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\begin{document}
\noindent
\ifsolutions
\begin{center}
\textbf{\Large{Math 251\\ \mbox{} \\
Final Exam - Fall 2015 - Solutions}}
\end{center}
\else
\begin{center}
\textbf{\Large{Math 251\\ \mbox{} \\
Final Exam - Fall 2015 - Problem List}}
\end{center}
\fi
\bigskip
\bigskip
Note: The wording of the problems here is not identical to that
on the exam as administered.
The total number of points was $100$,
so the point values are about half of what similar questions
would be on a midterm.
% \ifsolutions
% Also, for some problems several solutions are given
% (especially if students tried different methods),
% and a few comments on grading are given.
% \fi
\bigskip
\ifsolutions
{\textbf{Essentially no proofreading has been done!}}
\fi
\bigskip
\bigskip
\noindent
1. {(8 points; point values of parts as shown)}
A small spacecraft takes off from the surface of a
planet, reaches a maximum height, and then crashes.
Its position at time $t$ is given by $y (t) = 9 t^2 - 4t^3$,
where $y (t)$ is measured in kilometers (km) above the surface
and $t$ is measured in minutes (min).
Answer the following questions, being careful to give correct units
when called for.
\bigskip
(a) {(2 points)}
Find the speed of the spacecraft at time $t = 1$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Calculate $y' (t) = 18 t - 12 t^2$,
so the speed is $y' (1) = 18 - 12 = 6$.
Thus, the answer is $6 {\mbox{km/min}}$.
(The units are {\emph{required}}.)
\fi
\bigskip
(b) {(2 points)}
How long will it take for the spacecraft to reach its maximum height?
\ifsolutions
\bigskip
{\emph{Solution:}}
First find the maximum height,
which occurs when $y' (t) = 0$.
We have $y' (t) = 18 t - 12 t^2$,
so solve:
\[
18 t - 12 t^2 = 0
\]
\[
6 t (2 - 2 t) = 0
\]
\[
t = 0
\,\,\,\,\,\, {\mbox{or}} \,\,\,\,\,\,
t = \frac{3}{2}.
\]
We are told that the spacecraft starts out by rising,
reaches a maximum height, and then crashes,
to the answer is after $\frac{3}{2} {\mbox{min}}$.
(This can also be checked by applying the methods we have learned
to the function $y (t) = 9 t^2 - 4t^3$.)
\fi
\bigskip
(c) {(4 points)}
At time $t = 2$, what is the acceleration of the spacecraft?
Would a person in the craft feel himself speeding up or slowing
down at this moment?
Explain.
\ifsolutions
\bigskip
{\emph{Solution:}}
Calculate $y' (t) = 18 t - 12 t^2$,
so $y'' (t) = 18 - 24 t$,
and $y'' (2) - 18 - 48 = - 30$.
The numerical answer is therefore
$- 30 {\mbox{km/min$^2$}}$.
(The units are {\emph{required}}.)
The acceleration is negative,
and the velocity is $y' (2) = 18 \cdot 2 - 12 \cdot 2^2 = - 12$,
so the spacecraft is falling fast as time passes.
Thus, the person feels himself speeding up towards the ground.
Physically, strictly speaking,
this is ``slowing down'',
since position is measured in distance above the ground.
\fi
\bigskip
\bigskip
\noindent
2. {(6 points)}
The derivative of the function
$f (x) = ( x^3 + 3 x^2 + 3 x + 5 ) e^{- x}$
is given by
$f' (x) = - (x + 2) (x - 1)^2 e^{- x}$.
Find the critical
numbers of $f$ and for each one determine if it is a local minimum,
local maximum, or neither.
\ifsolutions
\bigskip
{\emph{Solution:}}
We find critical numbers by solving $f' (x) = 0$.
This happens when $x + 2 = 0$, when $(x - 1)^2 = 0$,
and when $e^{- x} = 0$.
Since $e^{- x}$ is never zero,
the critical numbers are at $x = - 2$ and $x = 1$.
To decide which critical numbers are local minimums or maximums,
we consider the factors of $f' (x)$:
\begin{itemize}
\item
On $(- \infty, \, - 2)$,
$-1 < 0$, $x + 2 < 0$, $(x - 1)^2 > 0$, and $e^{- x} > 0$,
so $f' (x) > 0$
and $f$ is nondecreasing.
\item
On $(- 2, \, 1)$,
$-1 < 0$, $x + 2 > 0$, $(x - 1)^2 > 0$, and $e^{- x} > 0$,
so $f' (x) < 0$
and $f$ is nonincreasing.
\item
On $(1, \infty)$,
$-1 < 0$, $x + 2 > 0$, $(x - 1)^2 > 0$, and $e^{- x} > 0$,
so $f' (x) < 0$
and $f$ is nonincreasing.
\end{itemize}
So $f$ has a local maximum at $x = - 2$,
but neither a local maximum nor a local minimum at $x = 1$.
\bigskip
Note:
The second derivative test doesn't help much here.
A calculation gives
$f'' (x) = (x - 1) (x^2 - 2 x - 5) e^{- x}$.
