%% MATH 131-01 Final Exam Solutions, Fall 2004
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%\textbf{Holy Cross College, Fall Semester, 2004} \\
%\textbf{MATH 131, Section 01, Final Exam} \\
%\textbf{Monday, December~13, 2:30~PM} \\
%Professor Little
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%\noindent{\bf Your Name}: \Name
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%\noindent
%{\it Instructions.} Clearly mark your answers, and show work on
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%space for scratch work. {\it You must show all work for full
%credit}, but please place answers in the boxes provided where appropriate.

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%\noindent
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%\[
%\mathscr{H\!A\!P\!P\!Y\ \  H\!O\!L\!I\!D\!A\!Y\!S}!
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\noindent
1. [5~points each]\quad Circle the number of the graph
   showing each of the following functions. 


\begin{itemize}
  \item{(a)} $f(x) = 3 - e^{-x}$\hfill {\it Answer:} V

  \item{(b)} $f(x) = x^3 - 2x^2 - x + 2$\hfill {\it Answer:} IV

  \item{(c)} $f(x) = 4\sin(\pi x)$\hfill {\it Answer:} I

  \item{(d)} $f(x) = 4 - 4\cos(\pi x)$\hfill {\it Answer:} III

\item{} Note: Graph II is $y = e^{-x}$.  V is what you get from II,
if that graph is reflected across $x$-axis and shifted
up by 3.
\end{itemize}

\centerline{\psfig{figure=1IV.ps,height=2.0in,width=2.0in,angle=-90}
$\quad\quad\quad$ \psfig{figure=1II.ps,height=2.0in,width=2.0in,angle=-90}}

\vskip 0.1truein

\centerline{\psfig{figure=1V.ps,height=2.0in,width=2.0in,angle=-90}
$\quad\quad\quad$ \psfig{figure=1III.ps,height=2.0in,width=2.0in,angle=-90}}

\vskip 0.1truein

\centerline{\psfig{figure=1I.ps,height=2.0in,width=2.0in,angle=-90}
}

\vfil\eject
\noindent 
2. [20~points] One of the functions given in the following table is linear
and the other is exponential.  Find formulas of the appropriate
type for each.

\[
\begin{array}{|c||r|r|r|r|r|}
\hline
x          & 1 & 2 & 3& 4& 5\Strut \\
\hline
f(x)       & 1.2 & 0.6 & 0.3 & 0.15 & 0.075 \Strut \\
\hline
g(x)       & -2.3& -0.6& 1.1 & 2.8 & 4.5\Strut \\
\hline
\end{array}
\]

\noindent
{\it Solution:}  $g(x)$ is linear, since the slopes of the lines
between all pairs of points in the table is $1.7$.  For instance
using the first two points:  $m = \frac{-0.6 + 2.3}{2-1} = 1.7$.
The equation for $g$ is obtained by the point-slope form: 
$y - (-2.3) = 1.7(x - 1)$, so $y = {1.7x - 4}$.

Since the problem said one is exponential, that means that
$f(x)$ is the exponential one.  We can find the equation for
$f(x) = ca^x$ as usual.  From the table with $x = 1,2$,
$ca^1 = 1.2$ and $ca^2 = 0.6$ so $\frac{ca^2}{ca} = a = \frac{0.6}{1.2} = \frac{1}{2}$.
Then from the first data point $c\frac{1}{2} = 1.2$, so $c = 2.4$.  
$g(x) = \frac{2.4}{2^x}$. 
\sep
\noindent
3.  
\begin{itemize}
\item{(a)} [15~points]  The depth of water in a tank oscillates sinusoidally
once every 4 hours.  The smallest depth is 2 feet and the maximum depth
is 5 feet, which occurs at $t = 0$.  Find a formula for the depth $d(t)$
if $t$ is the time in hours.
\vskip 20pt
\noindent
{\it Solution:}  The amplitude is $\frac{5 - 2}{2} = \frac{3}{2}$.  The
vertical shift is $\frac{5+2}{2} = \frac{7}{2}$.  Putting $t = 0$ at
the start of the period where there is a maximum means we want to use
$\cos$.  Finally the period is 4, so we get
$$d(t) = \frac{3}{2}\cos\left(\frac{\pi t}{2}\right) + \frac{7}{2}$$
\sep
\item{(b)} [5~points]  How fast is the depth changing at $t = 1.3$ hours?
Is it increasing or decreasing?

