%% Math 131, Fall, 2003 Course Review Sheet Answers
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\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Course Review Sheet Answers}
Professor Hwang \\
\end{center}

\be

\item What type of function (linear, exponential, power, trig) is the
  most appropriate for modeling short-term population growth (such as
  the US~population on a scale of a few years, or of a bacterial
  culture on a scale of ten minutes)? What about modeling long-term
  population growth? Write a few sentences to justify your answers.
  
  \soln In the short term, a linear function is sufficient because the
  rate of growth is nearly constant over short time intervals. Over
  the long term, an exponential function is more suitable. Populations
  grow at a rate proportional to their size, and an exponential
  function's derivative is proportional to the function itself. (In
  the very long term, additional considerations---lack of resources,
  epidemics, or medical advances, say---can make an exponential model
  inaccurate.)

  \sep

\item Let $f(x)=\frac{\sin x}{x}$ for $x\neq0$.
  \bd
\item{(a)} What value for~$f(0)$ makes~$f$ continuous at $x=0$?
\item{(b)} Show that $\lim\limits_{x\to0} f'(x)$ exists.
    
  \soln By l'H\^opital's rule, the value $f(0)=1$ makes~$f$ continuous
  at $x=0$. For part~(b), use the quotient rule to see that
  $f'(x)=\frac{x\cos x-\sin x}{x^2}$. L'H\^opital's rule gives
  \[
  \lim_{x\to0}\frac{x\cos x-\sin x}{x^2}
  =\lim_{x\to0}\frac{-x\sin x + \cos x-\cos x}{2x}
  =-\lim_{x\to0}\frac{\sin x}{2}=0.
  \]

\item{(c)} Answer the same two questions for $g(x)=\frac{e^x-1}{x}$
  
  \soln $g(0)=1$, $\lim\limits_{x\to0}g'(x)=\frac{1}{2}$.

  \ed

\sep

\item A stone is dropped from a high bridge. After $t$~seconds, the
  height of the stone is $y(t)=150-16t^2$ feet.

  \bd
  \item{(a)} Without using a graphing calculator, sketch the graph of
  height as a function of time. What is the domain of the
  function~$y$? 

  \item{(b)} Find the average speed of the stone (including units)
  over the time interval $0.5\leq t\leq 2$, and draw the secant line
  whose slope represents this average speed.

  \soln The stone hits when $y(t)=0$, or $t=\frac{5}{4}\sqrt{6}\simeq
  3.062$~seconds, so the domain is~$[0,\frac{5}{4}\sqrt{6}]$. The
  average speed of the stone on the inteval~$[0.5,2]$ is $-40$~ft/sec.
  \begin{center}
    \input{rsoln4a.eepic}
  \end{center}

  \ed

  \medskip

  \sep

\item A savings account is opened on January~1 with~\$1000 at a
  constant interest rate of~2\% per year, compounded continuously.

  \bd
\item{(a)} Find a formula for the account balance as a function of
  time, including units.

  \soln The balance is $B(t)=1000(1.02)^t$ dollars after $t$~years.

\item{(b)} How many years does it take for the balance to double? To
  grow to \$1,000,000?

  \soln About 35~years; about 349~years.

\item{(c)} Find the average rate of earnings for the first quarter
  of the third year; express your answer in dollars per day.
  
  \soln The third year starts when $t=2$ (!), so use $t=2$~and
  $t=2.25$ to compute the change in balance: \$5.163. (I do not plan
  to ask questions whose wording could mislead you in this way.)

\item{(d)} Use linear approximation to estimate the amount of
  interest that accrues on the first day of the tenth year.

  \soln First, $B'(t)=1000\ln(1.02)(1.02)^t$ dollars per year after
  $t$~years, or about $0.054254(1.02)^t$~dollars per day after
  $t$~years. The interest accrued on the first day of the tenth year
  is roughly $0.054254(1.02)^9\simeq0.065$

  \ed

  \vfil\eject

\item Let $f(x)=(x^2-1)^2$. Find the critical point(s) and inflection
  point(s) of~$f$ by making sign diagrams for the first and second
  derivatives. Classify the critical point(s) as local minima, local
  maxima, or neither. Use the information you find to make a careful
  sketch of the graph of~$f$ on the interval $[-2,2]$. Do not use a
  graphing calculator.
  