So $f'' (-2) = - 9 e^2 < 0$,
showing that there is a local maximum at $x = - 2$,
but $f'' (1) = 0$,
which doesn't help.
\fi
\bigskip
\bigskip
\noindent
3. {(8 points)}
A pendulum swings back and forth on the surface of the
earth.
The relation between the period $T$ and the length $l$ of the
pendulum can be modelled by
the equation $T = \sqrt{k l}$, where $T$ is measured in seconds,
$l$ is measured in meters, and $k = 4 \, {\text{s}^2}/{\text{m}}$.
Answer the following questions, being careful to include units when
appropriate.
\bigskip
(a) {(1 point)}
When $l$ is 25 meters, compute the period of the pendulum.
\ifsolutions
\bigskip
{\emph{Solution:}}
At $l = 25$,
we get
$T = \sqrt{(4 \, {\text{s}}^2/{\text{m}}) \cdot (25 \, {\text{m}})}
= \sqrt{100 \, {\text{s}}^2}
= 10 \, {\text{s}}$.
So the period is $10 \, {\text{s}}$.
(The units are {\emph{required}}.)
\fi
\bigskip
(b) {(4 points)}
Using part (a) as your base value, give a linear approximation
for the period of the pendulum when the length $l$ is changed to $27$
meters.
\ifsolutions
\bigskip
{\emph{Solution:}}
It is better to use function notation:
$T (l) = \sqrt{4 l} = 2 \sqrt{l} = 2 t^{1/2}$.
Now we do the linear approximation.
We have
$T' (l) = 2 \big( \frac{1}{2} t^{- 1/2} = t^{- 1/2}$.
So
\[
T (27)
\approx T (25) + T' (25) (27 - 25)
= 10 + \frac{1}{\sqrt{25}} (2)
= 10 + \frac{2}{5}.
\]
So the estimated is $10{.}4 \, {\text{s}}$.
(The units are {\emph{required}}.)
\fi
\bigskip
(c) {(3 points)}
Suppose you do a linear approximation for $l = 205$ meters,
based on the starting point $l = 200$.
Is the linear approximation greater or less than the actual
period?
Explain (perhaps using a picture).
\ifsolutions
\bigskip
{\emph{Solution:}}
We have $T'' (l) = - \frac{1}{2} t^{- 3/2}$.
This is negative for all $l$ in the domain,
so the graph of $T$ is concave down,
and hence below its tangent lines.
Therefore the linear approximation is greater than the actual
period.
(Picture missing.)
% ???
\fi
\bigskip
\bigskip
\noindent
4. {(4 points)}
Let $f$ and $g$ be functions such that:
\[
g (1) = 2, \,\,\,\,\,\, g' (1) = 1, \,\,\,\,\,\, f (1) = 2,
\andeqn f' (1) = 6
\]
and
\[
g (2) = 1, \,\,\,\,\,\, g' (2) = 5, \,\,\,\,\,\, f (2) = 3,
\andeqn f' (2) = - 1.
\]
Let $h (x) = f (g (x))$.
Find $h' (2)$.
(You will not need to use all the information provided.)
\ifsolutions
\bigskip
{\emph{Solution:}}
Using the chain rule,
\[
h' (2) = f' (g (2)) g' (2) = f' (1) \cdot 5 = 6 \cdot 5 = 30.
\]
\fi
\bigskip
\bigskip
\noindent
5. {(11 points; point values of parts as shown)}
Air Krill runs daily flights from the Eugene Airport to the
Falkland Islands.
When the company charges $\$400$ per
ticket, $24$ people fly.
Market research has shown each $\$10$ that the
fare is increased, $1$ fewer person flies.
Conversely, for each $\$10$ the
fare is lowered, $1$ more person flies.
\bigskip
(a) {(4 points)}
Let $c$ denote the price charged for each ticket.
(``$c$'' stands for ``cost''.)
Let $T (c)$ be
the number of tickets sold at that price,
and let $R (c)$ be the revenue at that price.
Write down formulas for $T (c)$ and $R (c)$.
\ifsolutions
\bigskip
{\emph{Solution:}}
We have
\[
T (c)
= 24 + \left( - \frac{1}{10} \right) (c - 400)
= 64 - \frac{c}{10}.
\]
Therefore
\[
R (c)
= c T (c)
= c \left( 64 - \frac{c}{10} \right)
= 64 c - \frac{c^2}{10}.
\]
\fi
\bigskip
(b) {(3 points)}
How much should Air Krill charge per ticket to maximize total revenue?
\ifsolutions
\bigskip
{\emph{Solution:}}
We want to maximize $R (c)$
We need to determine the appropriate interval.
We need to have $c \geq 0$.
Also, the number of passengers must be nonnegative,
so $64 - \frac{c}{10} \geq 0$,
so $c \leq 640$.
So we need to maximize the function
$R (c) = 64 c - \frac{c^2}{10}$
over the interval $[0, \, 640]$.
Compute:
\[
R' (c)
= 64 - \frac{c}{5}.
