%% Math 131-01 Midterm 3 solutions, Fall 2003

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\begin{center}

\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Section 01, Midterm 3 Solutions} \\
Professor Hwang
\end{center}

\be

\item{} [5 points each]\quad Evaluate the following limits:

\bd

\item{(a)} $\ds\lim_{x\to0}\frac{1-\cos x}{x^2}
  = \lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}
  = \frac{1}{2}$

\item{(b)} $\ds\lim_{x\to\infty}\frac{\ln(e^x+1)}{x}
  =\lim_{x\to\infty}\frac{e^x}{e^x+1}
  =\lim_{x\to\infty}\frac{e^x}{e^x}=1$

\ed

\sep

\item{} [10 points]\quad A small portion of the graph $y=f(x)$ (with
  horizontal and vertical scales as shown) is given below. Use linear
  approximation to determine which of the following functions might be
  depicted.
  \[
  f(x)=e^x\qquad f(x)=\sin x\qquad f(x)=\cos x
  \]

  \begin{center}
    \input{midterm3-01a.eepic}
  \end{center}
  
  \soln The function~$f$ shown satisfies $f(0)=1$ and $f'(0)=1$. Since
  $\sin 0=0$ and $\cos'0=-\sin 0=0$, $f$~cannot be either trig
  function. However, $f(x)=e^x$ is a possibility, because $f(0)=e^0=1$
  and $f'(0)=e^0=1$.

\sep

\item{} [10 points each]\quad Both parts of this question refer to the
  function $f(x)=4x^3-x^4$.

  \bd

  \item{(a)} Find and classify the critical point(s) of~$f$.

    \soln We have $f'(x)=12x^2-4x^3=4x^2(3-x)$. The sign diagram
    for~$f'$ is:
    \begin{center}
      \input{midsoln3-01a.eepic}
    \end{center}
    Thus $x=3$ is a \textbf{local maximum}, while $x=0$ is
    \textbf{neither} a minimum nor a maximum.

\vfil\eject

  \item{(b)} Find the absolute maximum and minimum values of~$f$ on
  the interval~$[1,4]$.

  \soln The only critical point of~$f$ in the interval~$[1,4]$ is~$3$,
  so we only need to compare the values of~$f$ at $x=1$, $3$, and~$4$:
  \[
  f(1)=3,\qquad f(3)=(4-3)\cdot3^3=27,\qquad f(4)=4^4-4^4=0.
  \]
  The absolute minimum value is~0, and the absolute maximum is~27.

  \ed

\sep

\item{} [20~points]\quad A spherical balloon is being inflated at
  $20~\textrm{in}^3/\textrm{sec}$. When the radius is 6~inches, at
  what rate is the radius increasing? The volume of a sphere of
  radius~$r$ is $V=\frac{4}{3}\pi r^3$.

  \soln Differentiating gives
  $\ds\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. When $r=6$, we have
  \[
  \frac{dr}{dt}=\frac{1}{4\pi r^2}\cdot\frac{dV}{dt}
  =\frac{20}{4\pi\cdot36}=\frac{5}{36\pi}
  \simeq 0.044\frac{\textrm{in}}{\textrm{sec}}
  \]

  \sep

\item{} [20~points]\quad Let $a>0$. A rectangle is inscribed under the
  curve $y=\sqrt{x}$:
  \begin{center}
    \input{midterm3-01b.eepic}
  \end{center}
  Find the value of~$x$, with $0<x<a$, for which the area of the
  rectangle is largest. Show all work for full credit.

  \soln The area of a rectangle is
  $\textrm{width}\times\textrm{height}$, so
  $A(x)=(a-x)\sqrt{x}=ax^{1/2}-x^{3/2}$. To maximize~$A$, we search
  for critical points:
  \[
  A'(x)=\textstyle\frac{1}{2}ax^{-1/2}-\frac{3}{2}x^{1/2}
  =\frac{1}{2}ax^{-1/2}(a-3x).
  \]
  The only critical point in the interval~$(0,a)$ is $x=\frac{a}{3}$,
  and a sign diagram confirms that this point gives the maximum area.

  \vfil\eject

\item{} [10~points]\quad Find the inflection point(s) of the function
  $f(x)=x^{2/3}-\frac{1}{5}x^{5/3}$. 

  \soln Compute the first two derivatives:
  $f'(x)=\frac{2}{3}x^{-1/3}-\frac{1}{3}x^{2/3}$, and 
  \[
  f''(x)=\textstyle-\frac{2}{9}x^{-4/3}-\frac{2}{9}x^{-1/3}
  =-\frac{2}{9}x^{-4/3}(1+x)=-\frac{2(1+x)}{9x^{4/3}}.
  \]
  The potential inflection points are $x=0$~and $x=-1$. To check which
  (if either) is an inflection point, we analyze the sign of~$f''$.
  Note that $x^{4/3}>0$ if $x\neq0$, so the sign of the second
  derivative is the same as the sign of~$-\frac{2}{9}(1+x)$ (except
  at~$x=0$, which we need not test). From this observation, the second
  derivative changes sign at $x=-1$ but not at $x=0$. Thus $x=-1$ is
  the only inflection point of~$f$.

  \sep

\item{} [10~points]\quad As a car brakes to a stop, the speed
  (measured in feet per second) after $t$~seconds is given by the
  table below. 
  \[
    \begin{array}{|c||c|c|c|c|}
      \hline
      \vp t & 0  &  1 &  2 & 3 \\
      \hline
      \vp v & 66 & 40 & 18 & 0 \\
      \hline
    \end{array}
  \]
  On the basis of this information, give the most accurate estimate of
  the distance required for the car to come to a stop.

  \soln Because the speed of the car is decreasing, the left-hand sum
  is an over-estimate and the right-hand sum is an underestimate. Each
  time interval is one second in duration, so the sums are
  \[
  \textrm{Left} =(66+40+18)\textstyle\frac{\textrm{ft}}{\textrm{sec}}
  \times 1~\textrm{sec}=124~\textrm{feet},\qquad
  \textrm{Right}=(40+18+0)=58~\textrm{feet}.
  \]
  Our best guess is the average, $91$~feet.

\ee

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