  \soln Done in class. Critical points are $x=\pm1$ (local minima) and
  $x=0$ (local maximum), the inflection points are $x=\pm1/\sqrt{3}$. 
  Calculate the function values at these points, then interpolate,
  using your knowledge of the concavity of the graph.

  \sep

\item The period of a satellite's orbit is proportional to the
  3/2~power of the radius of the orbit, measured from the center of
  the earth. 

  \bd
  \item{(a)} If an orbit's radius is increased by~10\%, by about how
  much does the period change? Use linear approximation, and show all
  work.

  \soln Let $T$~be the period of the orbit in minutes, $r$~the radius
  of the orbit in miles. (You should decide on the names of variables,
  \emph{including units}, as the first step of every word problem you
  do!) We are given that $T=kr^{3/2}$ for some positive constant~$k$. 
  
  Linear approximation says that a small change~$\Delta r$ in the
  radius gives rise to a small change $\Delta
  T\simeq\frac{dT}{dr}\Delta r=\frac{3}{2}kr^{1/2}\Delta r$ in the
  period. Dividing this equation by the original relation between
  $r$~and $T$ gives the fractional change in~$T$ given a fractional
  change in~$r$:
  \[
  \frac{\Delta T}{T}=\frac{3}{2}\frac{\Delta r}{r}.
  \]
  If the radius is increased by~10\% ($\Delta r/r=0.1$), the period
  increases by about~15\%.

  \item{(b)} The space shuttle orbits at an altitude of 150~miles, and
  takes 90~minutes to complete one orbit. Estimate the new orbital
  period if the altitude is increased to 200~miles. The radius of the
  earth is 3900~miles.
  
  \soln The initial radius is $3900+150=4050$~miles, and the change in
  the orbital radius is $200-150=50$~miles, so the fractional change
  in the radius is $\Delta r/r=50/4050\simeq0.0123$, or about~1.23\%. 
  The change in orbital period is about
  $90\cdot\frac{3}{2}(\Delta r/r)=1\frac{2}{3}$ minutes, so the new
  period is $91\frac{2}{3}$~minutes.
  \ed

  \vfil\eject

\item Water drains at 3~cubic feet per minute from a rain collector
  whose shape is an inverted cone of radius 6~feet and height 4~feet.
  Assuming the tank starts full, how rapidly is the water level
  falling initially?  How rapidly is the level falling when the depth
  of the water is 1~foot? Express your answers in feet per minute,
  feet per second, and inches per second. The volume of a cone of
  radius~$r$ and height~$h$ is $V=\frac{1}{3}\pi r^2h$.

  \soln Let $h$~be the depth of water in feet, and let~$r$ be the
  radius of the surface of the water at depth~$h$. For this cone,
  $r=1.5h$, so when the depth is~$h$, the volume in the tank is
  \[
  V=\frac{1}{3}\pi r^2h=\frac{3}{4}\pi h^3,\qquad\text{so}\quad
  \frac{dV}{dt}=\frac{9}{4}\pi h^2\frac{dh}{dt}.
  \]
  We are given that $dV/dt=-3$, so when the tank is full ($h=4$), we
  have $dh/dt=-4/(3\pi h^2)=-0.0265$~ft/min. When the depth is~1, the
  rate of decrease is $dh/dt=-0.4244$~ft/min. Divide by~60 to get
  ft/sec, and divide by~$60/12=5$ to get in/sec.

  \sep

\item The cross-sectional area of a tree increases by about the same
  amount each year, since the supply of nutrients is roughly constant,
  and the cross-sectional area is proportional to the volume of new
  tissue added. 

  \bd

  \item{(a)} Do the annual growth rings get thicker or thinner as the
  tree ages?
  
  \soln Thinner. (This is an average, idealized statement. Differing
  amounts of rain and sun play an important role in the amount of
  annual growth.)

  \item{(b)} A five-year-old tree has radius 2~inches. Approximately
  how thick will the next ring be? That is, what will be the radius
  of the tree when it is 6~years old? (Suggestion: Use related rates.)
  