\]
This is zero when $c = 320$.
To verify that this is a maximum,
the second derivative test is easiest.
Calculate:
$R'' (c) = - \frac{1}{5}$,
which is negative everywhere,
so any critical number is a global maximum.
Therefore revenue is maximized at a ticket cost of $\$320$.
\bigskip
Next easiest is the first derivative test.
Factor:
\[
R' (c) = \frac{320 - c}{5},
\]
which is positive for $c < 320$ and negative for $c > 320$.
So any critical number is a global maximum.
Therefore revenue is maximized at a ticket cost of $\$320$.
\bigskip
For the closed interval method, compare:
\[
R (0) = 0,
\]
\[
R (320)
= 64 (320) - \frac{320^2}{10}
= \frac{640 \cdot 320 - 320 \cdot 320}{10}
= \frac{320^2}{10}
> 0,
\]
and
\[
R (640) = 64 (640) - \frac{640^2}{10} = 0.
\]
Therefore revenue is maximized at a ticket cost of $\$320$.
\fi
\bigskip
(c) {(2 points)}
Suppose that the Air Krill airplanes can only seat $30$ people.
Write down an inequality involving $T (c)$ that represents this, and
deduce a corresponding inequality for $c$.
\ifsolutions
\bigskip
{\emph{Solution:}}
The statement says $T (c) \leq 30$.
Thus:
\[
64 - \frac{c}{10} \leq 30
\]
\[
\frac{c}{10} \geq 34
\]
\[
c \geq 340.
\]
\fi
\bigskip
(d) {(2 points)}
Does the condition in (c) change the answer for (b), and if so what is
the new answer?
Explain.
\ifsolutions
\bigskip
{\emph{Solution:}}
We must now maximize $R (c)$ over the interval
$[340, \, 640]$.
The function has no critical numbers in this interval,
so the maximum must occur at an endpoint.
The easiest method is to observe,
as in the second approach to part~(b),
that $R' (c) < 0$
on $[340, \, 640]$,
so the maximum must be at $c = 340$.
\bigskip
The alternative method is to
compare values at the endpoints:
\[
R (340)
= 64 (340) - \frac{340^2}{10}
= \frac{640 \cdot 340 - 340 \cdot 340}{10}
= \frac{300 \cdot 320}{10}
> 0,
\]
and $R (640) = 0$ (as in part~(b)).
Since $R (340) > R (640)$,
revenue is maximized at $c = 340$.
\fi
\bigskip
\bigskip
\noindent
6. {(3 points/part)}
Find the exact values of the following limits (possibly
including $\infty$ or $-\infty$), or explain why they do not exist.
Show supporting work.
\bigskip
(a)
${\displaystyle{ \lim_{x \to 0} \, \frac{x^4}{e^{7 x^4} - 1} }}$.
\ifsolutions
\bigskip
{\emph{Solution~1:}}
The limit has the indeterminate form
``${\displaystyle{ \frac{0}{0} }}$''.
Therefore we may use L'Hospital's Rule.
The chain rule gives
${\displaystyle{ \frac{d}{dx} \left( e^{7 x^4} \right) }}
= 24 x^3 e^{7 x^4}$,
so
\[
\lim_{x \to 0} \frac{x^4}{e^{7 x^4} - 1} =
\lim_{x \to 0} \frac{4 x^3}{28 x^3 e^{7 x^4}},
\]
if the second limit exists.
To evaluate the second limit,
cancel $4 x^3$ from the numerator and denominator, getting:
\[
\lim_{x \to 0} \frac{4 x^3}{28 x^3 e^{7 x^4}}
= \lim_{x \to 0} \frac{1}{7 e^{7 x^4}} = \frac{1}{7 e^{7 \cdot 0^4}}
= \frac{1}{7 e^0} = \frac{1}{7}.
\]
Therefore
\[
\lim_{x \to 0} \frac{x^4}{e^{7 x^4} - 1} = \frac{1}{7}.
\]
Note: If you don't cancel the factors $4 x^3$, the limit still has
the indeterminate form ``${\displaystyle{ \frac{0}{0} }}$''
You can therefore use L'Hospital's Rule again.
To get an answer this way,
you will in fact need to use L'Hospital's Rule
a total of $3$ more times, which will involve computing the 4th
derivative of $e^{7 x^4}$.
This is quite messy, so messy that I will not give details here.
\bigskip
{\emph{Solution~2:}}
We note that $x^4 \to 0^+$ as $x \to 0$.
Substituting $h = x^4$, we therefore get
\[
\lim_{x \to 0} \frac{x^4}{e^{7 x^4} - 1}
= \lim_{h \to 0^+} \frac{h}{e^{7 h} - 1}
= \frac{1}{ \lim_{h \to 0^+} \frac{e^{7 h} - 1}{h} }.
\]
With $f (t) = e^{7 t}$, the denominator in this last expression is
\[
\lim_{h \to 0^+} \frac{e^{7 h} - 1}{h}
= \lim_{h \to 0^+} \frac{f (h) - f (0)}{h} = f' (0).
