%% Math 131 Midterm 2, Fall 2004
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%\markright{Math 131, Midterm 2}

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\begin{document}
\centerline{MATH 131, section 1 -- Solutions for Practice Questions for Exam 2}
\centerline{October 17, 2007}

\begin{enumerate}
\item{}  All parts of this question refer to the parametric
curve $x = 3 \cos(2 t)$, $y = 5 \sin(2 t)$.

\begin{enumerate}
\item[a.] Eliminate the parameter $t$ and find a Cartesian equation for
this curve.
\vskip 10pt
\item[] {\it Solution:}  We have 
$\displaystyle{\frac{x}{3} = \cos(2t)}$ and $\displaystyle{\frac{y}{5} = \sin(2t)}$,
so 
\[
\left(\frac{x}{3}\right)^2 + \left(\frac{y}{5}\right)^2 = \cos^2(2t) + \sin^2(2t) = 1.
\]
The equation is $\displaystyle{\frac{x^2}{9} + \frac{y^2}{25} = 1}$, which we recognize as the 
equation of an {\it ellipse} centered at $(0,0)$, with semimajor axis 5 (along the $y$-axis),
and semiminor axis 3 (along the $x$-axis).
\vskip 10pt
\item[b.] What portion of the curve is traced out for $0\le t \le \frac{\pi}{2}$, 
and in which direction is the curve being traced?
\vskip 10pt
\item[] {\it Solution:}  As $t$ increases from $0$ to $\frac{\pi}{2}$, 
$\cos(2t)$ is decreasing from $1$ to $-1$, and $\sin(2t)$ is increasing
from $0$ to $1$ at $t = \frac{\pi}{4}$, then decreasing back to $0$ at $t = \frac{\pi}{2}$.
Therefore,
$x$ is decreasing from $3$ to $-3$ and $y$ is increasing up to $5$, then 
decreasing back to $0$.  The portion of the curve traced out is the top half
of the ellipse and we are moving {\it counterclockwise}.
\vskip 10pt
\item[c.] What would change in your answer to part b if the curve above was
replaced by $x = 3 \cos(-2 t)$, $y = 5 \sin(-2 t)$?
\vskip 10pt
\item[] {\it Solution:} We would be moving {\it clockwise} around the {\it bottom half} of 
the ellipse rather than counterclockwise around the top half.  
\end{enumerate}
  
\item{} Use the sum, product, and/or quotient 
  rules to compute the following derivatives. You may use any correct
  method, but must show work and simplify your answers for full
  credit.

\begin{enumerate}
  \item[a.] $\ds\frac{d}{dx}\Bigl(5x\sqrt{x}-\frac{2}{x^3}+11x-4\Bigr)$

  \vskip 10pt
  \item[] {\it Solution:}  By the power and sum rules, 
\[
\ds\frac{d}{dx}\Bigl(5x\sqrt{x}-\frac{2}{x^3}+11x-4\Bigr) = 
\frac{15}{2} x^{1/2} + 6x^{-4} + 11.
\]
\vfil\eject  
  \item[b.] $\ds\frac{d}{dt} (t^2e^{t})$
\vskip 10pt
  \item[] {\it Solution:}  By the product rule, 
\[
\frac{d}{dt} (t^2e^{t}) = t^2 e^t + 2t e^t = (t^2 + 2t)e^t.
\]
  
  \item[c.] $\ds\frac{d}{dz}\frac{z^2 - 2z + 4}{z^2 + 1}$
\vskip 10pt
 \item[] {\it Solution:}  By the quotient rule, 
\begin{eqnarray*}
\frac{d}{dz}\frac{z^2 - 2z + 4}{z^2 + 1} &=& \frac{(z^2+1)(2z-2)-(z^2-2z+4)(2z)}{(z^2+1)^2}\\
&=& \frac{2z^2-6z-2}{(z^2+1)^2}.
\end{eqnarray*}
\item[d.] $\ds\frac{d}{dx}\Bigl(\frac{\pi^2+\tan(e^\pi)-2x^e}{4}\Bigr)$
\vskip 10pt
 \item[] {\it Solution:}  Don't be fooled -- most of this function 
is constant.  The only part that is not is the $-\frac{x^e}{2}$, so 
the derivative is 
\[
\ds\frac{d}{dx}\Bigl(\frac{\pi^2+\tan(e^\pi)-2x^e}{4}\Bigr) 
= \frac{d}{dx}\left(\frac{\pi^2}{4} + \frac{\tan(e^\pi)}{4} - \frac{x^e}{2}\right)
= 0 + 0 -\frac{ex^{e-1}}{2} = -\frac{ex^{e-1}}{2}.
\]
\end{enumerate}