\vskip 20pt
\noindent
{\it Solution:}  The question is asking for the derivative of $d(t)$
at $t = 1.3$:  $d'(t) = -\frac{3\pi}{4}\sin\left(\frac{\pi t}{2}\right)$,
so $d'(1.3) = -\frac{3\pi}{4}\sin\left(\frac{1.3\pi}{2}\right) \doteq -2.1$
feet per hour.  Since this is negative, the depth is decreasing at $t = 1.3$.
\end{itemize}
\eject

\noindent
4.  Compute the following limits [5~points each].  Any legal method is OK.
\begin{itemize}
\item{(a)} {\it Solution:}  
\begin{eqnarray*}\lim_{x\to 1} \frac{x^2 + 1}{x - 3} & = & \frac{\lim_{x\to 1} x^2 + 1}{\lim_{x\to 1} x - 3}\\
&=& \frac{2}{-2}\\
&=& -1\\
\end{eqnarray*} 
\sep
\item{(b)} {\it Solution:} This limit is indeterminate of the form $0/0$, so we 
use L'Hopital's Rule -- twice!
\begin{eqnarray*}
\lim_{x\to 2} \frac{x^2 - 4x + 4}{\cos(x-2) - 1} &=&
\lim_{x\to 2} \frac{2x - 4}{-\sin(x-2)}\\
&=&\lim_{x\to 2} \frac{2}{-\cos(x-2)}\\
&=& -2\\
\end{eqnarray*}
\sep
\item{(c)} This one can also be done by L'Hopital,
or by some algebraic ``trickery'':
\begin{eqnarray*}
\lim_{x\to \infty} \frac{5x^2 - x + 21}{3x^2 + x + 1}
&=& \lim_{x\to \infty} \frac{5x^2 - x + 21}{3x^2 + x + 1}\cdot \frac{1/x^2}{1/x^2}\\
&=&\lim_{x\to \infty} \frac{5 - 1/x + 21/x^2}{3 + 1/x + 1/x^2}\\
&=&\frac{5}{3}\\
\end{eqnarray*}
\end{itemize}
\eject

\noindent
5. 
 \begin{itemize}
\item{(a)} [5~points] State the limit definition of the derivative.
{\it Solution:}
$$f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}$$
(provided that this limit exists).

\sep

\item{(b)} [10~points] Use the definition to compute $f'(x)$ for
$f(x) = \sqrt{x}$.

\vskip 20pt
\noindent
{\it Solution:}  
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\\
      &=& \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
      &=& \lim_{h\to 0}\frac{(x+h) - x}{h(\sqrt{x+h}+\sqrt{x})}\\
      &=& \lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\\
      &=& \frac{1}{2\sqrt{x}}\\
\end{eqnarray*}
\sep

\item{(c)} [10~points] Find the equation of the tangent line
to the graph $y = \sqrt{x}$ at the point $[4,2]$.
\vskip 20pt

\noindent
{\it Solution:}  From part B, the derivative of $\sqrt{x}$ at $x = 4$ is 
$\frac{1}{2\sqrt{4}}=\frac{1}{4}$.  This is the slope of the tangent line,
which has the equation $y - 2 = \frac{1}{4}(x - 4)$, or $y = \frac{1}{4}x + 1$.
\end{itemize}
\eject

\noindent
6.  Compute the following derivatives using the derivative rules.
You need not simplify. [5~points each]
\begin{itemize} 
\item{(a)}  $f(t) = t^3 - \frac {1}{{\root 4\of t}} + \pi^t$.
\vskip 20pt
\noindent
{\it Solution:}  By the power and exponential rules,
$$f'(t) = 3t^2 + \frac{1}{4} x^{-5/4} + \pi^t \ln(\pi)$$

\sep

\item{(b)}  $\ds g(x) = \frac{x^2 - 2}{\cos(x) + 1}$

\vskip 20pt
\noindent
{\it Solution:}  By the quotient rule,
$$g'(x) = \frac{(\cos(x) + 1)(2x) - (x^2 - 2)(-\sin(x))}{(\cos(x)+1)^2}$$

\sep

\item{(c)} $\ds h(z) = \ln(4z^2 + 2e^{\arctan(z)})$

\vskip 20pt
\noindent
{\it Solution:}  By the chain rule,
$$h'(z) = \frac{8z + \frac{2e^{\arctan(z)}}{1+z^2}}{4z^2 + 2e^{\arctan(z)}}$$

\sep

\item{(d)} Find $\frac{dy}{dx}$ if $x^2 y - 2y^3 = 3$.

\vskip 20pt
\noindent
{\it Solution:}  Differentiating implicitly, $x^2 \frac{dy}{dx} + 2xy - 6y^2\frac{dy}{dx} = 0$.
Solving for $\frac{dy}{dx}$, we have
$$\frac{dy}{dx} = \frac{-2xy}{-6y^2 + x^2}$$
\end{itemize}
\eject

\noindent
7.  Consider the family of curves defined by $y = f(x) = x^4 + 2ax^2$, where
$a$ is any fixed real number.  

\begin{itemize}
\item{(a)}  [10~points]  Find the {\it critical points} of $f$, construct 
a sign diagram for $f'(x)$ in the case that $a < 0$.  Which of your 
critical points are local maxima and which are local minima?

\vskip 20pt
\noindent
{\it Solution:}  We have $f'(x) = 4x^3 + 4ax = 4x(x^2 + a)$.  When $a < 0$,
this has three solutions, so three critical points: 
$x = -\sqrt{-a}, 0, \sqrt{-a}$. (Note: $a < 0$ means
$-a > 0$, so $\sqrt{-a}$ does exist in the real numbers in this case.)  The 
sign diagram should show $f' < 0$ for $x < - \sqrt{-a}$, $f' > 0$
for $-\sqrt{-a} < x < 0$, $f' < 0$ for $0 < x < \sqrt{-a}$ and 
$f' > 0$ for $x > \sqrt{-a}$.  This means that $x = \pm \sqrt{-a}$ are
local minima and $x = 0$ is a local maximum (First Derivative Test).  

