%% Math 131-03 Final Exam Solutions, Fall 2003

\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,amsfonts,mathrsfs}
\usepackage{epic,eepic,pstcol,comment}
%%\usepackage{floatflt}

\usepackage{math131}

\pagestyle{myheadings}
\markright{Math 131-03, Final Exam Solutions}

%% \setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}

\def\Answer#1{$\phantom{.}$\hfill
        #1\quad\fbox{$\phantom{\Big|Put answer here}$}}
\def\ShortAnswer#1{$\phantom{.}$\hfill
        #1\quad\fbox{$\phantom{\Big|answer}$}}

\def\Blank{$\underline{\phantom{Please write your name in this
        space}}$} 
\def\Strut{$\vphantom{\Big|}$}

\begin{document}

\begin{center}

\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Section 03, Final Exam Solutions} \\
Professor Hwang
\end{center}

\be

%% Differentiation freebie

\item{} [10~points each]\quad Let 
  $f(x)=e^{-x}\sin x=\ds\frac{\sin x}{e^x}$.

\bd

\item{(a)} Use the product rule to compute~$f'(x)$. Simplify your
  answer.

  \soln $f'(x)=e^{-x}(\cos x)+(-e^{-x})(\sin x)
  = e^{-x}(\cos x-\sin x)$
  
\item{(b)} Use the quotient rule to compute~$f'(x)$. Simplify your
  answer.

  \soln $\ds f'(x)=\frac{e^x(\cos x)-(\sin x)e^x}{(e^x)^2}
  = \frac{\cos x-\sin x}{e^x}$

\ed

Note well that the two methods give the same answer.

\sep

%% Linear approximation

\item{} [20 points]\quad A small portion of a graph $y=f(x)$ (with
  horizontal and vertical scales as shown) is given below. Determine
  (with justification) which of the following functions might be
  depicted, and circle your choice(s).
  \[
  f_1(x)=\sin x,\qquad\qquad
  f_2(x)=\ln (1+x),\qquad\qquad
  f_3(x)=e^x-1
  \]
  \begin{center}
    \input{final-03f.eepic}
  \end{center}

  \soln From the picture, we must have $f(0)=0$ (the graph passes
  through the origin) and $f'(0)=1$ (since the slope is~$1$). A
  short calculation shows that
  \begin{align*}
    f_1(0)&=\sin 0=0 && f_2(0)=\ln 1=0 && f_3(0)=e^0-1=0 \\
    f_1'(0)&=\cos 0=1 && f_2'(0)=\frac{1}{1+0}=1 &&
    f_3'(0)=e^0=1
  \end{align*}
  Consequently, any of the three functions could be depicted.

\vfil\eject

%% Left- and right-hand sums, approximation, accuracy

\item{} [10~points each]\quad The graph of $f(t)=2e^{-t^2/2}$ on the
  interval $[0,2]$ is shown.

\bd

\item{(a)} On the graph, sketch the left- and right-hand
  rectangles for 4~intervals of equal length, and shade the area
  representing the difference of the upper bound and the lower bound.

  \begin{center}
    \input{fsoln-03a.eepic}
  \end{center}

  
\item{(b)} Suppose left- and right-hand sums with $n$~intervals are
  used to compute upper and lower bounds for $\int_0^2 f(t)\,dt$. Find
  the smallest number,~$n$, of intervals so that
  \[
  \textrm{upper}-\textrm{lower}\leq0.02=\textstyle\frac{2}{100}
  \]

  \soln We use the formula
  $\textrm{upper}-\textrm{lower}=|f(b)-f(a)|\cdot\frac{b-a}{n}$,
  which applies to a decreasing function on~$[a,b]$, using
  $n$~intervals of equal length:
  \begin{align*}
    \textrm{upper}-\textrm{lower} &=
    |f(2)-f(0)|\cdot\frac{2-0}{n} \\
    &= |(2e^{-2}) - (2e^0)|\cdot\frac{2}{n} \\
    &= 4(1-e^{-2})\cdot\frac{1}{n}\leq \frac{2}{100}.
  \end{align*}
  Cross-multiplying, we find that $n\geq 200(1-e^{-2})\simeq 172.93$,
  so $n=173$~is the smallest number of intervals that ``works''.