\item{} The graph of a function~$f$ is shown
  below with several points marked. Find all the marked points at
  which the following are true, and give explanations for your answers.
  \begin{center}
    \input{midterm2a.eepic}
  \end{center}

\begin{enumerate}
  \item[a.] $f$~is discontinuous.
\vskip 10pt
 \item[] {\it Solution:}  Points $B,C$, and $G$.  Note that 
both one-sided limits exist and are equal at $x = B$, but the limit
is not the same as the value of the function there (a removable discontinuity).
At $C$, $f$ has an infinite discontinuity (vertical asymptote).  At $G$, 
$f$ has a jump discontinuity (both one-sided limits exist but they are unequal).  
\vskip 10pt
  \item[b.] $f$~is continuous, but the graph of $f$~has a vertical
    tangent line. 
\vskip 10pt
 \item[] {\it Solution:} This happens at point $A$ only.
\vskip 10pt
  \item[c.] $f$~is continuous, but the graph of $f$~has no tangent line.
 \vskip 10pt
 \item[] {\it Solution:} This happens at points $E$ (a ``corner'') and $F$ (a cusp).
\end{enumerate}

\vskip 10pt
\item{} Compute the indicated limits. Show all work for full credit.

  \begin{enumerate}

  \item[a.]
     $\ds\lim_{x\to 1}\frac{3x^2-5x-2}{x^2-4x+4}$
\vskip 10pt
 \item[] {\it Solution:}  We have 
\[
\lim_{x\to 1} 3x^2-5x-2 = 3(\lim_{x\to 1} x)^2 - 5\lim_{x\to 1} x - 2 = 3 - 5 - 2 = -4
\]
and  
\[
\lim_{x\to 1} x^2-4x+4 = (\lim_{x\to 1} x)^2 - 4\lim_{x\to 1} x + 4 = 1 - 4 + 4 = 1
\]
by the limit sum and product rules.  Since the limit of the denominator is not zero, 
the limit quotient rule then says
\[
\lim_{x\to 1}\frac{3x^2-5x-2}{x^2-4x+4} = \frac{\ds\lim_{x\to 1} 3x^2-5x-2}{\ds\lim_{x\to 1} x^2-4x+4}
= \frac{-4}{1} = -4.
\]
\item[b.] 
     $\ds\lim_{x\to 2}\frac{3x^2-5x-2}{x^2-4x+4}$
\vskip 10pt
 \item[] {\it Solution:} We cannot apply the technique used in part a here 
since now
\[
\lim_{x\to 2} 3x^2-5x-2 = 3(\lim_{x\to 2} x)^2 - 5\lim_{x\to 2} x - 2 = 12 - 10 + 2 = 0.
\]
and
\[
\lim_{x\to 2} x^2-4x+4 = (\lim_{x\to 2} x)^2 - 4\lim_{x\to 2} x + 4 = 4 - 8 + 4 = 0.
\]
For $x \ne 2$, we have after factoring and cancellation:
\[
\frac{3x^2-5x-2}{x^2-4x+4} = \frac{(3x+1)(x-2)}{(x-2)(x-2)} = \frac{3x+1}{x-2}.
 \]
Since the denominator is still going to 0 as $x \to 2$ while the numerator is
approaching 7, this limit {\em does not exist}.  In terms of the one-sided limits,
in fact,
\[
\lim_{x\to 2^-} \frac{3x+1}{x-2} = -\infty, \text{ while } \lim_{x\to 2^+} \frac{3x+1}{x-2} = +\infty.
\]

\item[c.] 
     $\ds\lim_{x\to\infty}\frac{3x^2-5x-2}{x^2-4x+4}$
\vskip 10pt
 \item[] {\it Solution:}  Divide the top and bottom by the fastest growing power
of $x$ that appears and then take the limit:
\begin{eqnarray*}
 \lim_{x\to\infty}\frac{3x^2-5x-2}{x^2-4x+4} &=& \lim_{x\to\infty}\frac{(3x^2-5x-2)\frac{1}{x^2}}{(x^2-4x+4)\frac{1}{x^2}}\\
&=& \lim_{x\to\infty}\frac{3-\frac{5}{x}-\frac{2}{x^2}}{1-\frac{4}{x}+\frac{4}{x^2}}\\
&=& 3/1\\
&=& 3.
\end{eqnarray*}
(Recall, this computation would show that the graph $\ds y = \frac{3x^2-5x-2}{x^2-4x+4}$
has a horizontal asymptote at $y = 3$ as $x \to +\infty$ and also as $x \to -\infty$.) 
\item[d.] 
     $\ds\lim_{x\to 2^+}\frac{|x-2|}{x^2 - 5x + 6}$
\vskip 10pt
 \item[] {\it Solution:}  Note $x^2 - 5x + 6 = (x - 2)(x - 3)$.  Hence the function is
\[
\frac{|x - 2|}{(x - 2)(x - 3)} = \frac{|x - 2|}{x - 2}\cdot \frac{1}{x - 3} = 
\begin{cases}
   \frac{1}{x - 3} & \text{if }  x > 2 \\
   \frac{-1}{x - 3} & \text{if } x < 2\\
\end{cases}.
\]
This shows 
\[
\lim_{x\to 2^+} \frac{|x-2|}{x^2 - 5x + 6} = \lim_{x\to 2^+} \frac{1}{x - 3} = -1.
\]
\end{enumerate}

