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\noindent
\textbf{Math 105 Exam II answers.}

\vspace{0.2in}

\begin{questions}

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

\question Solve the following for $x$.

\begin{parts}

\part $e^{3x-2} = 1$

\noindent 
Apply ln to both sides: on the left, you get $\ln(e^{3x-2})=3x-2$, and 
on the right you get $\ln(1)=0$.  So $3x-2=0$, so
\boxanswer{$x=\frac{2}{3}$}.

\part $\ln (5 e^{-2x}) = 0$ 

\noindent 
By the properties of logarithms, the left side can be rewritten as
$\ln (5 e^{-2x}) = \ln(5)+\ln(e^{-2x}) = \ln(5)+(-2x)$.  Now set this
equal to 0 and solve for $x$: \boxanswer{$x=\frac{\ln(5)}{2}$} $\approx 
0.805$.

\end{parts}

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

\question Evaluate the following; your answer should be an integer
or fraction.

\begin{parts}

\part $\log_{7}(\frac{1}{49})$

\noindent
$\log_{7}(\frac{1}{49}) = \log_{7}((7)^{-2}) = \boxanswer{-2}$.

\part $e^{\ln (2^{3})}$

\noindent 
$e^{\ln (2^{3})} = 2^{3} = \boxanswer{8}$.

\end{parts} 

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

\question What is the half-life of a substance with a decay
constant of $0.635$?

\noindent 
By the formula from class, the half-life is $\ln(2)$ divided by the
decay constant: \boxanswer{$\ln(2)/0.635$} $\approx 1.09$.


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

\question A substance decays exponentially, so that the amount
after $t$ days is given by the formula $y=A e^{-kt}$.  If exactly
$\frac{1}{3}$ of the initial amount remains after four days, what is
the decay constant $k$ of the substance?

\noindent
If you start with an amount $A$, then after 4 days, you will have
$\frac{A}{3}$ remaining.  So take the equation $\frac{A}{3} =
Ae^{-k4}$ and solve for $k$: cancel $A$ from each side to get
$\frac{1}{3} = e^{-4k}$.  Apply $\ln$ to both sides to get
$\ln(\frac{1}{3}) = -4k$.  Solve for $k$: $k=-\frac{\ln(1/3)}{4}$.
Using properties of logarithms, $\ln(1/3)=-\ln(3)$, so you can rewrite 
this as \boxanswer{$k=\frac{\ln(3)}{4}$} $\approx 0.275$.

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

\question You owe a loan shark $\$800$; he charges interest at a
rate of $35\%$.  What is the formula for the amount
of money owed after $t$ years in these cases?

\begin{parts}

\part If the interest is compounded weekly (52 times a year)?

\noindent 
\boxanswer{$800 (1 + \frac{0.35}{52})^{52t}$}

\part If the interest is compounded continuously?  

\noindent 
\boxanswer{$800 e^{0.35t}$}

\end{parts}

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

\question You deposit $\$2000$ into an account paying $7\%$,
compounded continuously.  How long will it take for the account to
grow to $\$3000$?

\noindent 
Solve this equation for $t$: $3000=2000e^{0.07t}$.  Divide by 2000 and 
apply $\ln$: $\ln(3/2) = 0.07t$, so
\boxanswer{$t=\frac{\ln(3/2)}{0.07}$} $= \frac{\ln(3) - \ln(2)}{0.07}
\approx 5.79$.

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

\question What is the definition of the derivative of the function
$f(x)$? (We are looking for the formula involving a limit and $h$ and
\dots)

\noindent 
\boxanswer{$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)
- f(x)}{h}$}

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

\question Compute the derivative $f'(x)$ of the function $f(x) =
3x^{2} - \frac{2}{x} + \sqrt[3]{x}$ using any rule we have had 
so far.

\noindent 
Rewrite $\frac{2}{x}$ as $2x^{-1}$, and rewrite $\sqrt[3]{x}$ as
$x^{\frac{1}{3}}$.  Then the answer is \boxanswer{$\displaystyle f'(x)
= 6x + \frac{2}{x^{2}} + \frac{1}{3} x^{-\frac{2}{3}}$}

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

\question Consider the function $f(x)=7x^{2}$.

\begin{parts}

\part What is the slope of the secant line to this curve
between the points $(1,7)$ and $(1+h,7(1+h)^{2})$?  Simplify as much as
possible.

\noindent 
\[
\frac{7(1+h)^{2}-7}{h} = \frac{7(1+2h+h^{2}) - 7}{h} =
\frac{14h+7h^{2}}{h} = \boxanswer{14+7h}.
\]

\part What is the slope of the tangent line to this curve at the point
$(1,7)$?

\noindent 
It's the limit of the answer for part (a), as $h$ goes to zero:
\boxanswer{14}.

\end{parts}

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

\question For what \emph{two} values of $x$ does the curve
$y=\frac{1}{3}x^{3}+\frac{1}{2}x^{2} -20x + 3$ have slope zero?

\noindent 
Take the derivative, set it equal to zero, and solve for $x$:
$x^{2}+x-20=0$.  Solve this using the quadratic formula (or by
factoring) to get \boxanswer{$x=4$} and \boxanswer{$x=-5$}.

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

\question Let $y=x^{1/2} - 3$.  What is the equation of the
line tangent to this curve at the point $(1,-2)$?

\noindent 
Take the derivative and plug in $x=1$ to find the slope: $y' =
\frac{1}{2}x^{-1/2}$, so the slope is $\frac{1}{2}$.  It goes through
the point $(1,-2)$, so the equation is $y-(-2) = \frac{1}{2}(x-1)$, or 
\boxanswer{$y=\frac{1}{2}x - \frac{5}{2}$}.

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

\question Let $f(x) = x^{4}-6x^{2}+7x+1$.  Compute $f''(x)$ using
any rule we have had so far.

\noindent 
$f'(x) = 4x^{3} - 12 x + 7$, so \boxanswer{$f''(x) = 12x^{2} - 12$}.

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

\question You throw a ball into the air at a rate of 80 ft./sec.,
so that its position (height) after $t$ seconds is given by the
formula $s(t) = -16t^{2} + 80t$.  What is the velocity of the ball
after 2 seconds?

\noindent 
Take the derivative of $s(t)$ and plug in $t=2$: $s'(t) = -32 t + 80$, 
so the velocity after 2 seconds is \boxanswer{$s'(2)=16$}.

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

\question A particle moves along the $x$-axis so that its position
after $t$ seconds is $s(t) = 3t^{3} + 2t + 1$.

\begin{parts}

\part What is the average velocity of the particle between times
$t=0$ and $t=1$?

\noindent 
Average velocity is (distance traveled) divided by time elapsed:
$\frac{s(1)-s(0)}{1-0}$.  $s(1)=6$ and $s(0)=1$, so this is
$\frac{6-1}{1-0} = \boxanswer{5}$.

\part What is the instantaneous velocity of the particle when
$t=1$?

\noindent
Instantaneous velocity is $s'(t)$.  In this case, that is
$s'(t)=9t^{2}+2$.  Plug in $t=1$ to get the answer:
\boxanswer{$s'(1)=11$}.

\part What is the acceleration of the particle when $t=1$?

\noindent 
Acceleration is given by $s''(t)=18t$.  Plug in $t=1$ to get 
the answer: \boxanswer{$s''(1)=18$}.

\end{parts}

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

\end{questions}

\end{document}