  \soln Let $t$~be the age of the tree in years, $r$~the radius in
  inches. We are given that $r^2=kt$ for some positive constant~$k$,
  so $2r(dr/dt)=k$. (Now we see quantitatively why the rings get
  thinner as $r$~gets larger.) Using $k=r^2/t$ and solving for
  $dr/dt$, we find that $dr/dt=r/(2t)$. If $r=2$~and $t=5$, we expect
  the next ring (whose thickness is roughly~$dr/dt$) to be about
  $1/5=0.2$~inches thick. (It's not a bad exercise to estimate the
  amount of growth algebraically; the result is about 0.19~in.)

  \vfil\eject

  \item{(c)} About how old will the tree be when its radius is
  10~inches?

  \soln Since $r=2$~when $t=5$, we find that $k=4/5=0.8$, so
  $r^2=0.8t$. Setting $r=10$ and solving for~$t$, we find that the
  tree has radius 10~inches when it is 125~years old.

  \ed
  
  \sep

\item A car drives along a straight highway. The horizontal position
  of a point on the tread of a tire is given by
  $x(t)=80t-1.25\sin(64t)$ feet after $t$~seconds. Find the
  maximum forward speed of a point on the tread. At what point of the
  rotation does this maximum forward speed occur? Show that the point
  comes to a stop once per revolution.
  \begin{center}
    \input{review4a.eepic}
  \end{center}

  \soln The forward speed is $x'(t)=80-80\cos(64t)$, which is at most
  160~ft/sec (when the point is above the axle). Once per revolution
  we have $\cos(64t)=1$ (when the point is directly below the axle,
  i.e., touching the ground), and at these times $x'(t)=0$.

  \sep

\item A rumor spreads in a population of $M$~people. After $t$~days,
  $P$~people have heard the rumor. To a good approximation,
  \[
  \frac{dP}{dt}=kP(M-P)
  \]
  for a positive constant~$k$ that measures how frequently the rumor
  is repeated. 

  \bd

  \item{(a)} Using the fact that $dP/dt$ is the number of people per
  day who hear the rumor for the first time, explain in words why the
  equation above holds.
  
  \soln At time~$t$, $P$~people have heard the rumor and $M-P$~people
  have not. The rate of spread is equal to the number of incidents
  (per unit time) in which one of the $P$~people who have heard tells
  the rumor to one of the $M-P$~people who have not heard. The number
  of such meetings is jointly proportional to $P$~and $M-P$.

\vfil\eject

  \item{(b)} How many people have heard the rumor when the rate of
  spread is largest?
  
  \soln Think of the differential equation as saying $dP/dt=f(P)$,
  with $f(P)=kP(M-P)=k(MP-P^2)$. The graph of~$f$ is a parabola that
  opens downward and that crosses the horizontal axis at $P=0$~and
  $P=M$. The maximum of~$f$ therefore occurs halfway between, at
  $P=M/2$. Since the rate of speard is~$f(P)$, the rate of spread is
  most rapid when half the people have heard (and half have not
  heard). 

  \item{(c)} Compute $d^2P/dt^2$; show that the graph of~$P$ as a
  function of~$t$ is concave up when $P<M/2$, and is concave down when
  $P>M/2$. Does this confirm your answer to part~(b)? Explain.

  \soln The chain rule gives
  \[
  \frac{d^2P}{dt^2}=k(M-2P)\frac{dP}{dt}=k^2(M-2P)P(M-P).
  \]
  A sign diagram for the right hand-side reveals that $d^2P/dt^2>0$ if
  $0<P<M/2$, while $d^2P/dt^2<0$ if $M/2<P<M$. Again we see that the
  rate of spread increases to a maximum ($dP/dt$~is increasing until
  $P=M/2$) and then decreases, so the maximum rate of spread happens
  when $P=M/2$.

  \ed

  \sep

\item A spherical balloon is inflated at a constant rate of
  $25~\textrm{in}^3\!/\textrm{sec}$. 

  \bd
  \item{(a)} When the radius is $4$~inches, how rapidly is the radius
  increasing?

  \item{(b)} Assuming the balloon never pops, what is the limit
  of~$dr/dt$ as $t\to\infty$?

  \soln Letting $r$~be the radius in inches and $V$~be the volume in
  cubic inches, we have
  \[
  V=\frac{4}{3}\pi r^3,\qquad \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.
  \]
  Since we are given $dV/dt=25$, we find that $dr/dt=25/(4\pi r^2)$
  in/sec. When $r=4$, $dr/dt\simeq0.124$~in/sec. As $r\to\infty$,
  $dr/dt\to0$. 

