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\centerline{\textbf{\large Exam \#1 solutions $\cdot$ Wednesday, September 17, 2008}}
\vspace*{.2in}
\centerline{MATH 124 $\cdot$ Calculus I $\cdot$ Section 26 $\cdot$ Fall 2008}
\vspace*{0.2in}

\emph{Note:  Some of my solutions are wordy, for the sake of explanation.  All
I expect of you is computation, except for questions which specifically ask for
verbal responses.}

John Kerl (\texttt{kerl.john.r@gmail.com}).

% ----------------------------------------------------------------
\textbf{Problem 1.}
Let
$$
	f(y) = 3y+7
$$
and
$$
	g(y) = y^2 e^{-2y}.
$$

\textbf{Part (a).}
Find $g(f(y))$.

\emph{Solution:}
$$
	(3y+7)^2 e^{-6y-14}.
$$

\textbf{Part (b).}

\emph{Solution:}
$$
y^2 e^{-2y} (3y+7).
$$

% ----------------------------------------------------------------
\textbf{Problem 2.}
Solve for $z$ exactly:
$$
	6 \cdot 3^z = 7 \cdot 2^z.
$$

\emph{Solution:}
\begin{align*}
	6 \cdot 3^z &= 7 \cdot 2^z \\
	\ln(6 \cdot 3^z) &= \ln(7 \cdot 2^z) \\
	\ln(6) + \ln(3^z) &= \ln(7) + \ln(2^z) \\
	\ln(6) + z \ln(3) &= \ln(7) + z \ln(2) \\
	z (\ln(3)-\ln(2)) &= \ln(7) - \ln(6) \\
	z &= \frac{\ln(7) - \ln(6)}{\ln(3)-\ln(2)}.
\end{align*}

% ----------------------------------------------------------------
\textbf{Problem 3.}
An unknown brownish substance weighing 1.5 pounds was placed in your roommate's
refrigerator.  It was then observed to increase in weight by 7 percent per day.
Find the number of days (to the nearest tenth of a day) it will take for the
substance to weigh 6 pounds.

\emph{Solution:}
\begin{align*}
	P(t) &= 1.5 (1.07)^t \\
	6 &= 1.5 (1.07)^t \\
	4 &= 1.07^t \\
	\ln(4) &= \ln(1.07^t) \\
	\ln(4) &= t \ln(1.07) \\
	t &= \frac{\ln(4)}{\ln(1.07)} \approx 20.5.
\end{align*}

% ----------------------------------------------------------------
\textbf{Problem 4.}
The following data represent profit $P$ (in dollars) versus quantity $q$ of
wristwatches sold at a nearby store.
$$
\begin{tabular}{|r|r|}
\hline $q$   & $P$ \\
\hline
\hline $140$ & $45$ \\
\hline $180$ & $75$ \\
\hline $220$ & $105$ \\
\hline $260$ & $135$ \\
\hline
\end{tabular}
$$

\textbf{Part (a).}
Write down a linear equation for $P$ as a function of $q$.  Simplify your work
as much as possible.  Then use the equation to find $P(90)$.

\emph{Solution:}
\begin{align*}
	P &= m(q-a) + b \qquad \textrm{(point-slope form)}\\
	a &= 140,
	b = 45,
	m = \frac{75-45}{180-140} = \frac{3}{4} \\
	P &= \frac{3}{4}(q-140) + 45 \\
	&= \frac{3q}{4} - 105 + 45 \\
	&= \frac{3q}{4} - 60. \\
	P(90) &= \frac{3\cdot 90}{4} - 60 = 67.5 - 60 = 7.5.
\end{align*}

\textbf{Part (b).}
Verbally describe the significance of the horizontal intercept, from a business
point of view.

\emph{Solution:}
A horizontal intercept of a function is an input value which gives zero output.
Here, input is $q$ and output is $P$.  The horizontal intercept is the quantity
$q$ of wristwatches which result in zero profit.  One might think of this as
the minimum quantity needed to break even.

\textbf{Part (c).}
Find the horizontal intercept.

