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\centerline{Mathematics 136, section 2 -- Advanced Placement Calculus}
\centerline{Solutions for Exam 1 -- October 1, 2009}
\vskip 10pt
\noindent
I.  All parts of this question refer to the graph $y = f(x)$ plotted in Figure 1 on the next page
\begin{enumerate} 
\item[A.] (5)  Do $\displaystyle{\lim_{x\to 2^+} f(x)}$ and
$\displaystyle{\lim_{x\to 2^-} f(x)}$ exist?  If so, say what they are; if not say why not.
\item[]  \emph{Solution:}  Yes, both one-sided limits exist: 
$\displaystyle{\lim_{x\to 2^+} f(x)} = 2$ and $\displaystyle{\lim_{x\to 2^-} f(x)} = 1/2$.
\item[B.] (5)  Does $f(x)$ appear to be differentiable at $x = 3$?   Why or why not? 
\item[] \emph{Solution:}  No, there is apparently a cusp point on the graph at $x = 3$, 
so there is no single well-defined tangent line.
\item[C.] (10) On separate axes, plot the graphs $y = 2f(x+1)$ and 
$y = f(2x) + 1$.  Label which graph is which.
\item[] \emph{Solution:}  $y=2f(x+1)$ is shifted one unit left and stretched vertically
by a factor of $2$.  $y = f(2x) + 1$ is compressed horizontally by a factor of $2$ 
and shifted up one unit. See graphs in Figures 2 and 3 on the next page, 
and pay special attention to the 
scales on the two axes.
\end{enumerate}


\vskip 10pt
\noindent
II.  At time $t = 0$, a steak at temperature $0^\circ$ F
is taken out of the freezer and left to thaw in a kitchen heated to 
$70^\circ$ F.  The steak's temperature increases fastest at first, then 
at a slower and slower rate as it nears the room temperature; it never goes over
$70^\circ$ F.
\begin{enumerate}  
\item[A.] (10)  Sketch a graph of the temperature of the steak as a function of time.
(For full credit, your graph should show the proper slopes and concavities.)
\item[] \emph{Solution:}  The graph should go through $(t,T) = (0,0)$, be increasing 
and concave down for all $t > 0$ and tend to a horizontal asymptote at $T = 70$.
Any graph showing all these properties gets full credit.
\item[B.] (5)  A formula for the temperature as a function of $t = $ time in hours
is $T = 70 - 70 e^{-2t}$.  How long does it take for the temperature to reach $65^\circ$?
\item[] \emph{Solution:}  We must solve for $t$ in $65 = 70 - 70 e^{-2t}$.  
Isolating the $e^{-2t}$ and taking logarithms, this gives:
\begin{align*}
70 e^{-2t} &= 5\\
e^{-2t} & = \frac{1}{14}\\
-2t &= \ln\left(\frac{1}{14}\right)\\
t &= \frac{-\ln\left(\frac{1}{14}\right)}{2} \doteq 1.3 \text{ hours }.
\end{align*}
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=2in,height=2in]{I.ps}
\caption{Figure for Problem I}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=2in,height=2in]{ICS1.ps}
\caption{$y = 2f(x+1)$ in question I C}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=2in,height=2in]{ICS2.ps}
\caption{$y = f(2x)+1$ in question I C}
\end{center}
\end{figure}