\]
We know $f' (t) = 7 e^{7 t}$, so the limit is $7 e^0 = 7$.
Therefore
\[
\lim_{x \to 0} \frac{x^4}{e^{7 x^4} - 1} = \frac{1}{7}.
\]
Note: We could also evaluate
${\displaystyle{ \lim_{h \to 0} \frac{e^{7 h} - 1}{h} }}$
using L'Hospital's Rule (once).
\fi
\bigskip
(b)
${\displaystyle{ \lim_{x \to 1} \, \frac{x - 2}{x^2 - x - 6} }}$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Both the numerator and denominator are continuous at $1$,
and the denominator is not zero there.
Therefore the limit can be evaluated by simply substituting $x = 1$.
That is,
\[
\lim_{x \to 1} \frac{x - 2}{x^2 - x - 6}
= \frac{1 - 2}{1^2 - 1 - 6} = \frac{1}{6}.
\]
Note that L'Hospital's Rule doesn't apply, since the limit does not
have an indeterminate form.
If you try to use it anyway, you get
\[
\lim_{x \to 1} \frac{1}{2 x - 1} = 1,
\]
which is the wrong answer.
\fi
\bigskip
(c)
${\displaystyle{ \lim_{x \to 0} \, \frac{x^2}{\cos(2 x) - 1} }}$.
\ifsolutions
\bigskip
{\emph{Solution:}}
The limit has the indeterminate form
``${\displaystyle{ \frac{0}{0} }}$''.
Therefore we may use L'Hospital's Rule.
The chain rule gives
${\displaystyle{ \frac{d}{dx} \left( \cos(2 x) - 1 \right) }}
= - 2 \sin (2 x)$,
so
\[
\lim_{x \to 0} \frac{x^2}{\cos(2 x) - 1}
= \lim_{x \to 0} \frac{2 x}{- 2 \sin (2 x)},
\]
if the second limit exists.
This limit also has the indeterminate form
``${\displaystyle{ \frac{0}{0} }}$''.
Therefore we may use L'Hospital's Rule again.
Using the chain rule again,
we get
\[
\lim_{x \to 0} \frac{2 x}{- 2 \sin (2 x)}
= \lim_{x \to 0} \frac{2}{- 4 \cos (2 x)},
\]
if the second limit exists.
Clearly
\[
\lim_{x \to 0} \frac{2}{- 4 \cos (2 x)}
= \frac{2}{- 4 \cos (2 \cdot 0)}
= - \frac{1}{2}.
\]
So
\[
\lim_{x \to 0} \frac{x^2}{\cos(2 x) - 1} - \frac{1}{2}.
\]
\fi
\bigskip
(d)
${\displaystyle{ \lim_{x \to \infty} \frac{3 x + 1}{5 x^2 - 9} }}$.
\ifsolutions
\bigskip
{\emph{Solution:}}
This has the indeterminate form ``$\frac{\infty}{\infty}$'',
so work is needed.
We factor out $x^2$ from both the numerator and denominator,
and then use the limit laws:
\begin{align*}
\lim_{x \to \infty} \frac{3 x + 1}{5 x^2 - 9}
& = \lim_{x \to \infty}
\frac{x^2 \left(\frac{3}{x} + \frac{1}{x^2} \right)}
{x^2 \left(5 - \frac{9}{x^2} \right)}
= \lim_{x \to \infty}
\frac{\frac{3}{x} + \frac{1}{x^2}}
{5 - \frac{9}{x^2}} \\
& = \frac{\lim_{x \to \infty} \frac{3}{x}
+ \lim_{x \to \infty} \frac{1}{x^2}}
{5 - \lim_{x \to \infty} \frac{9}{x^2}}
= \frac{0 + 0}{5 - 0}
= 0.
\end{align*}
\bigskip
{\emph{Alternate solution:}}
Here is a different way to arrange essentially the same calculation:
\begin{align*}
\lim_{x \to \infty} \frac{3 x + 1}{5 x^2 - 9}
& = \lim_{x \to \infty}
\frac{\left( \frac{1}{x^2} \right) \left( 3 x + 1 \right)}
{\left( \frac{1}{x^2} \right) \left( 5 x^2 - 9 \right)}
= \lim_{x \to \infty}
\frac{\frac{3}{x} + \frac{1}{x^2}}
{5 - \frac{9}{x^2}} \\
& = \frac{\lim_{x \to \infty} \frac{3}{x}
+ \lim_{x \to \infty} \frac{1}{x^2}}
{5 - \lim_{x \to \infty} \frac{9}{x^2}}
= \frac{0 + 0}{5 - 0}
= 0.
\end{align*}
\bigskip
{\emph{Second alternate solution.}}
(This solution can only been used after L'Hopital's rule
has been treated in the course.)
We have $\lim_{x \to \infty} (3 x + 1) = \infty$
and $\lim_{x \to \infty} (5 x^2 - 9) = \infty$.
The limit above therefore has the indeterminate form
``$\frac{\infty}{\infty}$'', so more work is needed.
