%% Math 131-03 Midterm 2 solutions, Fall 2003

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\begin{document}

\begin{center}

\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Section 03, Midterm 2 Solutions}
\end{center}

\begin{enumerate}

\item{} [5 points each]\quad Compute the derivatives of the given
  functions. 

\bd

\item{(a)} $f(t)=(t^2+1)^5$

  \soln $f'(t)=5(t^2+1)^4(2t)=10t(t^2+1)^4$

\hrulefill

\item{(b)} $g(x)=x^3\ln x$

  \soln $g'(x)=x^3\frac{1}{x}+3x^2\ln x = x^2(1+3\ln x)$

\hrulefill

\item{(c)} $z(u)=\ds\frac{u}{e^u-1}$

  \soln $z'(u)=\ds\frac{(e^u-1)-ue^u}{(e^u-1)^2}$

\hrulefill

\item{(d)} $s(t) = te^{t^2}$

  \soln $s'(t)=t(2te^{t^2})+e^{t^2} = e^{t^2}(1+2t^2)$

\hrulefill

\item{(e)} $b(x) = \ln(e^x-e^{-x})$

  \soln $b'(x)=\ds\frac{e^x + e^{-x}}{e^x - e^{-x}}$

\hrulefill

\item{(f)} $k(s)=\ds\frac{s^2+\ln s}{s}$

  \soln $k(s)=s+s^{-1}\ln s$, so 
  \[
  k'(s)=1+s^{-1}s^{-1}+(-s^{-2})\ln s = 1+s^{-2}(1-\ln s)
  \]

\ed

\eject

\item{} [10 points]\quad Find the tangent line to the curve
  $xy^2=xy+4$ at the point~$(2,2)$.

  \soln Implicit differentiation gives
  \[
  x\cdot2y\frac{dy}{dx}+y^2 = x\frac{dy}{dx}+y.
  \]
  At $(x,y)=(2,2)$, we have $8(dy/dx)+4=2(dy/dx)+2$, or
  $dy/dx=-1/3$. The equation of the tangent line is
  \[
  y-2=-\frac{1}{3}(x-2),\qquad\text{or}\quad
  y=-\frac{1}{3}x+\frac{8}{3}.
  \]

\hrulefill

\item{} [10 points]\quad Suppose
  $\ds\frac{dx}{dt}=\sqrt{1+e^{2x}}$. Find~$\ds\frac{d^2x}{dt^2}$ as a
  function of~$x$. 

  \soln By the chain rule,
  \[
  \frac{d^2x}{dt^2} 
  = \frac{1}{2}(1+e^{2x})^{-1/2}\cdot2e^{2x}\frac{dx}{dt}
  = \frac{1}{2}(1+e^{2x})^{-1/2}\cdot2e^{2x}(1+e^{2x})^{1/2}
  = e^{2x}.
  \]

\eject

\item{} In both parts of this question, let $f(x)=x^2$.

\bd

\item{(a)} [10~points]\quad Compute the derivative~$f'(x)$ \emph{from
    the definition}.

  \soln Compute the limit of the difference quotient:
  \begin{align*}
    f'(x) &= \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\
    &= \lim_{h\to0} \frac{(x+h)^2-x^2}{h} \\
    &= \lim_{h\to0} \frac{x^2+2xh+h^2-x^2}{h} \\
    &= \lim_{h\to0} 2x+h =2x.
  \end{align*}

\item{(b)} [10~points]\quad Find the equation of the line obtained by
  zooming in on the graph at the point~$(1,1)$, and sketch the line
  in the grid provided.

\soln By part~(a), we have $f'(1)=2$, so the tangent line has
equation $y-1=2(x-1)$, or $y=2x-1$.

\hfill
\input{midsoln2a-03.eepic}

\ed

\eject

\item{} [5~points each]\quad The position of a car (measured in feet)
  after $t$~seconds is:
  \[
    \begin{array}{|c||c|c|c|}
      \hline
      \vp t    & 0 & 1 & 1.5 \\
      \hline
      \vp s(t) & 0 & 80 & 130 \\
      \hline
    \end{array}
  \]

  \bd

  \item{(a)} Estimate the speed of the car when $t=0.5$ and when
  $t=1.25$. Give units.

  \soln Let $v(t)$ denote the velocity of the car at time~$t$. Then
  \[
  v(0.5)\simeq\frac{s(1)-s(0)}{1-0}=80,\qquad
  v(1.25)\simeq\frac{s(1.5)-s(1)}{1.5-1} = \frac{130-80}{0.5}=100.
  \]
  Both answers are in feet per second.

  \item{(b)} When $t=1$, does the car appear to be accelerating or
    decelerating? Write a sentence or two in support of your answer.

  \soln The car appears to be accelerating: The velocity increased
  from 80~ft/sec to 100~ft/sec between $t=0.5$~and $t=1.25$.

  \ed

%%\eject
\hrulefill

\item{} [10~points]\quad Let $f(x) = e^{-x^2/2}$. Find the interval(s)
  on which $f$~is both decreasing and concave down.

  \soln The chain and product rules give $f'(x)=-xe^{-x^2/2}$ and
  \[
  f''(x)=-x(-xe^{-x^2/2})-e^{-x^2/2}=e^{-x^2/2}(x^2-1).
  \]
  Since $e^{-x^2/2}$ is positive for every real~$x$, the derivative is
  positive for $x<0$~and negative for $x>0$, while the second
  derivative is negative for $-1<x<1$. Consequently, $f$~is decreasing
  on the interval~$[0,\infty)$, and is concave down on the
  interval~$[-1,1]$, hence is both decreasing and concave down on the
  interval $[0,1]$.

\eject

\item{} [5 points each]\quad (You must justify your answers for full
  credit.)  Define a function by
  \[
  f(x) = 
  \begin{cases}
    e^x & \text{ if $x<0$}, \\
    2-\cos x & \text{ if $x\geq0$}.
  \end{cases}
  \]

  \bd

  \item{(a)} Is~$f$ continuous at $x=0$? (Hint: One-sided limits!)

  \soln Plugging in gives $f(0)=2-\cos 0=2-1=1$. The one-sided limits are
  \[
  \lim_{x\to0^-}f(x) = \lim_{x\to0}e^x=e^0=1,\qquad
  \lim_{x\to0^+}f(x) = \lim_{x\to0}2-\cos x=2-\cos 0=1.
  \]
  Since the one-sided limits are equal, we find that
  $\lim\limits_{x\to0}f(x)=1=f(0)$, so $f$~is continuous at~$0$.

  \item{(b)} Is~$f$ differentiable at $x=0$?

  \soln Differentiating the formulas above gives
  \[
  f'(x) = 
  \begin{cases}
    e^x & \text{ if $x<0$}, \\
    \sin x & \text{ if $x>0$}.
  \end{cases}
  \]
  Note that the formula says nothing about what happens at~$x=0$. The
  one-sided limits of the derivative are
  \[
  \lim_{x\to0^-}f'(x) = \lim_{x\to0}e^x=e^0=1,\qquad
  \lim_{x\to0^+}f'(x) = \lim_{x\to0}\sin x=\sin 0=0.
  \]
  Since these are unequal, $f$~is not differentiable at $x=0$.

  \ed

\end{enumerate}

\end{document}