\vskip 10pt
\noindent
III.
\begin{enumerate}  
\item[A.]  (5) What is the exact mathematical definition of the 
\emph{derivative} of a function $f(x)$ at $x = a$?  
\item[] \emph{Solution:}  The derivative is 
\[
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h},
\]
provided that the limit exists.
\item[B.]  (10) Use the definition (not the ``shortcut rules'') 
to compute the derivative of 
$f(x) = \sqrt{x}$ at a general $x > 0$.
\item[] \emph{Solution:}  If $f(x) = \sqrt{x}$,
\begin{align*}
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{h}{h(\sqrt{x+h} + \sqrt{x})}\\
&= \lim_{h\to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\\
&= \frac{1}{2\sqrt{x}}.
\end{align*}
\item[C.]  (5) What is the equation of the tangent line to $y = \sqrt{x}$
at the point $(4,2)$?
\item[] From part B, the slope of the tangent line is $f'(4) = \frac{1}{4}$ and 
the equation is $y - 2 = \frac{1}{4}(x - 4)$ by the point-slope form.  The equation
can be rewritten as $y = \frac{1}{4}x + 1$.  
\end{enumerate}
\vskip 10pt
\noindent
IV.  Compute derivatives using the ``shortcut rules'':
\begin{enumerate}
\item[A.]  (5) $f(x) = \sin(x) + x^{5/3} - e^x$.
\item[] \emph{Solution:}  $f'(x) = \cos(x) + \frac{5}{3} x^{2/3} - e^x$.
\item[B.]  (5) $g(x) = \displaystyle{\frac{x^2 - x}{x^4 + 1}}$
\item[] \emph{Solution:}  By the quotient rule, 
\begin{align*}
g'(x) &= \frac{(x^4+1)(2x - 1) - (x^2 - x)(4x^3)}{(x^4+1)^2}\\
&= \frac{-2x^5+3x^4+2x-1}{(x^4+1)^2}.
\end{align*}
\item[C.]  (10) $h(x) = \sin^{-1}(\ln(x))$, and give the domain of $h(x)$.
\item[] \emph{Solution:}  By the derivative rule for inverse sine and the chain rule:
$$
h'(x) = \frac{1}{\sqrt{1 - (\ln(x))^2}}\cdot \frac{1}{x}.
$$
The domain of $h(x)$ is the set of positive $x$ for which $-1 \le \ln(x) \le 1$, 
so $e^{-1} \le x \le e$. 
\item[D.]  (10) Find $\frac{dy}{dx}$ by implicit differentiation if  
$\cos(x + y^2) + \tan^{-1}(x) y = 4$.
\item[]  Using implicit differentiation (with the chain and product rules),
$$
-\sin(x + y^2)(1 + 2yy') + \frac{y}{1+x^2} + \tan^{-1}(x) y' = 0.
$$
Solving for $y' = \frac{dy}{dx}$,
\begin{align*}
y' &= \frac{\sin(x+y^2) - \frac{y}{1+x^2}}{\tan^{-1}(x) - 2y\sin(x+y^2)}\\
&= \frac{(1+x^2)\sin(x+y^2) - y}{(1+x^2)(\tan^{-1}(x) - 2y\sin(x+y^2))},
\end{align*}
(after clearing denominators on the top).
\end{enumerate}
\vskip 10pt
\noindent
V.  (15)  100 cubic inches of cookie dough is being rolled out in a kitchen.
The dough has the shape of a circular cylinder at all times, but the length 
of the cylinder increases and the radius decreases as times goes on.  How fast is the radius
of the dough cylinder changing if the length is increasing at 1 inch per second when
the length is 10 inches?  
\vskip 10pt
\noindent
\emph{Solution:}  The volume of a cylinder is $V = \pi r^2 h$, and both $h$ (the ``length'') 
and $r$ are changing with time $t$.  The volume is constant, though.  So differentiating
with respect to $t$ in the volume equation we have:
\begin{equation}
\label{Vdiff}
0 = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}.
\end{equation}
We are given $\frac{dh}{dt} = 1$ inch per second when $h = 10$.  At that time, 
$100 = 10 \pi r^2$, so $r = \sqrt{10/\pi} \doteq  1.8$ inches.  Substituting,
into (\ref{Vdiff}), we have 
$$
0 = 2\pi \cdot \sqrt{10/\pi} \cdot 10 \cdot \frac{dr}{dt} + \pi \cdot 10/\pi \cdot 1.
$$
So 
$$\frac{dr}{dt} = - \frac{10}{20\sqrt{10}\sqrt{\pi}} = -\frac{1}{2\sqrt{10\pi}} \doteq -.09$$
(the radius is decreasing at about $.09$ inch per second). 

\vskip 10pt
\noindent
{\it Extra Credit} (10)  The graph $y = e^x$ is shifted 1 unit to the left.  Can the same 
transformed graph be obtained by a vertical stretching or shrinking of $y = e^x$?  If you 
say yes, give the stretching or shrinking factor.  If you say no, explain why not.
\vskip 10pt
\noindent
\emph{Solution:}  The answer is \emph{yes}.  Shifting to the left gives $y = e^{x+1} = e\cdot e^x$
by laws of exponents.  So the shifted graph could also be obtained by stretching
vertically by a factor of $e$(!)
\end{document}