\emph{Solution:}
\begin{align*}
	P &= \frac{3q}{4} - 60 \\
	0 &= \frac{3q}{4} - 60 \\
	\frac{3q}{4} &= 60 \\
	q &= 80.
\end{align*}

\textbf{Part (d).}
Verbally describe the significance of the vertical intercept, from a business
point of view.

\emph{Solution:}
The vertical intercept of a function is the output which results from a
zero input.  Here, $P$ is the profit (which may be negative) obtained when zero
wristwatches are sold.  One may think of this as the overhead cost.

% ----------------------------------------------------------------
\textbf{Problem 5.}
Let
$$
	f(x) = \frac{x^2 - kx + 3}{x-3}.
$$
Find $k$ such that $\lim_{x \to 3} f(x)$ exists.

\emph{Solution:}
As discussed in greater detail in class, we know the denominator goes to
zero at $x=3$.  We want to find $k$ so that the numerator also goes to zero
at $x=3$.
\begin{align*}
	3^2 - 3k + 3 &= 0 \\
	3k &= 12 \\
	k &= 4.
\end{align*}
Using this $k$, we have
\begin{align*}
	f(x) &= \frac{x^2 - 4x + 3}{x-3}
	= \frac{(x-1)(x-3)}{x-3}.
\end{align*}
The graph of this function is the line $y=x-1$ with a hole in it at $x=3$.

I missed an opportunity to make this a two-part question.  Part (a) would be as
above; part (b) would be to find $\lim_{x\to 3} f(x)$.  This would be $x-1$
evaluated at $x=3$, namely, 2.  Even though the function isn't defined at
$x=3$, we can fimd the limiting value of the function as $x$ \emph{approaches}
3 --- which is what limits are all about.

% ----------------------------------------------------------------
\textbf{Problem 6.}
On December 9, 1999 the Mars Polar Lander entered the Martian atmosphere.  The
following is a reconstruction of its altitude $A(t)$ versus time $t$ in
seconds, measured in seconds from the start of the final landing phase.  At an
altitude of 40 meters, the landing rockets switched off prematurely, letting
the lander free-fall to the Martian surface --- after which time the lander was
never heard from again.

\vspace*{0.2in}
\includegraphics{figures/marsfig.eps}
\vspace*{0.2in}

\textbf{Part (a).}
Using the graph, estimate $A(24)$.  Describe verbally what this quantity means.

\emph{Solution:} $A(24) \approx 50$.  At 24 seconds into the final landing
phase, the lander was 50 meters above the Martian surface.

\textbf{Part (b).}
Using the graph, estimate $A^{-1}(20)$.  Describe verbally what this quantity
means.

\emph{Solution:}
$A^{-1}(20) \approx 28.5$.  Since $A$ maps seconds to meters, $A^{-1}$
maps meters (here, 20) to seconds (here, 28.5).  So, the time when the lander
was 20 meters above the surface was 28.5 seconds.

\textbf{Part (c).}
Using the graph, estimate the time of impact.

\emph{Solution:}
$A(t) = 0$ at $t \approx 30$ seconds.

\textbf{Part (d).}
Using the graph, estimate the lander's speed at time of impact.

\emph{Solution:}
Draw a tangent line to the graph at $t=30$ and estimate its slope (rise over
run).  Rise is about $-8$ meters per run of $0.5$ seconds, or $-16$ meters per
second.  (This is not a gentle landing.)

In physics courses, velocity is taken to be signed, and speed is taken to be
unsigned.  That is, speed is absolute value of velocity.  I really wanted you
to do rise over run and get $-16$ m/s, and I should have asked for impact
velocity.  Since I asked for speed, I accepted $16$ m/s as an answer.

A few of you said that the speed at impact was 0 meters per second.  Based on
the way I worded the question, I felt compelled to give you full credit.  What
I was looking for, though, was the impact speed.

% 40 m:  2.4 m/s
% Then free fall at 3.7 m/s^2

% A(t) = at^2 + bt + c on both pieces: before and after 26.

% Before 26:
% A(20) = 100
% A(26) =  40
% A'(26) = -2.5  (rocket-assisted descent)

% After 26:
% A(26) = 40     (height continuity)
% A'(26) = -2.5  (velocity continuity)
% A''(26) = -3.7 (Martian free-fall)

% Solve for a,b,c on both pieces.  Results are in figures/marsfig.m.