\sep 
\item{(b)}  [10~points] Repeat part a, but assume now that $a > 0$.  
\vskip 20pt
\noindent
{\it Solution:}  As before, $f'(x) = 4x^3 + 4ax = 4x(x^2 + a)$.  But
when $a > 0$, the equation $x^2 + a = 0$ has no real solutions.  So
there is only one critial point: $x = 0$.  $f' < 0$ for $x < 0$ and 
$f' > 0$ for $x > 0$.  So $x = 0$ is a local minimum.

\sep

\item{(c)} [10~points] How many different {\it inflection points} does
the graph $y = f(x)$ have if $a < 0$?  Explain.

\vskip 20pt
\noindent
{\it Solution:}  When $a < 0$, $f''(x) = 12 x^2 + 4a = 0$ if $x = \pm \sqrt{-a/3}$.
The sign of $f''$ changes at each, so there are {\it two} inflection points in this case. 
\end{itemize}

\sep

\noindent
8. [15~points]  A cubical block of dry ice (solid $CO_2$) is 
evaporating and losing volume at the rate of $10~cm^3/min$.  How 
fast are the sides of cube shrinking when the block has volume 
$125~cm^3$?  Give the units of your answer.

\vskip 20pt
\noindent
{\it Solution:}  Let $x$ denote the side of the cube, so the volume is 
$V = x^3$.  Differentiating with respect to $t$, we get
$\frac{dV}{dt} = 3x^2\frac{dx}{dt}$.  We know $\frac{dV}{dt} = -10$ when
$V = 125$. At that time, $x = (125)^{1/3} = 5$, so 
$$\frac{dx}{dt} = \frac{-10}{3\cdot 5^2} = \frac{-2}{15}\doteq -.13$$
(units:  cm/min).
\eject
\noindent
9.  [20~points] Ship A travels along the path given by the parametric
curve $x = t, y = t^2$.  At the same time, ship B travels along
the curve $x = t, y = 4t-5$.  At what time are the two ships
{\it closest to one another}?
\vskip 20pt
\noindent
{\it Solution:}  At each time $t$, we can find the distance
between the ships using the distance formula for points in the 
plane:
$$d(t) = \sqrt{(x_A(t) - y_A(t))^2 + (y_A(t) - y_B(t))^2}
=\sqrt{(t - t)^2 + (t^2 - 4t + 5)^2} = |t^2 - 4t + 5|$$
But note that $t^2 - 4t + 5 = (t - 2)^2 + 1 > 0$ for all $t$.
So, we get $d(t) = t^2 - 4t + 5$.  We want the time when the
ships are closest, so we are looking for the minimum of 
$d(t)$.  To find this, take $d'(t) = 2t - 4$ and set $= 0$.
This says $t = 2$.  

\vskip 20pt
\noindent
{\it Note:}  Many people made tables of the distance
at whole number values for $t$: $t = 0,1,2,3$, etc.,
and concluded that 
the minimum distance ocurred at $t = 2$ from that
information.  While the conclusion is correct, 
the reasoning is not.  How do you know
that the distance doesn't reach an even smaller
value for some $t$ between $1$ and $2$, or
$t$ between $2$ and $3$?  So this solution did
not receive full credit, even though the answer
was correct.
\sep
\noindent
10.  
\begin{itemize}
\item{(a)} [7.5~points]  Compute the left- and right-hand sums
for the function $f(t) = e^{-t^2}$ on the interval $[0,1]$, using
$n = 4$ equal subdivisions.
\vskip 20pt
\noindent
{\it Solution:} $\Delta t = \frac{1 - 0}{4} = .25$.
$$LHS = e^0 \Delta t + e^{-(.25)^2}\Delta t + e^{-(.5)^2} \Delta t + e^{-(.75)^2}\Delta t
\doteq .82$$
and
$$LHS = e^{-(.25)^2}\Delta t + e^{-(.5)^2} \Delta t + e^{-(.75)^2}\Delta t + e^{-1}\Delta t
\doteq .66$$

\vskip 20pt
{\it Note:}  A lot of people were computing values of this function
incorrectly.  I couldn't tell exactly what went wrong, but I think
it was probably a calculator issue.  So I didn't assess a large
penalty if your method was OK, but the values were incorrect.
\sep
\item{(b)} [7.5~points] Consider the graph $y = f(t)$ given below.
Compute the average value $\overline{y}$ of $f$ on the 
interval $0\le t \le 5$, given that the area marked $A$ 
is 25 square units, and the areas marked $B$ and $C$ are both 16 square units.

\vskip 20pt
\noindent
{\it Solution:}  Since the areas $A$ and $C$ are below the $t$-axis,
in computing $\int_0^5 f(t)\ dt$, we count them with negative signs.
So the average value is
$$\overline{y} = \frac{1}{5 - 0}\int_0^5 f(t)\ dt = \frac{1}{5}(-25 + 16 - 16) = -5.$$
\end{itemize}
\end{document}
