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\begin{document}
\begin{center}

\textbf{College of the Holy Cross} \\
\textbf{MATH 133, Calculus With Fundamentals 1}\\
\textbf{Solutions for Final Examination -- Friday, December 15}
\end{center}

\vskip 10pt
\noindent
I.  The  graph $y = f(x)$ is given in light blue.  Match each equation with one of the 
numbered magenta graphs.
\vskip 10pt
\begin{center}
\includegraphics[width=3truein,height=2truein]{FI.png} 
\end{center} 

\begin{itemize}
\item[(5) A)] $y = 2 f(x + 6)$ is plot number: $\underline{\qquad\qquad}$
\item[] {\it Solution:}  2 (shifted left and stretched vertically)
\item[(5) B)] $y = \frac{1}{3} f(x) $ is plot number: $\underline{\qquad\qquad}$
\item[] {\it Solution:}  4 (compressed vertically)
\item[(5) C)] $y = f(x) + 3$ is plot number: $\underline{\qquad\qquad}$
\item[] {\it Solution:} 1 (shifted vertically)
\item[(5) D)] $y = f(x - 4)$ is plot number: $\underline{\qquad\qquad}$
\item[] {\it Solution:} 3 (shifted right)
\item[(5) E)] $y = -f(x + 4)$ is plot number: $\underline{\qquad\qquad}$
\item[] {\it Solution:} 5 (reflected across $x$-axis and shifted left)
\end{itemize}