\ed

\sep%%vfil\eject

%% Optimization

\item{} [20~points]\quad A rectangular poster will consist of a
  printed message, of area 96~square inches, surrounded by a border.
  The border is to be 1~inch wide along the sides and bottom, and
  2~inches wide along the top. What dimensions of the poster minimize
  its total area?

  \begin{center}
    \input{final-03g.eepic}
  \end{center}

  \soln Let $x$~and $y$ be the width and height of the \emph{printed
  area}, so that $xy=96$ and the total area of the poster is
  \[
  A=(x+2)(y+3)=xy+3x+2y+6=3x+\frac{192}{x}+102.
  \]
  Differentiating,
  \[
  \frac{dA}{dx}=3-\frac{192}{x^2}=0,\qquad\text{or}\quad
  x=8,\ y=12.
  \]
  The total width is 10~inches ($=x+2$), the total height is 15~inches
  ($=y+3$).

  \sep

%% Limits, continuity, l'Hopital's rule

\item{} Let $\ds f(x)=\frac{\cos x-1}{x}$ for $x\neq0$.

\bd

\item{(a)} [5~points]\quad What value for~$f(0)$ makes~$f$ continuous
  at $x=0$?

  \soln We want $\lim\limits_{x\to0} f(x)$. Since this limit
  is indeterminate, try l'H\^opital's rule:
  \[
  \lim_{x\to0}\frac{\cos x-1}{x}=\lim_{x\to0}-\frac{\sin x}{1}
  =-\frac{\sin 0}{1}=0.
  \]

\item{(b)} [10~points]\quad Compute $f'(x)$ for $x\neq0$.

  \soln By the quotient rule,
  \[
  f'(x)=\frac{x(-\sin x)-(\cos x-1)}{x^2}
  =\frac{-x\sin x-\cos x+1}{x^2}.
  \]

\vfil\eject

\item{(c)} [5~points]\quad Find $\lim\limits_{x\to0}f'(x)$.

  \soln Again, the limit is indeterminate. L'H\^opital's rule gives
  \[
  \lim_{x\to0}f'(x)=\lim_{x\to0}\frac{-x\sin x-\cos x+1}{x^2}
  =\lim_{x\to0}\frac{-x\cos x-\sin x+\sin x}{2x}
  =\lim_{x\to0}-\frac{\cos x}{2}=-\frac{1}{2}.
  \]

  \ed

  \sep

%% Definite integrals and signed areas, average value

\item{} [10 points each]\quad Each part of this question refers to the
  function pictured below. The areas enclosed are $A=25$, $B=16$, and
  $C=16$.

  \begin{center}
    \input{final-03b.eepic}
  \end{center}

\bd

\item{(a)} Find $\ds\int_0^3 f(x)\,dx$ (show all work for full credit)
  
  \soln The integral is the signed area enclosed by the graph, so
  \[
  \int_0^1 f(x)\,dx=25,\qquad \int_1^3 f(x)\,dx=-16.
  \]
  Splitting the interval of integration at $x=1$ gives
  \[
  \int_0^3 f(x)\,dx = \int_0^1 f(x)\,dx + \int_1^3 f(x)\,dx
  = 25-16 = 9.
  \]

\item{(b)} Find the average value of~$f$ on the interval~$[0,5]$ \\
  Suggestion: Start by writing down the average value in terms of an
  integral. 

  \ed

  \soln The average value of~$f$ on the interval~$[a,b]$ is
  $\ds\frac{1}{b-a}\int_a^b f(x)\,dx$. Here,
  \[
  \text{Average value} = \frac{1}{5}\int_0^5 f(x)\,dx
  = \frac{1}{5}(25 - 16 + 16) = 5.
  \]

  \vfil\eject

%% Related rates

\item{} The top of a 13-foot ladder rests against a vertical wall, and
  the foot of the ladder slides away from the wall at 2~ft/sec. Let
  $x$~be the distance from the wall to the foot of the ladder in feet,
  and $y$~the height of the top of the ladder in feet. Both parts of
  this question concern the instant at which the base of the ladder is
  7~ft from the wall.

  \begin{center}
    \input{final-03c.eepic}
  \end{center}

  \bd
  
\item{(a)} [5~points]\quad How high is the top of the ladder? Express
  your answer \textbf{symbolically}.

  \soln By the Pythagorean theorem, $x^2+y^2=13^2$. Since we are given
  that $x=7$, we have $y=\sqrt{13^2-7^2}=\sqrt{169-49}=\sqrt{120}$.