In this case, L'Hospital's Rule applies.
We get:
\[
\lim_{x \to \infty} \frac{3 x + 1}{5 x^2 - 9}
= \lim_{x \to \infty} \frac{3}{5 x},
\]
provided the last limit exists.
This limit is zero because
$\lim_{x \to \infty} 3 = 3$
and $\lim_{x \to \infty} 5 x = \infty$.
Therefore
\[
\lim_{x \to \infty} \frac{3 x + 1}{5 x^2 - 9}
= 0.
\]
\fi
\bigskip
\bigskip
\noindent
7. {(7 points)}
At noon Horton is $10$ miles due west of Gertrude.
Horton walks north and Gertrude walks west;
neither walks at a steady rate.
Both have devices
that can continuously measure the distance between them.
After one hour, Horton has walked $3$ miles
and Gertrude has walked $6$ miles.
Horton's device tells him that he is at that moment
walking at $5$ miles per hour,
and the distance between him and Gertrude
is increasing at a rate of $2$ miles per hour.
How fast is Gertrude walking at this moment?
Include units in your answer.
\ifsolutions
\bigskip
{\emph{Solution:}}
Note: There is no picture in this file.
% A picture may be provided separately.
Let $h (t)$ be the distance (in miles) Horton has walked
north at time~$t$
(with time measured in hours), and let $g (t)$ be the
distance (in miles) Gertrude has walked west at time~$t$.
Let $t = 0$ represent noon on the day in question.
Let $z (t)$ be the distance (also in miles) between
Horton and Gertrude at time~$t$.
Then the information given says that:
\[
h (1) = 3, \,\,\,\,\,\, h' (1) = 5, \,\,\,\,\,\,
g (1) = 6, \andeqn z' (1) = 2.
\]
% (See the end of the solution for other possible choices
% for measuring distance.)
We want to find $g' (1)$.
At time~$t$,
So the Pythagorean Theorem gives
$z (t)^2 = h (t)^2 + [10 - g (t)]^2$.
Differentiating, we get
\[
2 z (t) z' (t) = 2 h (t) h' (t) + 2 [10 - g (t)] [- g' (t)].
\]
(Don't forget to use the chain rule!)
Put $t = 1$ and (for simplicity) divide by $2$:
\[
z (1) z' (1) = h (1) h' (1) + [10 - g (1)] [- g' (1)].
\]
Next, substitute values.
(Note that this can only be done {\emph{after}} differentiating!)
We need
\[
z (1) = \sqrt{h (0)^2 + [10 - g (t)]^2} = \sqrt{3^2 + (10 - 6)^2} = 5,
\]
and we then get
\[
5 \cdot 2 = 3 \cdot 5 + (10 - 6) \cdot (- g' (1)).
\]
So
\[
10 = 15 - 4 g' (1),
\]
and
\[
g' (1) = \frac{10 - 15}{- 4} = \frac{5}{4}.
\]
So Gertrude is walking west at $\frac{5}{4}$ miles per hour.
(The units are necessary!)
\bigskip
Here are descriptions of some alternatives.
First, one could solve for $g (t)$.
This is messy (involving square roots),
and the details are omitted.
\bigskip
Second, one could measure distances differently.
If one imagines the Horton's location at noon to be the
origin,
with Horton on the $y$-axis and Gertrude on the $x$-axis,
then it is natural to let $y (t)$ be how far Horton has gone north
and let $x (t)$ be how far east Gertrude is from that point.
With these conventions,
the distance between them is given by $z (t)^2 = x (t)^2 + y (t)^2$,
and
the information given says that:
\[
x (0) = 4, \,\,\,\,\,\, y (0) = 3, \,\,\,\,\,\,
y' (0) = 5, \andeqn z' (0) = 2.
\]
Differentiation
gives $2 z (t) z' (t) = 2 x (t) x' (t) + 2 y (t) y' (t)$.
The algebra will be a bit different through the rest of the
calculation, but the final result will be the same.
\bigskip
Finally, you could do everything in physicists' notation.
I will only show the first version.
The equation relating the quantities is
$z^2 = h^2 + [10 - g]^2$.
Differentiating (using the chain rule, because
everything is a function of~$t$!), we get
\[
2 z \frac{dz}{dt}
= 2 h \frac{d h}{d t} + 2 (10 - g) \left( - \frac{d g}{d t} \right),
\]
so
\[
z \frac{dz}{dt} = h \frac{d h}{d t} - (10 - g) \frac{d g}{d t}.
\]
Substituting values (implicitly putting $t = 1$, and using
$z = 5$ at $t = 1$, as above):
\[
5 \cdot 2 = 3 \cdot 5 - (10 - 6) \cdot \frac{d g}{d t}.
\]
So ${\displaystyle{\frac{d g}{d t} = \frac{5}{4}}}$.
\fi
\bigskip
\bigskip
\noindent
8. {(6 points)}
Find the equation of the tangent line to the curve
$x^2 + x y + y^2 = 3$ at the point $(1, 1)$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Let's write it with $y$ as an explicit function $y (x)$ of~$x$:
\[
x^2 + x y (x) + y (x)^2 = 3.
