%% Math 131-01 Midterm solutions, Fall 2003

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\begin{document}

\begin{center}

\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Section 01, Midterm 1 Solutions} \\
Professor Hwang
\end{center}

\begin{enumerate}

\item{} Compute the following limits. 

\bd

\item{(a)} [5 points]\quad $\ds\lim_{x\to2}\frac{x^2-x-6}{x-3}$

\soln Plugging in $x=2$ gives
$\ds\frac{2^2-2-6}{2-3}=\frac{-4}{-1}=\framebox{\vp\quad4\quad}$

\hrulefill

\item{(b)} [10 points]\quad $\ds\lim_{x\to3}\frac{x^2-x-6}{x-3}$

\soln Setting $x=3$ gives the formal answer~$0/0$, so we must cancel a
common factor:
\[
\lim_{x\to3}\frac{x^2-x-6}{x-3}
=\lim_{x\to3}\frac{(x+2)(x-3)}{(x-3)}
=\lim_{x\to3}(x+2) = \framebox{\vp\quad5\quad}
\]

\ed

\sep

\item{}  [15 points]\quad Find the horizontal and vertical asymptotes
  of the function
\[
y=\frac{3x(x^2+3x+2)}{(x^2-1)(x+3)}
\]
Show all work, and clearly mark your answers.

\soln To find the horizontal asymptote (if any), we consider the
highest-degree terms in the numerator and denominator:
\[
y=\frac{3x(x^2+3x+2)}{(x^2-1)(x+3)}
=\frac{3x^3 + 9x^2 + 6x}{x^3+3x^2-x-3}.
\]
The ratio is $\ds\frac{3x^3}{x^3}=3$, so \framebox{$\vp y=3$} is a
horizontal asymptote.

To find vertical asymptotes, we set the denominator to~0, which gives
potential vertical asymptotes at $x=1$, $x=-1$, and $x=-3$. To see
which of these really are vertical asymptotes, we check to see if the
numerator is zero at each point. Setting $p(x)=3x(x^2+3x+2)$, we have
$p(-1)=-3(1-3+2)=0$, $p(1)=3(1+3+2)\neq0$, and $p(3)=9(9+9+2)\neq0$. 
In summary, \framebox{$\vp x=1$}~and \framebox{$\vp x=-3$} are
vertical asymptotes. 

An alternative method is to cancel all common factors in the original
function. Any remaining zeros of the denominator are definitely
vertical asymptotes:
\[
y=\frac{3x(x^2+3x+2)}{(x^2-1)(x+3)}
=\frac{3x(x+1)(x+2)}{(x+1)(x-1)(x+3)}
=\frac{3x(x+2)}{(x-1)(x+3)},
\]
Again we find $x=1$~and $x=-3$. (Note that this method may be more
difficult to apply in practice, but it always works, while setting the
numerator to~0 can give ``false negatives''.)

\sep

\item{} Barium-140 has a half-life of 12.75~days. 

\bd

\item{(a)} [10~points]\quad Find a formula $Q=Q_0e^{kt}$ for the
  amount of Barium-140 remaining after $t$~days.

\soln We seek the value of~$k$ for which $0.5=e^{12.75k}$. Taking
natural logarithms,
\[
k=\frac{\ln 0.5}{12.75}\simeq -0.054364484,\qquad
\framebox{$\Strut Q=Q_0e^{-0.054364484t}$}
\]

\sep

\item{(b)} [10~points]\quad How many days does it take for 95\%~of a
  sample of Barium-140 to decay? (Watch round-off error!)

\soln When $95\%$~of the sample has decayed, the fraction remaining is
$0.05=e^{kt}$. This occurs when
\[
t=\frac{1}{k}\ln 0.05\simeq \framebox{$\vp 55.1\text{ days}$}
\]

\ed

\sep

\item{} [10~points]\quad The desert temperature~$H$ varies
  sinusoidally from a low of~$40^\circ\textrm{F}$ at 5~AM to a high
  of~$80^\circ\textrm{F}$ at 5~PM. Write a formula for~$H$ as a
  function of~$t$, with $t$~measured in hours from 5~PM. You may use
  the graph below for reference.

\begin{center}
\input{midsoln1a-01.eepic}
\end{center}

\soln Because the temperature starts at maximum displacement from
average, we seek a formula $H=D+A\cos(Bt)$, with $t$~measured in hours
from 5~PM. The average temperature is $D=60\fahr$, while the amplitude
(the maximum displacement from average) is $A=20\fahr$. Since one full
period of the temperature function spans 24~hours, we have $24B=2\pi$,
or $B=\pi/12$. These observations give
\[
\framebox{$\vp\textstyle H=60+20\cos(\frac{\pi t}{12})$}
\]

\sep

\item{} [20~points]\quad According to Kepler's law, the period of an
  earth satellite is proportional to the $\frac{3}{2}$~power of the
  radius of its orbit.  The moon's orbital period is 29~days, and the
  radius of its orbit is $238,000$~miles. Find the radius of the orbit
  of a satellite whose period is 24~hours.
  
  \soln Let $P$~be the orbital period of a satellite in \textbf{days},
  and let~$r$ be the radius of the orbit in \textbf{thousands of
    miles}. (These choices of unit are not the only ones possible, but
  are useful in keeping the numbers smaller.) We are given that
  $P=kr^{3/2}$ for some constant~$k$, and also that $29=k(238)^{3/2}$.
  The question asks for the value of~$r$ for which $1=kr^{3/2}$. Note
  that we do not need to solve for~$k$: Just divide one equation by
  the other!  This gives
\[
\frac{1}{29}=\frac{r^{3/2}}{(238)^{3/2}}
=\Bigl(\frac{r}{238}\Bigr)^{3/2},\qquad\text{or}\quad
r=238\Bigl(\frac{1}{29}\Bigr)^{2/3}\simeq25.2,
\]
which corresponds to an orbit of radius
\framebox{$\vp25,200~\text{miles}$} There is not much point giving an
answer to greater accuracy, because the moon's orbit is not exactly
238,000~miles in radius, nor is the period exactly 29~days. The main
lessons are to choose units that keep the numbers small, use the same
units throughout, and solve algebraically to the extent possible.

\sep

\item{} An automobile costs \$25,000, and depreciates in value by
  20\%~per year.

\bd

\item{(a)} [10~points]\quad Compute the average rate of depreciation
  for the first year of ownership. Express your answer in dollars per
  day.

\soln The value of the car in dollars is given, after $t$~years, by
the formula $V(t)=25,000(0.8)^t$. After 1~year, the value is~\$20,000,
a depreciation of~\$5,000. The average rate of depreciation over the
first year is
\[
\frac{\$5,000}{365~\text{days}}=\framebox{$\vp\$13.70/\text{day}$}
\]

\sep

\item{(b)} [10~points]\quad Estimate the actual depreciation for the
  first day of ownership. Express your answer in dollars, rounded to
  the nearest penny.

\soln Using the formula found in part~(a), the value of the car after
1~day is roughly
\[
\textstyle V(\frac{1}{365})=25,000(0.8)^{1/365}\simeq\$24,984.72.
\]
This represents a depreciation of \framebox{$\vp\$15.28$}

\ed

\end{enumerate}

\end{document}