\item{(b)} [10~points]\quad How rapidly is the top of the ladder
  dropping? For full credit, express your answer 
  \textbf{both symbolically and numerically}, including units. No
  credit for answers off by more than a factor of~10.

  \ed

  \soln Differentiating $x^2+y^2=13^2$ with respect to~$t$ (time
  measured in seconds),
  \[
  2x\frac{dx}{dt}+2y\frac{dy}{dt}=0,\qquad
  \text{or}\quad
  \frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}
  =-\frac{7}{\sqrt{120}}\cdot2=-\frac{14}{\sqrt{120}}\simeq = -1.28
  \]
  feet per second.

  \sep

%% Derivatives and monotonicity
  
\item{} [10~points each]\quad Let $a>0$ be constant, and define
  $f(x)=(a-x)\sqrt[3]{x}$ for $x$~real.

\bd

\item{(a)} Compute $f'(x)$ and make a sign diagram. Find the critical
  point(s) of~$f$, and classify as local maximum, local minimum, or
  neither.

  \soln Multiply out first: $f(x)=ax^{1/3}-x^{4/3}$, so
  \[
  f'(x)=\textstyle\frac{a}{3}x^{-2/3}-\frac{4}{3}x^{1/3}
  =\frac{1}{3}x^{-2/3}(a-4x).
  \]
  We find that $f'(x)=0$ when $x=a/4$, and that $f'(x)$~does not exist
  when $x=0$. These are the two critical points of~$f$. Properties of
  exponents guarantee that $x^{-2/3}>0$ for $x\neq0$, so the sign
  of~$f'(x)$ is the same as the sign of $(a-4x)$, namely is positive
  when $x<a/4$ and negative when $x>a/4$. The point $x=0$ is neither a
  local minimum nor maximum, while $x=a/4$ is a local maximum.

\item{(b)} Compute $f''(x)$ and make a sign diagram. Find the
  inflection point(s) of~$f$, if any.

  \soln Differentiating the \emph{expanded} first derivative,
  \[
  f''(x)=\textstyle\frac{2a}{9}x^{-5/3}-\frac{4}{9}x^{-2/3}
  =\frac{2}{9}x^{-5/3}(a+2x).
  \]
  Note that the largest negative power of~$x$ has been factored out in
  each case, and that only the expanded form is differentiated (to
  avoid the product rule). The potential inflection points are $x=0$
  (where $f''(x)$~does not exist) and $x=-a/2$ (where $f''(x)=0$). It
  is easiest to make separate sign diagrams for the corresponding
  factors: $x^{-5/3}$~changes sign at $x=0$, while $(a+2x)$ changes
  sign at $x=-a/2$. Consequently, $f''(x)$~changes sign at both values
  of~$x$, so both give rise to inflection points.

  \ed

  \sep

%% Implicit differentiation

\item{} [10~points each]\quad A curve is defined implicitly by the
  equation $x^3-3xy^2=2$.

\bd

\item{(a)} Find $\ds\frac{dy}{dx}$ in terms of $x$~and $y$

  \soln Implicit differentiation and the product rule give
  \[
  3x^2-3y^2-3x\Bigl(2y\frac{dy}{dt}\Bigr)=0,\qquad\text{or}\quad
  \frac{dy}{dx}=\frac{x^2-y^2}{2xy}.
  \]

\item{(b)} Use part~(a) to find the equation $y=mx+b$ of the line
  tangent to the curve at the point $(x_0,y_0)=(2,1)$, and sketch
  this line.

  \soln Plugging in $(x,y)=(2,1)$ gives the slope of the tangent
  line:
  \[
  m=\frac{(2)^2-(1)^2}{2(2)(1)}=\frac{3}{4}.
  \]
  The point-slope form is $y-1=\frac{3}{4}(x-2)$, or
  $y=\frac{3}{4}x-\frac{1}{2}$.
  
  \vspace*{-1.5in} \parbox[t]{3.125in}{To sketch the line, note that
    the curve has horizontal tangents when $x^2=y^2$, or $x=\pm y$.
    Plugging back into the original equation, $x^3-3xy^2=-2x^3=2$, or
    $x^3=-1$. The only real solution is $x=-1$, which gives the two
    points $(-1,1)$~and $(-1,-1)$. Since the two points with
    horizontal tangent are separated by two grid squares, each square
    is $1\times1$, the origin is as marked, and the tangent line (with
    the point of tangency) is as indicated.}
  \input{fsoln-03e.eepic}

  \ed

\ee

\end{document}