\]
Differentiate both sides with respect to~$x$,
using the chain rule on the last term on the left:
\[
2 x + y (x) + x y' (x) + 2 y (x) y' (x) = 0.
\]
Now substitute $x = 1$ and $y (x) = 1$:
\[
2 \cdot 1 + 1 + 1 \cdot y' (x) + 2 \cdot 1 \cdot y' (x) = 0.
\]
\[
3 + 3 y' (x) = 0
\]
\[
y' (x) = - 1.
\]
This is the slope of the tangent line.
Therefore the equation of the tangent line is
\[
y - 1 = (- 1) (x - 1),
\]
which is simplified to $y = - x + 2$.
\bigskip
For those who prefer the other notation, here it is written with
${\displaystyle{ \frac{d y}{d x} }}$.
Differentiate with respect to~$x$ as before:
\[
2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0.
\]
Now substitute $x = 1$ and $y = 1$:
\[
2 \cdot 1 + 1 + 1 \cdot \frac{d y}{d x}
+ 2 \cdot 1 \cdot \frac{d y}{d x}
= 0.
\]
\[
3 + 3 \frac{d y}{d x} = 0
\]
\[
\frac{d y}{d x} = - 1.
\]
This is the slope of the tangent line.
Therefore the equation of the tangent line is
\[
y - 1 = (- 1) (x - 1),
\]
which is simplified to $y = - x + 2$.
\fi
\bigskip
\bigskip
\noindent
9. {(4 points/part)}
In each part below, find the derivative of the given
function.
\bigskip
(a)
$f (x) = \sin (x^2) + (\sin (x))^2$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Use the chain rule on both terms,
getting
\[
f' (x)
= \cos (x^2) \frac{d}{d x} \big( x^2 )
+ 2 \sin (x) \frac{d}{d x} \big( \sin (x) )
= 2 x \cos (x^2) + 2 \sin (x) \cos (x).
\]
\fi
\bigskip
(b)
$g (x) = \tan (5 x) + \ln (7) x^3 + \pi^2$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Use the chain rule on the first term,
getting
\[
g' (x)
= \sec^2 (5 x) \cdot 5 + \ln (7) \cdot 3 x^2
= 5 \sec^2 (5 x) + 3 \ln (7) x^2.
\]
(The expression $\pi^2$ is a {\emph{constant}},
so its derivative is zero.)
\fi
\bigskip
(c)
$q (t) = e^{3 t} \sqrt{t}$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Rewrite as $q (t) = e^{3 t} t^{1/2}$.
Use the product rule,
using the chain rule on the first factor:
\[
q' (t)
= \frac{d}{d t} \big( e^{3 t} ) t^{1/2}
+ e^{3 t} \frac{d}{d t} \big( t^{1/2} )
= 3 e^{3 t} t^{1/2} + e^{3 t} \left( \tfrac{1}{2} \right) t^{-1/2}.
\]
You can (but don't have to) rearrange this more nicely as
\[
q' (t)
= t^{-1/2} e^{3 t} \left( 3 t + \tfrac{1}{2} \right).
\]
\fi
\bigskip
(d)
$h (s) = {\displaystyle{ \frac{7 s + 3}{3 s - 7} }}$.
\ifsolutions
\bigskip
{\emph{Solution:}}
Use the quotient rule,
getting:
\[
h' (s)
= \frac{ \frac{d}{d s} \big( 7 s + 3 \big) (3 s - 7)
- ( 7 s + 3 ) \frac{d}{d s} \big( 3 s - 7 \big)}{(3 s - 7)^2}
= \frac{ 7 (3 s - 7) - ( 7 s + 3 ) \cdot 3}{(3 s - 7)^2}
= \frac{- 58}{(3 s - 7)^2}.
\]
\fi
\bigskip
\bigskip
\noindent
10. {(2 points/part)}
The picture below is the graph of the {\emph{DERIVATIVE}}
$y = f' (x)$ for a certain function $f$.
{\textbf{CAUTION:}} You are given the graph of the {\emph{derivative}}
$f' (x)$, {\emph{not}} the graph of $f (x)$, but you are asked
questions about $f (x)$.
The points referred to in parts~(a) and (b) are marked on
the graph with dots.
\vspace{1ex}
\centerline{
%\includegraphics[width=12cm][scale=0.5]{M2Graph}
\includegraphics[scale=0.5]{M2Graph}
}
\vspace{1ex}
% {\epsfxsize=12cm \epsffile{M2Graph.eps} }
% \bigskip
\ifsolutions
\bigskip
For reference, the formula I used is
$f' (x) = \frac{1}{195} (x - 7) x (x + 4) (x + 7)$.
For reference in the solutions below, here is the graph of the
function $f$ whose derivative is shown above and which satisfies
$f (6) = -17$.
(Any other function with the given derivative differs from this
one by a constant, as we saw in Section~4.2 of the book.)
You can check, by drawing tangent lines, that the graph of the
derivative of this function does indeed have the shape shown
in the graph above.
