<?xml version="1.0" encoding="UTF-8"?>
<commit>
  <added type="array"/>
  <modified type="array">
    <modified>
      <diff>@@ -635,4 +635,22 @@ Prove that the total number of maxima and minima possessed by a third-order poly
 is at most two.
 \end{hwwithsoln}
 
+\begin{hwwithsoln}{pyramidal-tent}
+Euclid proved that the volume of a pyramid equals $(1/3)bh$, where $b$ is the area of its
+base, and $h$ its height. A pyramidal tent without tent-poles is erected by blowing air
+into it under pressure. The area of the base is easy to measure accurately, because the
+base is nailed down, but the height fluctuates somewhat and is hard to measure accurately.
+If the amount of uncertainty in the measured height is plus or minus $e_h$, find the
+amount of possible error $e_V$ in the volume.
+\end{hwwithsoln}
+
+\begin{hwwithsoln}{rocket-height}
+A hobbyist is going to measure the height to which her model rocket rises at the peak of
+its trajectory. She plans to
+take a digital photo from far away and then do trigonometry to determine the height,
+given the baseline from the launchpad to the camera and the angular height of the rocket
+as determined from analysis of the photo. Comment on the error incurred by the inability to
+snap the photo at exactly the right moment.
+\end{hwwithsoln}
+
 \end{hwsection}</diff>
      <filename>ch01/ch01.tex</filename>
    </modified>
    <modified>
      <diff>@@ -374,16 +374,6 @@ for terms up to order $x^2$.
 \end{hw}
 
 \begin{hw}
-Expand $(1+x)^{1/3}$ in a Taylor series around $x=0$. The value $x=28$ lies outside
-this series' radius of convergence, but we can nevertheless use it to
-extract the cube root of 28 by recognizing that $28^{1/3}=3(28/27)^{1/3}$.
-Calculate the root to four significant figures of precision, and check it
-in the obvious way.
-\end{hw}
-
-\pagebreak
-
-\begin{hw}
 In classical physics, the kinetic energy $K$ of an object of mass $m$ moving at velocity $v$
 is given by $K=\frac{1}{2}mv^2$. For example, if a car is to start from a stoplight and then
 accelerate up to $v$, this is
@@ -407,6 +397,14 @@ the classical expression was inferred.
 \pagebreak
 
 \begin{hw}
+Expand $(1+x)^{1/3}$ in a Taylor series around $x=0$. The value $x=28$ lies outside
+this series' radius of convergence, but we can nevertheless use it to
+extract the cube root of 28 by recognizing that $28^{1/3}=3(28/27)^{1/3}$.
+Calculate the root to four significant figures of precision, and check it
+in the obvious way.
+\end{hw}
+
+\begin{hw}
 Find the Taylor series expansion of $\log_2 x$ around $x=1$, and use it to evaluate
 $\log_2 1.0595$ to four significant figures of precision. Check your result by using the
 fact that 1.0595 is approximately the twelfth root of 2. This number is the ratio of</diff>
      <filename>ch05/ch05.tex</filename>
    </modified>
    <modified>
      <diff>@@ -226,6 +226,7 @@ $k$ are irrelevant, and can be set to 1. We then have
 Plugging in $r=\pm 1$, we get a positive result, which confirms that the concavity is upward.
 
 \hwsolnhdr{prove-n-extrema}
+
 Since polynomials don't have kinks or endpoints in their graphs, the maxima and minima must be points where the
 derivative is zero. Differentiation bumps down all the powers of a polynomial by one, so 
 the derivative of a third-order polynomial is a second-order polynomial. A second-order polynomial
@@ -234,6 +235,25 @@ quadratic formula. (If the number inside the square root in the quadratic formul
 there could be less than two real roots.) That means a third-order polynomial can have at most two
 maxima or minima.
 
+\hwsolnhdr{pyramidal-tent}
+
+Considering $V$ as a function of $h$, with $b$ treated as a constant, we have for the slope
+of its graph
+\begin{align*}
+  \dot{V} &amp;= \frac{e_V}{e_h} \qquad ,\\
+\intertext{so}
+  e_V &amp;= \dot{V}\cdot e_h \\
+      &amp;= \frac{1}{3}b e_h
+\end{align*}
+
+\hwsolnhdr{rocket-height}
+
+Thinking of the rocket's height as a function of time, we can see that goal is to measure
+the function at its maximum. The derivative is zero at the maximum, so the error incurred
+due to timing is approximately zero. She should not worry about the timing error too much.
+Other factors are likely to be more important, e.g., the rocket may not rise exactly
+vertically above the launchpad.
+
 \beginsolutions{2}
 
 \hwsolnhdr{fourth-power}</diff>
      <filename>ch99/hwans.tex</filename>
    </modified>
  </modified>
  <removed type="array"/>
  <parents type="array">
    <parent>
      <id>56dddfdee02d6df5d2a11dd5d1a89bc47775b067</id>
    </parent>
  </parents>
  <author>
    <name>Ben Crowell</name>
    <email>githubcrowell09@lightandmatter.com</email>
  </author>
  <url>http://github.com/bcrowell/calculus/commit/fe02eeeee2add45c71c28c7ed6464acc40a7abc0</url>
  <id>fe02eeeee2add45c71c28c7ed6464acc40a7abc0</id>
  <committed-date>2009-04-15T15:25:53-07:00</committed-date>
  <authored-date>2009-04-15T15:25:53-07:00</authored-date>
  <message>before renumbering chapters</message>
  <tree>4f827012c0818bdbd4c549cef4b826e922cf8c5e</tree>
  <committer>
    <name>Ben Crowell</name>
    <email>githubcrowell09@lightandmatter.com</email>
  </committer>
</commit>
