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\noindent
\textbf{Math 105 Exam I.}
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\noindent
\textbf{Math 105 Exam III answers.}

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\begin{questions}

%%%%%%%%%%%%%%%%

\question A company determines that when producing $x$ units, its
profit is given by this function: $\displaystyle P(x) = -\frac{1}{3}
x^{2} + 200x - 1800$.

\begin{parts}

\part[4] What is the marginal profit function?

\noindent Marginal profit is the derivative (with respect to
production level) of profit: \\
 \boxanswer{$MP(x)=-\frac{2}{3}x + 200$}.

\part[2] What is the marginal profit when the production level $x$ is
150?

\noindent Plug 150 in for $x$: \boxanswer{$MP(150)=100$}.

\part[2] What does that tell you about the change in profit as the
production level increases from 150 to 151?

\noindent \boxanswer{Profit increases by 100}.

\end{parts}

%%%%%%%%%%%%%%%%

\question[8] Use linear approximation to estimate $(1.03)^{2/3}$.

\noindent Let $f(x)=x^{2/3}$, so $f'(x)=\frac{2}{3} x^{-1/3}$.  Let
$x=1.03$ and $a=1$.  Then the linear approximation formula says 
\[
f(x) = (1.03)^{2/3} \approx f(a) + f'(a) (x-a) = 1^{2/3} + \frac{2}{3} 
1^{-1/3} (1.03 - 1) = 1 + \frac{2}{3}(0.03) = \boxanswer{1.02}.
\]

%%%%%%%%%%%%%%%%

\question[6] A piece of petrified wood is found at an archaeological site
and contains $87\%$ of the original carbon-14.  How old is the piece
of wood?  (The decay constant for carbon-14 is $k=0.000121$.)

\noindent From the basic carbon dating equation, $0.87 = e^{-0.000121
t}$.  Solve for $t$: \\
\boxanswer{$\displaystyle t=\frac{\ln
(0.87)}{-0.000121} = 1150$} years.

%%%%%%%%%%%%%%%%

\question[15] Find the derivatives of these functions:

\begin{parts}

\part $\ln (4x^{2/5})$

\noindent First, rewrite using properties of logs: $\ln (4x^{2/5} =
\ln(4) + \ln (x^{2/5}) = \ln(4) + \frac{2}{5}\ln(x)$.  Now
differentiate to get \boxanswer{$\displaystyle \frac{2}{5} \frac{1}{x}$}.
(Or instead you can use the chain rule with $f(x)=\ln(x)$ and
$g(x)=4x^{2/5}$.  The end result is $\frac{1}{4x^{2/5}}
\frac{8}{6}x^{-3/5}$, which can be simplified to get $\frac{2}{5x}$.)

\part $(x^{3}-3x+2)^{17}$

\noindent Use the chain rule, applied to the functions $f(x)=x^{17}$
and $g(x)=x^{3}-3x+2$: \boxanswer{$17 (x^{3}-3x+2) (3x^{2}-3)$}.

\part $\sqrt{6^{x}}$

\noindent One method is the chain rule, applied to the functions
$f(x)=\sqrt{x}$ and $g(x) = 6^{x}$: \boxanswer{$\displaystyle
\frac{1}{2} (6^{x})^{-1/2} (\ln 6) 6^{x}$}.  Another way: rewrite
$\sqrt{6^{x}}$ as $(6^{x})^{1/2} = 6^{x/2}$.  Now differentiate:
\boxanswer{$\displaystyle \frac{1}{2} (\ln 6) (6^{x/2})$}.  (These two 
answers may look different, but they are actually equal.)

\end{parts}

%%%%%%%%%%%%%%%%

\question[8] A square, with edge of length $x$, is expanding.  When the
edge is 5 ft., the edge is growing at 3 ft./min.  How fast is the area 
growing at this point?

\noindent The chain rule says that $\frac{dA}{dt} = \frac{dA}{dx}
\frac{dx}{dt}$.  Also, $A=x^{2}$, so $\frac{dA}{dx}=2x$, and we are
told that $\frac{dx}{dt} = 3$.  So $\frac{dA}{dt} = (2x) 3$.  Plugging 
in $x=5$ gives the answer: \boxanswer{30} square feet per minute..

%%%%%%%%%%%%%%%%

\question[10] You get a cup of coffee from the drive-through window of a
fast food restaurant.  The coffee starts at $190^{\circ}$---too hot to
drink---and the temperature in your car is a constant $70^{\circ}$.
After 1 minute, the coffee cools to $175^{\circ}$.  You like your
coffee at $150^{\circ}$; how long until it reaches this temperature?

\noindent Newton's law of cooling gives the equation $H=M+Ae^{kt}$.
Here $M=70$, and when $t=0$, $H=190$, so: $190=70+Ae^{0}=70+A$.  So
$A=190-70=120$.  The equation for $H$ is now $H=70+120e^{kt}$.  To
find $k$, we know that when $t=1$, $H=175$: $175=70+120e^{k}$.  Solve
for $k$: $105=120e^{k}$, so $\frac{105}{120}=e^{k}$, so
$k=\ln(\frac{105}{120}) = \ln(\frac{7}{8}) = -0.1335$.  Finally, set
$H=150$ and solve for $t$: $150=70+120 e^{\ln(7/8) t}$, so
$80=120e^{\ln(7/8)t}$, so $2/3 = e^{\ln(7/8)t}$.  Take the log of both 
sides: $\ln(2/3) = \ln(7/8)t$.  Therefore \boxanswer{$\displaystyle
t=\frac{\ln(2/3)}{\ln(7/8)} = 3.04$} minutes.