Also shown on the graph below
are the tangent lines at the points in parts (a) and (b),
and all the points (including those in Part~(c)) are marked on
the graph with dots.
\vspace{1ex}
\centerline{
\includegraphics[width=12cm]{M2SolnGraph}
}
\vspace{1ex}
% {\epsfxsize=12cm \epsffile{M2SolnGraph.eps} }
This function has the formula
\[
f (x) = \frac{1}{975} x^5 + \frac{1}{195} x^4 - \frac{49}{585} x^3
- \frac{98}{195} x + \frac{833}{325}.
\]
\fi
\bigskip
(a)
Is $f$ increasing, decreasing, or nearly flat at $x = -1$,
or is there not enough information provided to determine this?
Why?
\ifsolutions
\bigskip
{\emph{Solution:}}
You can read off the graph of $f'$ that $f' (-1) \approx 0.7 > 0$.
(The exact value is $\frac{48}{65} \approx 0.738462$.)
Therefore $f$ is increasing at $x = -1$.
The tangent line to the graph of $f$ at $x = -1$ is shown on the
graph above, and you can see that its slope is positive.
\fi
\bigskip
(b)
Is $f$ concave up or concave down at $x = 3{.}5$,
or does $f$ (nearly)
have an inflection point at $x = 3{.}5$, or is there not
enough information provided to determine this?
Why?
\ifsolutions
\bigskip
{\emph{Solution:}}
The graph shown is of the derivative $f'$ of $f$.
Therefore $f'' (3{.}5)$ is the slope (first derivative) of the
graph shown at $3{.}5$.
Clearly it is negative.
Thus $f'' (3{.}5) < 0$.
Therefore $f$ is concave down at $x = 3{.}5$.
The tangent line to the graph of $f$ at $x = 3{.}5$ is shown on the
graph above, and you can see it is above the graph of $f$.
\fi
\bigskip
(c)
At which values of $x$ in $[-8, 8]$ (the interval shown)
does $f$ have a local minimum?
Explain.
\ifsolutions
\bigskip
{\emph{Solution:}}
Local minimums occur at critical numbers, that is, numbers $c$
such that $f' (c) = 0$ or $f' (c)$ does not exist.
There are no numbers $c$ such that $f' (c)$ does not exist.
We can read off the graph of $f'$ that $f' (c) = 0$
for $c = -7$, $c = -4$, $c = 0$, and $c = 7$.
We can test which of these are local minimums in either of
two ways: the first derivative test or the second derivative test.
Here is the first derivative test.
(This explanation is much more detailed than was required on the exam.)
At $c = -7$, the values shown for $f' (x)$ on the graph of $f'$,
for $x < -7$ but $x$ close to $-7$, are positive,
while values shown for $f' (x)$,
for $x > -7$ but $x$ close to $-7$, are negative.
So $f$ is increasing before we get to $-7$, and decreasing afterwards;
therefore, $f$ has a local {\emph{maximum}} at $c = -7$.
Similarly, $f$ has a local {\emph{maximum}} at $c = 0$.
At $c = -4$, the values shown for $f' (x)$ on the graph of $f'$,
for $x < -4$ but $x$ close to $-4$, are negative,
while values shown for $f' (x)$,
for $x > -4$ but $x$ close to $-4$, are positive.
So $f$ is decreasing before we get to $-4$, and increasing afterwards;
therefore, $f$ has a local {\emph{minimum}} at $c = -4$.
Similarly, $f$ has a local {\emph{minimum}} at $c = 7$.
Thus, the answer to the problem is $x = -4$ and $x = 7$.
Here is the second derivative test.
(Again, this explanation is much more detailed than was
required on the exam.)
Since the graph shows $f'$, the second derivative $f'' (x)$,
which is the derivative of $f'$ at $x$, is the slope of the
tangent line to the graph shown at $x$.
The tangent lines at $x = -7$ and $x = 0$ clearly have negative slopes,
while those at $x = -4$ and $x = 7$ clearly have positive slopes.
Thus, the second derivative test tells us that $f$ has local minimums
at $x = -4$ and $x = 7$ (critical numbers where the second derivative
is positive).
(Also, there are local maximums at $x = -7$ and $x = 0$, since these
are critical numbers where the second derivative is negative.)
You can see from the graph of $f$ provided above that $f$ does indeed
have local minimums (marked with dots) at $x = -4$ and $x = 7$,
and local maximums at $x = -7$ and $x = 0$.
\fi
\bigskip
(d)
Is $f (0)$ positive, negative, or zero, or is there not enough
information to determine this?
Explain.
\ifsolutions
\bigskip
{\emph{Solution:}}
Since we only know~$f'$,
we only know $f$ up to a constant.
(For example, if $g (x) = f (x) + 117$ for all~$x$,
then $g' (x) = f' (x)$ for all~$x$.)
So we can't determine anything about $f (0)$ by itself
(only how it is related to the values of $f$ at other points).
\fi
\bigskip
\bigskip
\noindent
11. {(8 points)}
You want to build a box with a square base.