(See the end of this document if you are interested to know where
this graph came from.)

% ----------------------------------------------------------------
\textbf{Problem 7.}
Invert the following function:
$$
	f(x) = C a^x.
$$
(You may assume $C>0$ and $a>0$.)

\emph{Solution:}
\begin{align*}
	x &= C a^y \\
	\ln(x) &= \ln(C a^y) \\
	\ln(x) &= \ln(C) + \ln(a^y) \\
	\ln(x) &= \ln(C) + y\ln(a) \\
	y \ln(a) &= \ln(x) - \ln(C) \\
	y &= \frac{\ln(x) - \ln(C)}{\ln(a)} \\
	f^{-1}(x) &= \frac{\ln(x) - \ln(C)}{\ln(a)}\
\end{align*}

% ----------------------------------------------------------------
\textbf{Problem 8.}
Evaluate the following limit, using algebraic properties of limits:
$$
	\lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - x^2 - 2x}{h}.
$$

\emph{Solution:}
\begin{align*}
	\lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - x^2 - 2x}{h}
	&= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 2x+ 2h - x^2 - 2x}{h} \\
	&= \lim_{h \to 0} \frac{2xh + h^2 + 2h}{h} \\
	&= \lim_{h \to 0} 2x + h + 2 \\
	&= 2x + 2.
\end{align*}

% ----------------------------------------------------------------
\textbf{Problem 9.}
Numerically estimate the following limit to 3 decimal places:
$$
	\lim_{\theta \to 0} \frac{2\theta}{\sin(3\theta)}.
$$

\emph{Solution:}
Graphing and zooming in repeatedly, one should find $0.667$ to three
decimal places.  (The exact answer turns out to be $2/3$, as we will
see later in the course when we learn about l'H\^opital's rule.)

%% ================================================================
\newpage

\textbf{Side note:}  How the graph for question 6 was obtained.  This is an
illustration of using derivatives to solve for something interesting.  My
technique uses second derivatives, so this should make sense after we discuss
section 2.5.

I read the Wikipedia article on the Mars Polar Lander incident and gathered the
following facts:

\begin{itemize}

\item The final landing phase (backshell separation) was at 1400 meters
altitude, and at that point the lander's velocity was $-80$ m/s.

\item The landing rockets turned off at 40 meters altitude.  (They should not
have --- hence the splat.  Apparently a vibration sensor on the landing legs
detected vibrations due to descent, and misinterpreted them as having been due
to the legs touching the ground.)

\item I wanted the 40-meter incident to be visible on the graph so I did a
rough sketch of altitude vs. time, from 1400 meters down to zero.  Then I kept
the part with altitude from 100 meters down to zero.  Altitude 100 meters
occurred at about 20 seconds.

\item The lander's velocity at 12 meters altitude should have been $-2.4$
meters/second, had the landing rockets not turned off early.

\item I arbitrarily used $-2.5$ m/s as the lander's velocity at 40 meters.

\item Under constant acceleration, the height of the lander should be $at^2 +
bt + c$.  There are two quadratic polynomials, one for each piece of a
piecewise function:  one for the period when the landing rockets were on, so
the net acceleration (landing rockets minus gravity) was positive, and another
for the free-fall period when only gravity was involved.  That is,
$$
	A(t) = \begin{cases}
		a_1 t^2 + b_1 t + c_1, & 0 \le t \le 26; \\
		a_2 t^2 + b_2 t + c_2, & 26 \le t \le \textrm{impact time}.
	\end{cases}
$$
(You didn't need to know these equations to do the exam problem --- you just
used the graph they produced.)

\item Mars acceleration due to gravity (i.e. $g$) is $-3.7$ m/s${}^2$.
(Compare this to $-9.8$ m/s${}^2$ on Earth.  I'd weigh 55 lbs. on Mars compared
to 145 lbs. on Earth.  I'd feel unusually light, but a long fall would still
hurt.)

\end{itemize}

Summarizing, I had the following data points.