%%%%%%%%%%%%%%%%

\question[7] Let $\displaystyle y=\frac{x-2}{x+1}$.  Find the equation of
the line tangent to this curve at the point $(0,-2)$.

\noindent Use the quotient rule to find the derivative of $y$: $y' =
\frac{(x+1)1 - (x-2)1}{(x+1)^{2}} = \frac{x+1-x+2}{(x+1)^{2}} =
\frac{3}{(x+1)^{2}}$.  When $x=0$, $y'=3$, so the slope of the tangent 
line is $3$.  The line goes through the point $(0,-2)$, so the
equation of the line is \boxanswer{$y=3x-2$}.

%%%%%%%%%%%%%%%%

\question[8] Let $f(x)=xe^{-3x}$.

\begin{parts}

\part Where is $f(x)$ increasing?

\noindent The derivative of $f(x)$ is $f'(x)=e^{-3x} + x(-3e^{-3x}) =
e^{-3x}(1-3x)$.  This is zero when $1-3x=0$, i.e., when $x=1/3$.  This 
is the only critical point.  Plug in $x=0$ to see what happens when
$x<1/3$: $f'(0) = 1$.  This is positive.  Plug in $x=1$ to see what
happens when $x>1/3$: $f'(1) = e^{-3}(-2)$.  This is negative
(positive times negative is negative).  So \boxanswer{$f(x)$ is
increasing when $x<1/3$}.

\part Where is $f(x)$ decreasing? 
 
\noindent \boxanswer{$f(x)$ is decreasing when $x>1/3$}.

\end{parts}

%%%%%%%%%%%%%%%%

\question[8] Where does the function $f(x)=x^{3}-3x^{2}$ have critical
points?  For each of those critical points, is it a relative maximum,
a relative minimum, or neither?

\noindent The derivative is $f'(x)=3x^{2}-6x=x(3x-6)$.  This is never
undefined.  It is zero when \boxanswer{$x=0$} or when
\boxanswer{$x=2$}, so these are the critical points.  When $x<0$, then 
$f'(x)$ is positive (plug in $x=-1$, for example).  When $0<x<2$, then 
$f'(x)$ is negative.  Therefore $f'(x)$ switches from positive to
negative at $x=0$, so \boxanswer{$x=0$ is a relative maximum}.  When
$x>2$, then $f'(x)$ is positive, so \boxanswer{$x=2$ is a relative
minimum}.

%%%%%%%%%%%%%%%%

\question[8] A circle of radius $r$ has area $A=\pi r^{2}$.  The radius is 
changing: at time $t$, the radius is given by $r=f(t)=3t+2$.  What is
$\displaystyle \frac{dA}{dt}$, the rate of change of the area with
respect to time?  (Write your answer in terms of $t$, not $r$.)

\noindent The chain rule says that $\frac{dA}{dt} = \frac{dA}{dr}
\frac{dr}{dt}$.  Here, $\frac{dA}{dr} = 2\pi r$, and
$\frac{dr}{dt}=3$, so $\frac{dA}{dt} = (2 \pi r) (3) = 6\pi r$.  To
write in terms of $t$, substitute in $r=3t+2$:
\boxanswer{$\frac{dA}{dt} = 6\pi (3t+2) = 18\pi t + 12\pi$}.

%%%%%%%%%%%%%%%%

\question[14] Find the derivatives of these functions:

\begin{parts}

\part $\displaystyle \frac{e^{x^{4}+7}}{x+2}$ 

\noindent This is a combination of the quotient and chain rules: the
derivative is 
\[
\frac{(x+2)(e^{x^{4}+7})' - (e^{x^{4+7}})(x+2)'}{(x+2)^{2}}
= \frac{(x+2)(4x^{3})(e^{x^{4}+7}) - (e^{x^{4}+7})(1)}{(x+2)^{2}}.
\]
So the answer is \boxanswer{$\displaystyle
\frac{(x+2)4x^{3}e^{x^{4}+7} - e^{x^{4}+7}}{(x+2)^{2}}$}.

\part $x^{2} \ln (x^{2} +7x - 2)$

\noindent This is a combination of the product and chain rules: the
derivative is 
\[
(x^{2})' (\ln (x^{2} + 7x -2)) + x^{2} (\ln (x^{2} + 7x -2))' = 
2x \ln (x^{2} + 7x -2) + x^{2} \frac{1}{x^{2} + 7x - 2} (2x+7).
\] 
So the answer is \boxanswer{$\displaystyle 2x \ln (x^{2} + 7x -2) +
\frac{x^{2} (2x+7)}{x^{2} + 7x - 2}$}.

\end{parts}

%%%%%%%%%%%%%%%%

\end{questions}

\end{document}