Let $b$ be the
length of one side of the base, and let $h$ be the height, both
measured in meters.
Given the
constraints that $10 \leq b \leq 25$ and $2 b + h = 60$,
what are the biggest
and smallest volumes for such a box?
\ifsolutions
\bigskip
{\emph{Solution:}}
Since the base is square with side length~$b$,
its area is~$b^2$.
Therefore its volume is $V = b^2 h$.
So we are supposed to minimize and maximize $b^2 h$
subject to the constraints $2 b + h = 60$ and $10 \leq b \leq 25$.
We eliminate one variable from the formula for~$V$ using
the first constraint:
\[
2 b + h = 60
\]
\[
h = 60 - 2 b
\]
\[
V = b^2 h = b^2 (60 - 2 b).
\]
So we must minimize and maximize
the function $V (b) = b^2 (60 - 2 b)$
on the interval $[10, 25]$.
We find critical numbers.
To differentiate easily, rewrite
\[
V (b) = 60 b^2 - 2 b^3.
\]
Differentiate, and factor so we can solve $V' (b) = 0$:
\[
V' (b) = 120 b - 6 b^2
= 6 b (20 - b).
\]
This is zero when $b = 0$ or $b = 20$.
We reject $b = 0$ since it is not in the interval $[10, 25]$.
(I must see you do this!)
Now compare values:
\[
V (10) = 10^2 (60 - 2 \cdot 10) = 4000,
\]
\[
V (20) = 20^2 (60 - 2 \cdot 20) = 400 \cdot 20 = 8000,
\]
and
\[
V (25) = 25^2 (60 - 2 \cdot 25) = 625 \cdot 10 = 6250.
\]
So the maximum volume is $8000 {\text{m}}^3$
and the minimum volume is $4000 {\text{m}}^3$.
(The units are {\emph{necessary}}.)
\bigskip
One can use the first or second derivative test to decide
whether the critical number is a maximum or minimum.
But this is wasted effort,
since one eventually needs to find $V (10)$, $V (20)$,
and $V (25)$ anyway.
\fi
\bigskip
\bigskip
\noindent
12. {(3 points/part)}
The function $P (t) = (6 t + 1) e^{k (t - 1)}$ models the
population of a colony of bacteria at time~$t$,
where $P$ is measured in hundreds of bacteria
and $t$ is measured in hours.
Observations indicate that after one
hour there are $700$ bacteria, and at that time the colony is growing at
a rate of $200$
bacteria per hour.
\bigskip
(a)
Find~$k$.
\ifsolutions
\bigskip
{\emph{Solution:}}
We are told that $P (1) = 7$ and $P' (1) = 2$.
({\textbf{Caution! Read the problem!}}
$P (t)$ is measured in {\textbf{hundreds}} of bacteria!)
So
\[
7 = P (1) = (6 \cdot 1 + 1) e^{k (1 - 1)} = 7 e^0 = 7.
\]
This equation doesn't tell us anything:
the statement that at $t = 1$ there are $700$ bacteria
is redundant.
The statement about $P' (1)$ is more useful.
Using the product and chain rules, we get
\[
P' (t)
= 6 e^{k (t - 1)}
+ (6 t + 1) e^{k (t - 1)} \frac{d}{d t} \big( k (t - 1) \big)
= 6 e^{k (t - 1)} + (6 t + 1) e^{k (t - 1)} \cdot k.
\]
Substituting $t = 1$ gives
\[
P' (1)
= 6 e^{k (1 - 1)} + (6 \cdot 1 + 1) e^{k (1 - 1)} k
= 6 + 7 k.
\]
This is supposed to be~$2$,
so
$2 = 6 + 7 k$,
that is, $k = - \frac{4}{7}$.
\fi
\bigskip
(b)
What happens to the population of bacteria in the long run,
as $t \to \infty$?
Explain.
\ifsolutions
\bigskip
{\emph{Solution:}}
Using the information from part~(a),
we can rewrite the formula as
\[
P (t) = (6 t + 1) e^{(- 4 / 7) (t - 1)}.
\]
We are asked about $\lim_{t \to \infty} P (t)$.
Write this as
\[
\lim_{t \to \infty} \frac{6 t + 1}{e^{(4 / 7) (t - 1)}}.
\]
In this form, it has the indeterminate form
${\displaystyle{ \frac{\infty}{\infty} }}$.
Therefore we may use L'Hospital's Rule.
Thus
\[
\lim_{t \to \infty} \frac{6 t + 1}{e^{(4 / 7) (t - 1)}}
= \lim_{t \to \infty} \frac{6}{\frac{4}{7} e^{(4 / 7) (t - 1)}},
\]
if the second limit exists.
But this limit is zero,
since $\lim_{t \to \infty} e^{(4 / 7) (t - 1)} = \infty$.
So
\[
\lim_{t \to \infty} \frac{6 t + 1}{e^{(4 / 7) (t - 1)}} = 0.
\]
The population of bacteria goes to zero,
that is, the bacteria die out.
\fi
\end{document}