  \ed

  \sep

\item A cylindrical can of radius~$r$ and height~$h$ must hold 1~liter
  (1000~cc). Find the dimensions that minimize the surface area
  $A=2\pi r^2 + 2\pi rh$. The volume of a cylinder is $V=\pi r^2h$,
  and your answer should be given in both symbolic and numerical form.
  
  \soln We are given that $1000=\pi r^2h$, or $\pi rh=1000/r$. As a
  function of~$r$, the surface area is $A(r)=2\pi r^2 + 2000/r$.
  Taking the derivative, $A'(r)=4\pi r-2000/r^2$, which is zero
  exactly when $r^3=2000/(4\pi)$, or
  $r=\sqrt[3]{500/\pi}\simeq5.42$~in. The height would be $h=1000/(\pi
  r^2)\simeq10.83$~in.

  \sep

\item Consider the curve $x^3+y^3-6xy=0$. Compute the slope of the
  tangent line at~$(x,y)$. Find the equations of the tangent lines at
  the points $(3,3)$~and $(4/3,8/3)$.

  \soln Implicit differentiation gives $3x^2+3y^2y'(x)-6y-6xy'(x)=0$,
  or 
  \[
  y'(x)=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}.
  \]
  At $(3,3)$, the slope is~$-1$, while at~$(4/3,8/3)$ the slope
  is~$4/5$. The tangent lines are $y=6-x$ or $x+y=6$, and
  $y-8/3=(4/5)(x-4/3)$.

  \sep

\item Find the tangent line to the curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$
  at the point $(x_0,y_0)$. Show that the sum of the $x$- and the
  $y$-intercepts is~$c$, regardless of the point of tangency.
  
  \soln Implicit differentiation gives
  $\frac{1}{2}x^{-1/2}+\frac{1}{2}y^{-1/2}\frac{dy}{dx}=0$, so the
  slope of the tangent line at~$(x_0,y_0)$ is $m=-(y_0/x_0)^{1/2}$ and
  the equation is $y-y_0=m(x-x_0)$ for this value of~$m$. We want to
  find the sum of the $x$- and $y$-intercepts. Each intercept is found
  by setting a variable to~0 and solving for the other. Thus, the
  $y$-intercept is $y_0-mx_0=y_0+\sqrt{x_0y_0}$~and the $x$-intercept
  is $x_0-y_0/m=x_0+\sqrt{x_0y_0}$. The sum of the intercepts is
  \[
  (x_0+\sqrt{x_0y_0})+(y_0+\sqrt{x_0y_0})
  =x_0 + 2\sqrt{x_0y_0} + y_0 =
  (\sqrt{x_0}+\sqrt{y_0})^2=(\sqrt{c})^2=c,
  \]
  regardless of the point of tangency.

  \sep

\item Use calculus to show that the function
  $f(x)=\arctan(x)+\arctan(1/x)$ is constant. Explain this fact
  geometrically. What is the constant?

  \soln By the chain rule,
  \[
  f'(x)=\frac{1}{1+x^2} + \frac{-1/x^2}{1+1/x^2}
  =\frac{1}{1+x^2} - \frac{1}{1+x^2} =0
  \]
  for every $x\neq0$. A right triangle of base~1 and height~$x$ shows
  that the constant value of~$f$ is $\pi/2$: We have $\tan\theta_1=x$
  (or $\theta_1=\arctan x$) and $\tan\theta_2=\frac{1}{x}$ (or
  $\theta_2=\arctan\frac{1}{x}$), see figure on next page. 
  \begin{center}
    \input{rsoln4d.eepic}
  \end{center}

  \medskip

  \sep

\item Two cars start moving from the same point; the first travels
  south at 60~mph and the second travels west at 25~mph. At what rate
  is the distance between the cars increasing after two hours?
  
  \soln Let $t$~be time measured in hours, and consider coordinates
  with the origin at the cars' starting point, with the positive
  $y$-axis pointing \emph{south} (upside-down from normal maps) and
  with distances in miles.  The cars' respective positions are
  $c_1(t)=(0,60t)$ and $c_2(t)=(25t,0)$. By the Pythagorean theorem,
  the distance (in miles) between the cars after $t$~hours is
  \[
  L(t)=\sqrt{c_1^2(t)+c_2^2(t)}=t\sqrt{60^2+25^2}=65t.
  \]
  The distance between the cars increases at a steady speed of 65~mph,
  regardless of how much time has passed. (For extra practice, answer
  the same question assuming the second car departs one hour after the
  first car. In this event, the elapsed time \emph{does} matter.)

