\chapter{Related Rate Volume Problems}
\index{Related Rate Volume Problems} 
Suppose we have a container that is described by rotating some function $x=
g(y)$ around the $y$-axis between $y=c$ and $y=d$, and suppose that the
container is filled with water to a height $h$.  Then the volume of water
is given by the definite integral 
\[
V(h)=\int_c^{c+h} \pi g(y)^2dy.
\] 
We
write $V(h)$ to indicate that the volume of the water is a function of its
depth $h$.

\smallskip

In this situation one type of question we might want to answer is: given
the rate at which water is being pumped into the container, what is the
rate at which the water level is rising?  In symbols this says: given
$dV/dt$, find $dh/dt$.  Alternately, we might know $dh/dt$ (by observing
how fast the water level is rising) and want to find $dV/dt$.  These are
called ``related rate'' problems.  According to the chain rule, the two
rates $dV/dt$ and $dh/dt$ are related as follows: 
\[
\frac{dV}{ dt}=\frac{dV}{ dh}\cdot\frac{dh}{ dt}.
\]

\smallskip

Since $V(h)$ is given as an integral with $h$ in the upper limit of
integration, we can use the ``fundamental theorem of calculus'' to find
$dV/dh$, {\bf without ever having to actually compute an integral}.  To see
how to do this, let's review what the fundamental theorem says, using the
present notation.  Suppose we have a function defined by an integral
\[
      V=\int_c^{u(h)}f(y)dy.
\]
Then 
\[
       \frac{dV}{ du}=f(u(h))
\]
(this says that {\em
the derivative of an integral with respect to the upper limit of
integration is obtained by removing the integral sign and substituting the
upper limit of integration in place of the dummy variable in the
integrand}).  Thus
\[
    \frac{dV}{ dh} =\frac{dV}{ du}\cdot\frac{du}{dh}
                    =f(u(h))\frac{du}{dh}\,.
\]

\smallskip

In our situation with the volume of revolution we have $u(h)=h+c$, so that
$du/dh=1$, and the integrand $f(y)$ is $\pi g(y)^2$.  Thus, the fundamental
theorem tells us that 
\[
\frac{dV}{ dh}=\pi g(c+h)^2,\qquad\text{and~so}\qquad
\frac{dV}{ dt}=\pi g(c+h)^2\frac{dh}{ dt}.
\]

Another way to say this is as follows: {\it the ratio of volume rate to
height rate is the cross-sectional area}, i.e., the area of the top
surface of the liquid at the instant in question.  This principle also
applies when the liquid is filling a container whose cross-sections are
not necessarily circles.

\smallskip

\begin{example}
A container is formed by rotating the function $x=4\sin(\pi
y/12)$ around the $y$-axis between $y=0$ and $y=6$ (the units are feet).
At a certain instant the container is filled to a depth of 4 ft, and water
is being poured in at the rate of 5 ft${}^3$/sec.  How fast is the water
level rising?

\smallskip

It's generally a good idea (even if not absolutely essential) to \index{draw a
picture} draw a
picture of what we're dealing with.  As explained earlier % Reference?
 in the
section on sinusoidal functions, the graph of $x=4\sin(\pi y/12)$ is
obtained from the graph of $x=\sin y$ by stretching in the $x$-direction by
a factor of 4 and in the $y$-direction by a factor of $12/\pi$ (so that its
period expands from $2\pi$ to 24):


\begin{center}
%-figure CA -----------------------------
\begin{psfigure}{2in}{2in}{figCA.ps}
%\putcrosshair{0,0}
\puttext{-30., -25.}{\small$\pi$}
\puttext{0, -25.}{\small$x = \sin y$}
\puttext{-15,-55}{\small$1$}
\end{psfigure}
%------------------------------
\hspace{0.5 truein}
%-figure CB -----------------------------
\begin{psfigure}{2in}{2in}{figCB.ps}
%\putcrosshair{0,0}
\puttext{37.5, 10.}{\small$\displaystyle x = 4\sin\left(\frac{\pi}{12}
y\right)$}
%\puttext{-27.5, -5.}{\small$6$}
\puttext{-27.5, 40.}{\small$12$}
\puttext{7., -52.5}{\small$4$}
\end{psfigure}\\
%------------------------------
\end{center}

\medskip

\noindent
Taking this function from $y=0$ to
$y=6$ means taking the part until $x$ reaches its first maximum at the
point (4,6).  (See the picture at the right).  According to the above
formula (here $c=0$), we have 
\[
\frac{dV}{dt} =\pi\bigl(4\sin(\pi h/12)\bigr)^2\frac{dh}{dt}\,.
\] 
At the instant in question $h=4$ and
$dV/dt=5$, so we have 
\begin{gather*}
5=16\pi\sin^2(\pi/3)\frac{dh}{dt}
=16\pi(0.5\sqrt{3})^2\frac{dh}{dt}= 12\pi\frac{dh}{dt} \,,\\
\intertext{so that}
\frac{dh}{dt} = \frac{5}{12\pi} =
0.1326\frac{\rm ft}{\rm second}=1.59\frac{\rm inch}{\rm second}.
\end{gather*}
 
\smallskip

In this problem, notice that we never had to evaluate an integral.  If we
had instead been asked to find the volume of water in the container when
$h=4$, then we would have had to compute $\pi\int_0^4 16\sin^2(\pi
y/12)\,dy$.  At this point we do not have a technique for finding the
anti-derivative of $\sin^2$ (later we will get a method for doing this type
of integral).  The point is: there is a big difference between asking about
the {\it rates} at which the height and volume are increasing and asking
about the actual values of the height and volume.  In the latter case you
need to evaluate an integral, whereas in the former case you do not.

\end{example}

\bigskip

\newpage

%
