%% Math 131 Midterm 3, Fall 2004
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%\markright{Math 131, Midterm 3}

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\begin{document}

\begin{center}

\textbf{Math 131, Midterm 3 Solutions} \\
\end{center}

\vskip 20pt


\begin{enumerate}
  
\item{} [5~points each]\quad Match each parametric curve with its
  graph. (Each graph shows the square $-1\leq x\leq 1$, 
  $-1\leq y\leq 1$.)

  \bd

  \item{(a)} $x(t)=-1+2t$, $y(t)=1-2t$, $0\leq t\leq 1$\hfill
    $\bigcircle$~I\quad $\bigcircle$~II\quad $\bigcircle$~III\quad
    $\bigcircle$~IV\quad $\bigotimes$~V

    {\bf Explanation.} This is a line with slope $-2/2=-1$. 

    \vskip 0pt plus 0.5fil

    \sep

  \item{(b)} $x(t)=\cos(2t)$, $y(t)=\sin(2t)$, $0\leq t\leq \pi$\hfill
    $\bigotimes$~I\quad $\bigcircle$~II\quad $\bigcircle$~III\quad
    $\bigcircle$~IV\quad $\bigcircle$~V

    {\bf Explanation.} Points $(x,y)$ on the curve satisfy $x^2+y^2=1$, so they 
    lie on the unit circle.

    \vskip 0pt plus 0.5fil

    \sep

  \item{(c)} $x(t)=\cos t$, $y(t)=\frac{1}{2}\sin t$, $0\leq t\leq
    2\pi$\hfill $\bigcircle$~I\quad $\bigcircle$~II\quad
    $\bigcircle$~III\quad $\bigotimes$~IV\quad $\bigcircle$~V

    {\bf Explanation.} This is the standard parametrization of the unit 
    circle, but with the $y$-coordinate scaled by a factor of $1/2$, so it 
    must be an ellipse.

    \vskip 0pt plus 0.5fil

    \sep

  \item{(d)} $x(t)=1-2t$, $y(t)=-t$, $0\leq t\leq 1$\hfill
    $\bigcircle$~I\quad $\bigotimes$~II\quad $\bigcircle$~III\quad
    $\bigcircle$~IV\quad $\bigcircle$~V

    {\bf Explanation.} This is a line with slope $-1/-2=1/2$.

    \vskip 0pt plus 0.5fil

    \sep

  \ed

\begin{center}
  \input{midterm3a.eepic}\quad  \input{midterm3b.eepic}\quad
  \input{midterm3c.eepic}\quad  \input{midterm3d.eepic}\quad
  \input{midterm3e.eepic} 
\end{center}

\vfil\eject

\item{} [5~points each]\quad Compute the indicated limits. Show all
  work for full credit. 

  \bd

  \item{(a)} $\ds\lim_{x\to1}\frac{\cos(\frac{\pi}{2}x)}{x-1}$

    {\bf Answer.} Since $\cos(\pi/2)=0$, the numerator and denominator both approach zero. 
    By L'Hopital's Rule,
        \[
        \lim_{x\to1}\frac{\cos(\frac{\pi}{2}x)}{x-1}
        =\lim_{x\to1}\frac{-\frac{\pi}{2}\sin(\frac{\pi}{2}x)}{1}
        =-\frac{\pi}{2}
        \]


  \MyAnswer{Limit =}{  -\frac{\pi}{2}}
\sep
  \item{(b)} $\ds\lim_{x\to0}\frac{x^2+1}{x^2+3}$

{\bf Answer.} 
    \[
    \lim_{x\to0}\frac{x^2+1}{x^2+3}=\frac{\lim_{x\to0}x^2+1}{\lim_{x\to0}x^2+3}=\frac13
    \]

  \MyAnswer{Limit = }{\frac13}
\sep
  \item{(c)} $\ds\lim_{x\to\infty}\frac{x\ln x}{x^{1.01}}$
{\bf Answer.} First simplify.
    \[
    \lim_{x\to\infty}\frac{x\ln x}{x^{1.01}}=\lim_{x\to\infty}\frac{\ln x}{x^{0.01}}
    \]
    The numerator and denominator both approach $\infty$, so by L'Hopital's Rule,
        \[
        \lim_{x\to\infty}\frac{\ln x}{x^{0.01}}
        =\lim_{x\to\infty}\frac{\frac1x}{0.01x^{-0.99}}
        =\lim_{x\to\infty}\frac1{0.01x^{0.01}}=0
        \]

  \MyAnswer{Limit = }{0}
  \ed

\sep
\item{} Each part refers to the function $f(x)=2x^3-6x^2$.

  \bd
  
  \item{(a)}  [10~points]\quad Find and classify (local min/max, or
  neither) the critical points of~$f$.


    {\bf Answer.} The derivative of $f$ is $f'(x)=6x^2-12x=6x(x-2)$, so the critical points of 
    $f$ are $x=0$ and $x=2$. 
    
    To determine the type of each critical point, apply either the first or second derivative test. 
    
    First Derivative Test. Since $f'(-1)=18$, $f'(1)=-6$ and $f'(3)=18$, the sign of $f'$ changes 
    from $+$ to $-$ at $x=0$, so $f$ has a local maximum at $x=0$, and the sign of $f'$ changes from 
    $-$ to $+$ at $x=2$, so $f$ has a local minimum at $x=2$. 
    