Before 26 seconds:
\begin{itemize}
\item $A(20) = 100$.
\item $A(26) = 40$.
\item $A'(26) = -2.5$ m/s (velocity at 40 meters).
\end{itemize}
After 26 seconds:
\begin{itemize}
\item $A(26) = 40$ (continuity of height from the other piece of the piecewise function).
\item $A'(26) = -2.5$ m/s (continuity of velocity from the other piece of the piecewise function).
\item $A''(26) = -3.7$ m/s${}^2$ (acceleration due to Martian gravity).
\end{itemize}

In both cases, there are three equations in three unknowns.  So we can solve
for them.

Note that we have
\begin{align*}
	A(t)   &=   a_1 t^2 + b_1 t + c_1 \\
	A'(t)  &= 2 a_1 t   + b_1  \\
	A''(t) &= 2 a_1
\end{align*}
for the first piece, and similarly for the second piece (replacing $a_1$,
$b_1$, and $c_1$ with $a_2$, $b_2$, and $c_2$).

%Before 26 seconds:
%\begin{align*}
%	A(20)  &= 100:  & 400 a_1 + 20 b_1 + c_1 &= 100. \\
%	A(26)  &=  40:  & 676 a_1 + 26 b_1 + c_1 &=  40. \\
%	A'(26) &= -2.5: &  52 a_1 +    b_1       &= -2.5.
%\end{align*}
%After 26 seconds:
%\begin{align*}
%	A(26)   &=   40:  & 676 a_2 + 26 b_2 + c_2 &=  40. \\
%	A'(26)  &= -2.5:  &  52 a_2 + b_2          &= -2.5. \\
%	A''(26) &= -3.7: &    2 a_2                &= -3.7.
%\end{align*}

Before 26 seconds:
\begin{align*}
	\begin{array}{rrr} A(20)  \\ A(26) \\ A'(26) \\ \end{array}
	\begin{array}{ccc} =      \\ =     \\ =      \\ \end{array}
	\begin{array}{rrr} 100:   \\ 40:   \\ -2.5:  \\ \end{array}
%
	\begin{array}{rrr} 400a_1 \\ 676a_1 \\ 52a_1 \\ \end{array}
	\begin{array}{ccc} +      \\ +      \\ +     \\ \end{array}
	\begin{array}{rrr} 20b_1  \\ 26b_1  \\ b_1   \\ \end{array}
	\begin{array}{ccc} +      \\ +      \\ {}    \\ \end{array}
	\begin{array}{rrr} c_1    \\ c_1    \\ {}    \\ \end{array}
	\begin{array}{ccc} =      \\ =      \\ =     \\ \end{array}
	\begin{array}{rrr} 100    \\ 40     \\ -2.5  \\ \end{array}
\end{align*}
After 26 seconds:
\begin{align*}
	\begin{array}{rrr} A(26)  \\ A'(26) \\ A''(26) \\ \end{array}
	\begin{array}{ccc} =      \\ =      \\ =       \\ \end{array}
	\begin{array}{rrr} 40:    \\ -2.5:  \\ -3.7:   \\ \end{array}
%
	\begin{array}{rrr} 676a_2 \\ 52a_2  \\  2a_2   \\ \end{array}
	\begin{array}{ccc} +      \\ +      \\ {}      \\ \end{array}
	\begin{array}{rrr} 26b_2  \\   b_2  \\ {}      \\ \end{array}
	\begin{array}{ccc} +      \\ {}     \\ {}      \\ \end{array}
	\begin{array}{rrr} c_2    \\ {}     \\ {}      \\ \end{array}
	\begin{array}{ccc} =      \\ =      \\ =       \\ \end{array}
	\begin{array}{rrr} 40     \\ -2.5   \\ -3.7    \\ \end{array}
\end{align*}

Eliminating variables, I found
\begin{align*}
a_1 &= 1.25, &
b_1 &= -67.5, &
c_1 &= 950; \\
a_2 &= -1.85, &
b_2 &=  93.7, &
c_2 &= -1145.5.
\end{align*}

Since $2a_1$ and $2a_2$ are the accelerations, we find that the free-fall
acceleration was $-3.7$ m/s${}^2$, as expected, and the acceleration during
rocket-assisted descent was $2.5$ m/s${}^2$.  This is acceleration due to
gravity and landing rockets, so we can estimate that the landing rockets
themselves were contributing an acceleration of $6.2$ m/s${}^2$.

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