\vskip 10pt
\noindent
II.  A cup of hot chocolate is set out on a counter at $t = 0$.  The 
temperature of the chocolate $t$ minutes later is
$C(t) = 65 + 55 e^{-t/4}$ (in degrees F).
\vskip 10pt
\begin{enumerate}
\item[A)] (5) What is the temperature of the chocolate at $t = 0$?\\
\item[] {\it Solution:}  $C(0) = 65 + 55\cdot 1 = 120$ degrees F.
\item[B)] (10) What is the rate of change of the temperature at $t = 10$ minutes?\\
\item[] {\it Solution:}  The rate of change is 
$$C'(10) = 55 e^{-10/4} \cdot (-1/4) \doteq -1.13$$
(degrees per minute)
\item[C)] (10) How long does it take for the temperature to 
reach $100^\circ F$?\\
\item[] {\it Solution:} Solve the equation $100 = 65 + 55 e^{-t/4}$ for $t$:
$$t = -4\ln(35/55) \doteq 1.81$$
(minutes). 
\end{enumerate}
\vskip 10pt
\noindent
III.  Compute the following limits.  Any legal method is OK.  
\vskip 10pt
\begin{enumerate}
\item[(A)] (10) $\displaystyle{\lim_{x \to 3} \frac{x^2 + x - 12}{x^2 - 5x + 7}}$.\\
\item[] {\it Solution:}  This one is not an indeterminate form, since the denominator
is not zero at $x = 3$.  By the Limit Laws, the limit is 
$$ = \frac{3^2 + 3 - 12}{3^2 - 15 + 7} = 0.$$ 
\item[(B)] (10) $\displaystyle{\lim_{x \to 1} \frac{x - 1}{x^2 - 1}}$.
\item[] {\it Solution:}  This one is an indeterminate form, but we can factor the bottom
and cancel the $x - 1$ to get
$$\lim_{x \to 1} \frac{x-1}{x^2 - 1} = \lim_{x \to 1} \frac{1}{x + 1} = \frac{1}{2}.$$
\item[(C)] (10) $\displaystyle{\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}}$
\item[] {\it Solution:}  Multiply the top and the bottom by the conjugate radical,
simplify the top, cancel the $x$ between the top and the bottom, then evaluate:
\begin{eqnarray*}
\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} & = & 
\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x + 1} + 1}\\
&=& \lim_{x \to 0} \frac{(x+1) - 1}{x(\sqrt{x+1}+1)}\\
&=& \lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{2}.
\end{eqnarray*}
\item[(D)] (10) $\displaystyle{\lim_{x \to 0} \frac{\tan(x)}{x}}$
\item[] {\it Solution:}  Since $\tan(x) = \sin(x)/\cos(x)$, the limit product law says:
$$ = \lim_{x\to 0} \frac{\sin(x)}{x} \cdot \lim_{x\to 0} \frac{1}{\cos(x)} = 1\cdot 1 = 1.$$
\end{enumerate}
\vskip 10pt
\noindent
IV.  A) (10) {\it Using the limit definition}, and showing all necessary steps
to justify your answer, compute $f'(x)$ for $f(x) = 5x^2 + x$.
\vskip 10pt
\noindent
{\it Solution:}
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\\
&=& \lim_{h\to 0} \frac{5(x + h)^2 + (x + h) - 5x^2 - x}{h}\\
&=& \lim_{h\to 0} \frac{5x^2 + 10xh + 5h^2 + x + h - 5x^2 - x}{h}\\
&=& \lim_{h\to 0} \frac{10xh + h+ 5h^2}{h}\\
&=& \lim_{h\to 0} 10 x + 1 + 5h\\
&=& 10 x + 1.
\end{eqnarray*}
IV. (continued)  Using appropriate derivative rules, compute the derivatives
of the following functions.  You do not need to simplify your answers.
\begin{enumerate}
\item[B)] (5) $\displaystyle{g(x) = 4x^3 +  \sqrt{x} + \frac{2}{\sqrt[4]{x}} + e^2}$
\item[] {\it Solution:} Write $\sqrt{x} = x^{1/2}$ and $1/\sqrt[4]{x} = x^{-1/4}$.
By the power rule:
$$g'(x) = 12x^2 + \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-5/4}$$
\item[C)] (10) $\displaystyle{h(x) = \frac{\cos(x) + x}{\sin(x)}}$
\item[] {\it Solution:}  By the Quotient Rule, 
$$h'(x) = \frac{\sin(x)(-\sin(x) + 1) - (\cos(x) + x)\cos(x)}{\sin^2(x)}.$$
\item[D)] (10) $i(x) = \ln(x^3+e^x)$
\item[] {\it Solution:}  By the derivative rule for $\ln(u)$ (with the chain rule), 
$$i'(x) = \frac{1}{x^3 + e^x}\cdot (3x^2 + e^x).$$
\item[E)] (10) $j(x) = \tan^{-1}(3x + 7)$
\item[] {\it Solution:}  By the derivative rule for the inverse tangent and the chain rule, 
$$j'(x) = \frac{1}{1 + (3x + 7)^2}\cdot 3.$$
\end{enumerate}
\vskip 10pt
\noindent
V.  The following graph shows the {\it derivative}
$f'(x)$ for some function $f(x)$ defined on $0\le x \le 4$.
Note:  This \emph{is not} $y = f(x)$, it is $y = f'(x)$.  
\begin{center}
\includegraphics[width=3truein,height=2truein]{FV.png} 
\end{center} 
Using the graph, {\it estimate}
\begin{enumerate}
\item[A)] (5) The interval(s) on which $f(x)$ is increasing.
\item[] {\it Solution:}  These are the intervals where $f'(x) > 0$ so $(0,1)$ and $(2,3)$. 
\item[B)] (10) The critical numbers of $f(x)$ \emph{in the open interval} $(0,4)$.  Say
what the behavior of $f(x)$ is at each critical number (local max, local min, neither)  
\item[] {\it Solution:}  In the open interval $(0,4)$ $f$ has critical points
at $x = 1,2,3$.  $f'$ changes sign from $+$ to $-$ at $x = 1,3$, so those are
local maxima; $f'$ changes sign from $-$ to $+$ at $x = 2$, so that is 
a local minimum.  (All of this is based on the first derivative test.)
\item[C)] (10) The interval(s) on which $y = f(x)$ is concave down. 
\item[] {\it Solution:}  $f(x)$ is concave down on intervals where $f'(x)$ 
is decreasing, so roughly $(0.4,1.4)$ and $(2.4,3.4)$.  
\end{enumerate}
\vskip 10pt
\noindent
VI. A town wants to build a pipeline from a water station on a small
island 2 miles from the shore of its water reservoir to the town. 
One possible route is shown dotted in red. The town is 6 miles along the shore
from the point nearest the island.  It costs \$3 million   
per mile to lay pipe under the water and \$2 million per mile
to lay pipe along the shoreline.
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{FVI.png} 
\end{center} 
\begin{enumerate}
\item[A)] (5)  Give the cost $C(x)$ of constructing the pipeline
as a function of $x$.
\item[] {\it Solution:}  By the Pythagorean theorem, the distance over
water is $\sqrt{4 + x^2}$ and the distance over land is $6 - x$.  By the given 
information about cost per mile, we have 
$$C(x) = 3\sqrt{4 + x^2} + 2(6 - x)$$
(where the cost is given in units of millions of dollars).
\item[B)] Differentiate your cost function to find $C'(x)$.
\item[] {\it Solution:}  The derivative is 
$$3\cdot \frac{1}{2}\cdot (4 + x^2)^{-1/2}\cdot 2x - 2 = 
\frac{3x}{\sqrt{4 + x^2}} - 2.$$
\item[C)] Find all the critical points of $C(x)$.
\item[] {\it Solution:}  The function $C(x)$ has
a critical number for $x > 0$ at the positive solution of $C'(x) = 0$:
\begin{eqnarray*}
0 &=& \frac{3x}{\sqrt{4 + x^2}} - 2, \text{or}\\
3x &=& 2 \sqrt{4 + x^2} \\
9x^2 &=& 16 + 4x^2 \\
5 x^2 &=& 16 \\
x &=& \frac{4}{\sqrt{5}} \doteq 1.79.
\end{eqnarray*}
There is another critical point at $x \doteq -1.79$, but that would 
clearly give a longer pipeline costing more, so we can ignore it.  
\item[D)]  Where on the shoreline should the pipeline hit land to minimize
the costs of construction?  Say how you know this gives the minimum
cost.  
\item[] {\it Solution:} To find the minimum of $C(x)$, we
can restrict to $x$ in the closed interval $[0,6]$, since it clearly does no
good to take $x < 0$ or $x > 6$.  
We have $C(0) = 18$, $C(6) = 3\sqrt{40} \doteq 19.0$, and
$C\left(\frac{4}{\sqrt{5}}\right) \doteq 16.47$.  So 
the minimum cost is attained at $x = \frac{4}{\sqrt{5}}\doteq 1.79$ miles.
Alternatively, one could use the First or Second Derivative Test to show 
this is a local minimum and then argue from the fact that there is only 
one critical point for $x > 0$.  
\end{enumerate}
\vskip 10pt
\noindent
VII. (15)  A block of dry ice (solid $CO_2$) is 
evaporating and losing volume at the rate of $10$ cm${}^3$/min. 
It has the shape of a cube at all times.  How 
fast are the edges of cube shrinking when the block has volume 
$216$ cm${}^3$?  
\vskip 10pt
\noindent
{\it Solution:}  Call the side of the cube $x$.  Then $V = x^3$.
Taking time derivatives, we have $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. From the given information,
when $V = 216$, $x = 6$ and $\frac{dV}{dt} = -10$.  Therefore the rate of change of 
the side of the cube is 
$$\frac{dx}{dt} = \frac{-10}{3\cdot 6^2} = \frac{-5}{54} \doteq -.093$$
(units cm/min).  The side of the cube is decreasing at
about $.09$ cm/min.
\end{document}