\item{} Let $f(x)=x^3-x^2$. 

  \begin{enumerate}
\item[a.]  Find all intervals on which~$f$ is
    \textbf{decreasing}.
\vskip 10pt
\item[] {\it Solution:}  To answer this, we need $f'(x)$, since
$f$ is decreasing on an interval when $f'(x) < 0$ on that interval.
Here $f'(x) = 3x^2 - 2x = x(3x - 2)$.  The graph $y = f'(x)$ is a 
parabola opening up.  We see $f'(x) = 0$ at $x = 0$ and $x = 2/3$, so
$f'(x)$ is {\em negative} for $0 < x < 2/3$ and positive for
$x < 0$, $x > 2/3$.  {\em Answer}:  In interval notation,
$f$ is decreasing for $x \in (0,2/3)$.
\vskip 10pt
  \item[b.] Find all intervals on which~$f$ is
    \textbf{concave up}.
\vskip 10pt
 \item[] {\it Solution:}  To answer this, we need $f''(x)$, since
$f$ is concave up on an interval when $f''(x) > 0$ on that 
interval.  We have $f''(x) = 6x - 2$, so $f''(x) < 0$ for $x < 1/3$
and $f''(x) > 0$ for $x > 1/3$.  {\em Answer}: $f$ is 
concave up for $x \in (1/3, +\infty)$
\vskip 10pt
\item[c.] Find all intervals on which~$f$ is
    both \textbf{increasing}~and \textbf{concave down}.
\vskip 10pt
 \item[] {\it Solution:}  We want the intervals where $f'(x) > 0$ and $f''(x) < 0$.  
From the solutions for parts a, b, this is true only when $x < 0$.  {\em Answer}: 
$f$ is increasing and concave down for $x \in (-\infty, 0)$.  
  \end{enumerate}

\item{} Do not use the differentiation rules from Chapter~3 in this
  question.

  \begin{enumerate}

  \item[a.] State the limit definition of the
  derivative~$f'(x)$.
\vskip 10pt
 \item[] {\it Solution:} 
\[
f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h},
\]
if the limit exists. 

  \item[b.] Use the definition to
    compute the derivative function of $f(x)=\ds\frac{1}{3x}$.
\vskip 10pt
 \item[] {\it Solution:}  Using the definition from part a, 
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{ \ds{ \frac{1}{3(x+h)} - \frac{1}{3x} } }{h}\\
&=& \lim_{h\to 0} \frac{3x - 3(x+h)}{9hx(x+h)} \quad \text{(simplify fraction)}\\
&=& \lim_{h\to 0} \frac{-3h}{9hx(x+h)}\quad \text{(cancel like terms on top)}\\
&=& \lim_{h\to 0} \frac{-1}{3x(x+h)} \quad \text{(cancel the $3h$ top and bottom)}\\
&=& \frac{-1}{3x^2}.
\end{eqnarray*}

\item[c.] Find the equation of the line tangent to
    the graph $y=\ds\frac{1}{3x}$ at $x=2$.
\vskip 10pt
 \item[] {\it Solution:}  When $x = 2$, $\ds{f(2) = \frac{1}{6}}$, and
$\ds{f'(2) = \frac{-1}{12}}$ from part b.
So we calculate the point-slope form of the equation of the tangent line, 
which is 
$\ds{y - \frac{1}{6} = \frac{-1}{12}(x - 2)}$, or after simplifying,
\[
y = \frac{-1}{12} x + \frac{1}{3}.
\]
  \end{enumerate}

\item{} The total cost (in $ \$ $) of repaying a car loan at 
interest rate of $r \%$ per year is $C = f(r)$.  

   \begin{enumerate}

   \item[a.] What is the meaning of the statement $f(7) = 20000$?
\vskip 10pt
 \item[] {\it Solution:}  This means that if the interest rate is $7\%$ per year,
then the cost of repaying the loan is $\$20000$.

   \item What is the meaning of the statement $f'(7) = 3000$? What are the units
         of $f'(7)$? 
\vskip 10pt
 \item[] {\it Solution:}  This means that when $r = 7\%$ per year, the rate of
change of the cost
of repaying the loan is $\$3000$ dollars per $\%$ per year
(the units are dollars per $\%$ per year).  
   \end{enumerate}

  \vfil

\end{enumerate}

\end{document}
