\documentclass[12pt]{article}
%\usepackage{graphicx}
%\topmargin -.75in
\textheight 11in \textwidth 6.75in \oddsidemargin -.35in
\pagestyle{empty} \hoffset .5in
\input{pdfcolor}
\begin{document}
\begin{center}
{\bf\underline{\huge{{Quiz 3}}}}
\end{center}
\noindent Davis \hspace{4in} Name:
\noindent M211 \hspace{4in} Pledge:
\vspace{.2in}
\reversemarginpar
\noindent 1. Consider \marginpar{(10pts.)}a rod of length 1,
with density $\delta (x) = 1 + kx$, where $k$ is a positive
constant. Suppose the rod is lying on the positive $x$-axis with
one end at the origin.
\begin{description}
\item[a.] Find the center of mass as a function of $k$.
\Red $\overline{x} = \frac{\int_0^1 x(1+kx) dx}{\int_0^1 (1+kx)
dx} = \frac{\frac{1}{2} + \frac{k}{3}}{1+\frac{k}{2}}$ \Black
\item[b.] How far to the right can the center of mass be?
\Red The formula for center of mass above increases as $k$ gets
bigger, and the limit as $k \rightarrow \infty$ is $\frac{2}{3}$,
so the center of mass cannot be any further right than
$\frac{2}{3}$. \Black
\end{description}
%\vspace{2in}
\noindent 2. You are \marginpar{(10pts.)}digging a hole for a
rectangular swimming pool 20 feet by 10 feet by 5 feet deep. The
dirt is more compact the deeper you go, and the density is $\delta
(x) = x+5$ pounds per cubic foot when you are $x$ feet below
ground level. How much work do you need to do to dig this hole?
\Red Take a ``slab'' of dirt at depth $x$ and width $\Delta x$:
it has volume $20(10)\Delta x$ and weight $20(10)\Delta x (x+5)$.
This slab is moved $x$ feet up (measuring $x$ from ground level),
so the work done on this slab is $20(10)\Delta x (x+5) x$. When we
add the other slabs, we get an approximation of work done as
$\sum_i 20(10)\Delta x (x_i+5) x_i$ (I have put the subscript $i$
in the equation to indicate that the slabs come from different
depths). Letting the number of intervals go to $\infty$ turns
this into an integral, $\int_0^5 20(10) x(x+5) dx = 20,833.3$
foot-pounds of work.
\Purple (YOU MAY IGNORE THIS LAST PART IF YOU DON'T WANT TO WORRY
ABOUT MEASURING THE DISTANCE FROM THE BOTTOM OF THE POOL!)
If you
measured distance from the bottom of the pool, I think it is best
to introduce a different variable than $x$ to measure this so you
avoid confusion with the variable in the density function (I will
use $h$). The volume of a slab at height $h$ is still
$20(10)\Delta h$ (note that I have changed the width of the slab
to $\Delta h$). The weight per cubic foot at height $h$ is the
same as the weight per cubic foot at depth $5-h$, and we know the
formula for the weight per cubic foot at depth $x$ is $\delta(x) =
x+5$. Thus, in terms of $h$, the weight per cubic foot at depth
$5-h$ is $(5-h) + 5 = 10-h$. Multiplying by volume yields
$20(10)\Delta h (10-h)$ pounds for the slab. This slab is moved a
distance of $5-h$, yielding work of $20(10)\Delta h (10-h) (5-h)$.
We turn this into a Riemann sum and then an integral as before,
and we get the integral $\int_0^5 20(10)(10-h) (5-h)dh = 20,833.3$
foot-pounds of work.
%\vspace{2in}
\end{document}