    Second Derivative Test. The second derivative of $f$ is 
    $f''(x)=12x-12$. Since $f''(0)=-12$ is negative, $f$ has a local maximum at $x=0$, and 
    since $f''(2)=12$ is positive, $f$ has a local minimum at $x=2$. 
    
    \sep

  \item{(b)} [5~points]\quad Find the maximum and minimum values
    of~$f(x)$ if $-\frac{3}{2}\leq x\leq \frac{5}{2}$.


{\bf Answer.} To find the global maximum and minimum, evaluate $f$ at each critical 
point and at each endpoint:
\begin{align*}
  f(0)&=0\\
  f(2)&=-8\\
  f(-\frac32)&=-20.25\\
  f(\frac52)&=-6.25
\end{align*}

  \MyAnswer{Minimum value:}{-20.25}
  \MyAnswer{Maximum value:}{0}

  \sep

  \item{(c)} [5~points]\quad In the grid provided, sketch the graph
  $y=f(x)$ for $-1\leq x\leq3$. (Note that axis labels are provided.)
  For full credit, clearly indicate the critical and inflection
  point(s) in this interval, and label each such point with \emph{both
  coordinates}.

  {\bf Answer.} $f''(x)=12x-12=12(x-1)$, so $f''$ changes sign at $x=1$. 
  Therefore $f$ has an inflection point at $(1,-4)$. 
  The critical points have coordinates $(0,0)$ and $(2,-8)$.
  
  
  \ed

  \begin{center}
    \input{midterm3fsol.eepic}
  \end{center}

  \eject

\item{} [15~points]\quad Crusader Movie Rentals finds that they can
  rent 160~movies per night at \$1~per movie. For every dollar that
  the rental fee increases, 40~fewer movies are rented. What price
  should be charged to maximize the revenue (total rental income)?


{\bf Answer.} Let $p$ be the price in dollars per movie, and 
let $n$ be the number of movies that can be rented at that price. 
The revenue is then $R=np$. Let $x$ be the increase in the price from 1 dollar. 
Then $p=1+x$, and $n=160-40x$, so
    \[
    R(x)=(1+x)(160-40x)=160+120x-40x^2.
    \]
The domain of $R$ is $-1\leq x\leq 4$. Since 
    \[
    R'(x)=120-80x,
    \]
the critical point of $R$ is $x=\frac32=1.5$. Since $R(-1)=R(4)=0$ and 
$R(1.5)=250$, the revenue is maximized when $x=1.5$, i.e. when $p=2.5$.

  \MyAnswer{Revenue maximized when $p=$}{\$2.50}
\sep

\item{} [15~points]\quad Find the equation of the line tangent to the
  curve $xy-2x-3y+1=0$ at the point~$(-2,1)$.

{\bf Answer.} Use implicit differentiation:
    \[
    x\frac{dy}{dx}+1\cdot y-2-3\frac{dy}{dx}=0
    \]
    Now solve for $\frac{dy}{dx}$:
        \[
        \frac{dy}{dx}=\frac{2-y}{x-3}
        \]
    When $x=-2$ and $y=1$,
        \[
        \frac{dy}{dx}=\frac{1}{-5}=-\frac15.
        \]
    So the slope of the tangent line is $m=-\frac15$. Applying the 
    point-slope formula gives $y-1=-\frac15(x+2)$.
    
  \MyAnswer{Equation of tangent line at~$(-2,1)$:}{y-1=-\frac15(x+2)\text{ or }y=-\frac15x+\frac35}

  \eject

\item{} A boat is drawn into a dock by a rope over a small pulley. The
  pulley is five feet higher than the bow of the boat (see
  figure). Let~$\ell$ be the length of rope, $x$~the distance from the
  boat to the dock.

  \begin{center}
    \input{midterm3g.eepic}
  \end{center}

  \bd

  \item{(a)} [5~points]\quad Find an equation relating $\ell$~and $x$,
  and determine~$x$ when $\ell=13$.


{\bf Answer.} Use the Pythagorean Theorem.

  \MyAnswer{Equation relating $\ell$~and $x$:}{x^2+25=\ell^2}

  \MyAnswer{When $\ell=13$, $x=$}{12}

  \sep

  \item{(b)} [10~points]\quad Suppose the rope is drawn in at
  3~ft/sec. How fast is the boat moving when the length of the rope is
  13~feet? 


{\bf Answer.} We are given $\frac{d\ell}{dt}=-3$ and want to find $\frac{dx}{dt}$ when $\ell=13$. 
Differentiate the equation $x^2+25=\ell^2$ with respect to $t$:
    \[
    2x\frac{dx}{dt}=2\ell\frac{d\ell}{dt}
    \]
When $\ell=13$, $x=12$, so
\begin{align*}
    2(12)\frac{dx}{dt}&=2(13)(-3)\\
    \frac{dx}{dt}&=-3.25
\end{align*}
  \MyAnswer{Speed =}{3.25\text{ ft/sec}}

  \ed

  \eject

\end{enumerate}

\end{document}
