate}
\vskip 10pt
\item[IV.] Compute each of the following integrals.  Show all work.
\begin{enumerate}
\item[A)] (5)  $\displaystyle{\int_1^4 x^2 - 3\sqrt{x} + 4\ dx}$
\item[] {\it Solution:}  By the Evaluation Theorem, this gives:
\[ 
\left.\frac{x^3}{3} - 2 x^{3/2} + 4x\right|_1^4 = \frac{64}{3} - 16 + 16 - \frac{1}{3} + 2 - 4 = 19.
\]
\item[B)] (10)  $\displaystyle{\int \sec^2(2x) {\root 3 \of {\tan(2x) + 4}}\ dx}$ 
\item[] {\it Solution:}  Let $u = \tan(2x) + 4$.  Then $du = 2\sec^2(2x)\ dx$, and 
the form is 
\[
\frac{1}{2} \int u^{1/3}\ du = \frac{1}{2}\cdot \frac{3}{4} u^{4/3} + C 
= \frac{3}{8} (\tan(2x)+4)^{4/3} + C.
\]
\item[]  Integrate using the indicated method and the table (say which entry you
are using if you use one).
\vskip 10pt
\item[C)]  (10) Use parts:  $\displaystyle{\int x^2 e^{4x}\ dx}$  (full credit
only for showing all work; you can check your result by the table)
\item[]  {\it Solution:} Integrate by parts twice making $u$ the power of $x$
both times:
\begin{align*}
\int x^2 e^{4x}\ dx &= \frac{x^2 e^{4x}}{4} - \frac{2}{4} \int x e^{4x}\ dx\\
&=  \frac{x^2 e^{4x}}{4} - \frac{1}{2}\left( \frac{x e^{4x}}{4} - \frac{1}{4}\int e^{4x}\ dx\right)\\
&=  \frac{x^2 e^{4x}}{4} - \frac{x e^{4x}}{8} + \frac{e^{4x}}{32} + C.
\end{align*}
\item[D)]  (12.5) Use a trigonometric substitution:
$\displaystyle{\int \frac{\sqrt{25 - x^2}}{x}\ dx.}$
\item[] {\it Solution:}  From the form of the integrand, let $x = 5\sin\theta$,
so $dx = 5\cos\theta\ d\theta$.  Then the integral can be evaluated 
by using the trigonometric identity $\cos^2 \theta = 1 - \sin^2\theta$:
\begin{align*}
\int \frac{\sqrt{25 - x^2}}{x}\ dx &= \int \frac{5\cos\theta}{5\sin\theta} 5\cos\theta\ d\theta\\
&= 5 \int \frac{\cos^2\theta}{\sin\theta}\ d\theta\\
&= 5 \int \frac{1 - \sin^2\theta}{\sin\theta}\ d\theta\\
&= 5 \int \csc\theta - \sin\theta\ d\theta\\
&= 5\left(\ln|\csc\theta - \cot\theta| + \cos\theta\right) + C \quad \text{\# 15 in table}\\
&= 5\ln\left|\frac{5}{x} - \frac{\sqrt{25-x^2}}{x}\right| + \sqrt{25-x^2} + C.
\end{align*}
\item[E)]  (12.5) Use partial fractions:
\[
\int \frac{x + 4}{(x + 1)(x^2 + 16)} \ dx.
\]
\item[] {\it Solution:}  From the form of the denominator we set up the 
partial fractions:
\begin{align*}
\frac{2x + 19}{(x + 1)(x^2 + 16)} &= \frac{A}{x+1} + \frac{Bx + C}{x^2 + 16}\\
2x + 19 &= A(x^2 + 16) + (Bx + C)(x + 1).
\end{align*}
Equating coefficients gives $A + B = 0$, $B + C = 2$ and $16A + C = 19$.  
So $A = 1$, $B = -1$, and $C = 3$.  Integrating with \#15 in table, we obtain:
\[
\int \frac{dx}{x+1} + \int \frac{-x + 3}{x^2 + 16}\ dx
= \ln|x+1| - \frac{1}{2}\ln|x^2 + 16| + \frac{3}{4}\tan^{-1}(x/4) + C.
\]
\end{enumerate}
\item[V.]
\begin{enumerate}
\item[A)]  (15)  A fish swimming against a current of $u$ ft/sec (constant)
for a distance $L$ (constant) will expend energy 
$$E(v) = \frac{Lv^3}{v-u}$$
if it swims at $v$ ft/sec ($u < v$).  What $v$ minimizes the total energy expended?
\item[] {\it Solution:}  By the quotient rule, 
\[
E'(v) = \frac{(v-u)\cdot 3Lv^2 - Lv^3\cdot(1)}{(v-u)^2}, 
\]
and $E'(v) = 0$ when the numerator is zero, so $3L(v-u)v^2 = Lv^3$.  
Dividing through by $Lv^2$, $3(v-u) = v$, so 
$v = \frac{3u}{2}$.  This a minimum by the first derivative test.   
\item[B)]  (15)  Using the definition (\emph{not} the Evaluation Theorem) 
compute 
\[
\int_2^3 x^2 + 4x\ dx.
\]
The following formulas may be useful:
\[
\sum_{i=1}^n i = \frac{n(n+1)}{2} \qquad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.
\]
\item[] {\it Solution:}  For the right-hand Riemann sum, we have
$\Delta x = \frac{3 - 2}{n} = \frac{1}{n}$ and $x_i = 2 + \frac{i}{n}$.  Then
\begin{align*}
\int_2^3 x^2 + 4x\ dx &=
\lim_{n\to\infty} \sum_{i=1}^n \left( \left(2 + \frac{i}{n}\right)^2 + 4\left(2 + \frac{i}{n}\right)\right)\frac{1}{n}\\
&= \lim_{n\to\infty} \sum_{i=1}^n \left(\frac{i^2}{n^2} + \frac{8i}{n} + 12\right)\frac{1}{n}\\
&= \lim_{n\to\infty} \left(\frac{1}{n^3} \sum_{i=1}^n i^2 + \frac{8}{n^2} \sum_{i=1}^n i + 12\right)\\
&= \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{n^3} + \frac{8n(n+1)}{2n^2} + 12\\
&= \frac{1}{3} + \frac{8}{2} + 12 = \frac{49}{3}.
\end{align*}
(Check by Evaluation Theorem:
\[
\int_2^3 x^2 + 4x\ dx = \left.\frac{x^3}{3} + 2x^2\right|_2^3 = 9 + 18 - \frac{8}{3} - 8 = 
\frac{49}{3}.)
\]
\end{enumerate}
\end{enumerate}

\end{document}