  \sep

\item A street light is at the top of a 15~foot post. A person 6~feet
  tall walks along a straight sidewalk underneath at 5~feet/sec. When
  they are 40~feet from the post, how rapidly is the length of their
  shadow increasing?
  
  \soln Assume the person walks beneath the light at $t=0$, with time
  measured in seconds. Let $x$~be the distance (in feet) to the base
  of the light pole, and let $\ell$~be the length of their shadow
  (measured from the base of the pole). Use similar triangles to find
  the length of the person's shadow, see left-hand figure on next
  page, then use related rates to find $d\ell/dt$~when $x=40$.

  \sep

\item The top of a 10-foot ladder rests against a vertical wall, and
  the base slides away from the wall at 2~feet/sec.

  \bd
  \item{(a)} When the base of the ladder is 6~feet from the wall, how
  rapidly is the top of the ladder dropping? 

  \item{(b)} When the angle between the ladder and the wall
  is~$45^\circ$, how rapidly is the angle changing? (Be careful with
  units!)
  
  \soln With time~$t$ measured in seconds, let $y(t)$~be the height of
  the top of the ladder at time~$t$, $x(t)$~the distance from the wall
  to the base of the ladder, and $\theta(t)$~the angle made by the
  wall and the top of the ladder, see right-hand figure below. The
  related rates calculations are left to you. (Note carefully that
  angles are measured in \emph{radians}, so the trig functions have
  the expected derivatives; you must convert your answer to degrees
  per second.)

  \begin{center}
    \input{rsoln4e.eepic}
    \qquad    \qquad    \qquad
    \input{rsoln4f.eepic}
  \end{center}

  \ed

  \sep

\item A car traveling 60~mph brakes to a stop in 5~seconds. If we
  have measurements of the speed as a function of time, we can use
  left- and right-hand sums to estimate the stopping distance.

  \bd
  \item{(a)} Are the left-hand sums under- or over-estimates?
  Explain. 

  \soln Over-estimates: The speed at the left-hand endpoint is larger
  than the speed at subsequent times.

  \item{(b)} Assuming the speed measurements are taken at equal
  intervals, how many measurements would be required to estimate the
  stopping distance to within one foot, that is, to make the
  difference between upper and lower estimates smaller than 2~feet?

  \soln The change in speed is $88$~ft/sec, over a time interval of
  length $5$~sec. If $n$~intervals are used (that is,
  $(n+1)$~measurements), the difference between left- and right-hand
  sums is $|f(b)-f(a)|\cdot(b-a)/n=440/n$~ft. To make this smaller
  than 2~ft, we need at least $220$~intervals, or $221$~measurements. 
  \ed

  \vfil\eject

\item If we want to estimate $\int_0^1 e^{-x^2}\,dx$ to an accuracy
  of~$0.01$ using left- and right-hand sums with equally-spaced
  points, what is the smallest number of intervals we need?
  
  \soln We want~$n$ such that $|e^0-e^{-1}|/n$~is at most~$0.02$, or
  $(1-e^{-1})/n\leq0.02$, so $n\geq32$.

  \sep
  
\item Without evaluating the integral, determine whether $\int_0^\pi
  e^x\cos x\,dx$ is positive. (Hint: The interval contains two
  half-arches of cos. If you can, express the given integral in terms
  of $\int_0^{\pi/2}e^x\cos x\,dx$.)

  \soln The integral is negative. In fact,
  $\ds\int_0^\pi e^x\cos x\,dx
  =(1-e^{\pi/2})\int_0^{\pi/2} e^x\cos x\,dx$

  \sep

\item Let $f'$~be the function below. On the grid provided, carefully
  sketch the function~$f$, assuming $f(0)=0$. Hint: The graph of
  $f$~has no corners.
  \begin{center}
    \input{review4b.eepic}
  \end{center}
  \begin{center}
    \input{rsoln4c.eepic}
  \end{center}


\ee

\end{document}
