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\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Solutions to Worksheet on Linear Approximation
\end{center}

\noindent The area and volume of a sphere of radius~$r$ are given by
$A=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$.  Use linear approximation to
answer the following questions.

\soln The linear approximation formulas give
\begin{align*}
\Delta A&\simeq\frac{dA}{dr}\Delta r=2(4\pi r)\Delta r, &
\Delta V&\simeq\frac{dV}{dr}\Delta r=4\pi r^2\Delta r. \\
\intertext{Dividing these equations by the respective formulas for
  area and volume,}
\frac{\Delta A}{A}&\simeq 2\frac{\Delta r}{r}, &
\frac{\Delta V}{V}&\simeq 3\frac{\Delta r}{r}.
\end{align*}
The same formulas may be deduced immediately from the fact that area
and volume depend upon the radius by power formulas whose exponents
are respectively 2~and 3. In words, the second pair of formulas says
that a given percent change in the radius of a sphere gives rise to
(roughly) twice that percent change in the area and three times that
percent change in the volume.

\be

\item If the radius is increased by~1\%, the area changes by
  about~2\%.



\item If the radius is increased by~5\%, the area changes by
  about~10\%.



\item If the radius is increased from $4$~cm to $4.2$~cm, by exactly
  how much does the area change? Is this consistent with your results
  above? 
  
  \soln The change in radius is $\Delta r=4.2-4=0.2$~cm, a percent
  change of $\frac{\Delta r}{r}=\frac{0.2}{4}=0.05$, or~5\%. The exact
  change in area is
  \[
  \Delta A = 4\pi(4.2)^2 - 4\pi(4)^2 = 4\pi(4.2^2-4^2)
           = 4\pi(1.64)~\textrm{cm}^2,
  \]
  a percent change of exactly 
  $\frac{\Delta A}{A}=\frac{4\pi(1.64)}{4\pi(16)}=0.1025$,
  or~$10.25$\%. This is consistent with our rough approximation, which
  predicted a change of~10\%.



\hrulefill

\item If the radius is increased by~1\%, the volume changes by
  about~3\%.



\item If the radius is increased by~4\%, the volume changes by
  about~12\%.



\item If the radius is increased from $20$~cm to $20.8$~cm, by exactly
  how much does the volume change? Is this consistent with your
  results above?

  \soln A calculation similar to that above shows that
  \[
  \frac{\Delta r}{r}=\frac{0.8}{20}=0.04,\qquad
  \frac{\Delta V}{V}=\frac{20.8^3-20^3}{20^3}=\frac{998.912}{8000}
    =0.124864,
  \]
  or about $12.4$\%. This is consistent with our approximation,
  which predicted a~12\% change.



\hrulefill

\item If the volume is increased by~1\%, by about what percent does
  the area change?
  
  \soln The area of a sphere is proportional to the $2/3$ power of the
  volume (see below), so by the general reasoning mentioned earlier, a
  1\%~change in volume gives rise to a change of roughly~$2/3$\% in
  the area.



\item If the volume is increased by~3\%, by about what percent does
  the area change?

  \soln $3\times(2/3)=2$\%.



\item Describe at least two different approaches to the last two
  questions. 

  \soln One possibility is to solve for the radius of a sphere as a
  function of volume, then to plug into the area formula; since
  $r\propto V^{1/3}$, we find that $A\propto V^{2/3}$. The chain rule
  provides a second approach:
  \[
  \Delta A\simeq\frac{dA}{dV}\Delta V
  = \frac{dA}{dr}\frac{dr}{dV}\Delta V,
  \]
  then to find the percent change in area in terms of the percent
  change in volume. Yet another method is to use the formulas found
  above: 
  \[
  \frac{\Delta A}{A}\simeq 2\frac{\Delta r}{r},\quad
  \frac{\Delta V}{V}\simeq 3\frac{\Delta r}{r},\qquad\text{so}\quad
  \frac{\Delta A}{A}\simeq \frac{2}{3}\frac{\Delta V}{V}.
  \]

\ee

\eject

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