
\chapter{Related Rates}

Suppose we have two variables $x$ and $y$ (in most story problems the
letters will be different, but for now let's use $x$ and $y$) which are
both changing with time.  A ``related rates'' problem is a problem where we
know one of the rates of change at a given instant --- say, $dx/dt$ --- and
we want to find the other rate $dy/dt$ at that instant. \index{related rates}

\smallskip

If $y$ is written in terms of $x$, i.e., $y=f(x)$, then this is easy
to do using the chain rule:
\[
\frac{dy}{ dt}=\frac{dy}{ dx}\cdot\frac{dx}{ dt}.
\]
That is, find the derivative of $f(x)$, plug in the value of
$x$ at the instant in question, and multiply by the given value of
$\dot{x}=dx/dt$ to get $\dot{y}=dy/dt$.

\smallskip

But often $x$ and $y$ will be related in some other way, for example
$x=f(y)$, or $F(x,y)=$const, or perhaps $F(x,y)=G(x,y)$, where $F(x,y)$
and $G(x,y)$ are expressions involving both variables.  In all cases, you
can solve the related rates problem by taking $\frac{d}{ dt}$ of both sides,
plugging in all the known values (namely, $x$, $y$, and $\dot{x}$), and
then solving for $\dot{y}$.


\smallskip

To summarize, here are the steps in doing a \index{related rates problem} related rates problem when
there is a simple direct relation in the two variables $x,y$.

\begin{enumerate}

\item Decide what the two variables are.
\item Find an equation relating them.
\item Take $d/dt$ of both sides.
\item Plug in all known values at the instant in question.
\item Solve for the unknown rate.

\end{enumerate}

Sometimes it may not be convenient or desirable to find the equation between
$x,y$ directly. They might in turn be related to each other by a chain of variables and conditions. We will explain such cases in the examples below.


% Example 1
\begin{example}
 
The radius of a circular puddle is $3$ m, and it is increasing at the rate of
$1$ cm/min.  How fast is the puddle's area increasing?  \index{related rates: puddle problem}

\smallskip

Here the two things that change with time are: the radius $r$ and the area
$A$.  We know $dr/dt=1$ cm/min $=0.01$ m/min.  We want to find $dA/dt$.
Thus, in this problem $r$ is playing the role of $x$, and $A$ is playing
the role of $y$.  The relation between the two variables is: $A=\pi r^2$.
Taking $d/dt$ of both sides, we obtain
\[
\frac{dA}{ dt}=\pi\cdot 2r\frac{dr}{ dt}.
\]
Notice that, since we are taking derivatives with respect to $t$, the
derivative of $r^2$ is $2r\frac{dr}{ dt}$, by the rule for the derivative of
a power of a function (the $u^n$-rule).  In this case we alternately could
have used the chain rule for the variables $A$, $r$ and $t$:
\[
\frac{dA}{ dt}=\frac{dA}{ dr}\cdot\frac{dr}{ dt}=2\pi r\cdot\frac{dr}{ dt}.
\]
We now plug in the information given.  At the instant in question, $r=3$,
and $dr/dt=0.01$.  Thus, $A=2\pi\cdot 3\cdot 0.01=0.1885$ m${}^2$/min.


\smallskip

{\bf Note:} A crucial point to notice in this and similar problems is that,
even though a numerical value is given for $r$, $r$ must be regarded as a
variable, not a constant.  {\it The value of the variable at the instant in
question cannot be plugged into the equation until} {\bf after} {\it taking
$d/dt$ of the equation}.

\end{example}


% Example 2
\begin{example}

You are inflating a spherical balloon at the rate of $7$ cm${}^3$/sec.  How
fast is its radius increasing at the instant when $r=4$? How fast
is the area increasing at the same time? \index{related rates: balloon problem}

\smallskip

First we answer the first question.

Here the variables are the radius $r$ and the volume $V$.  We know $dV/dt$,
and we want $dr/dt$.  The two variables are related by means of the
equation $V=\frac{4}{ 3}\pi r^3$.  Taking $d/dt$ of both sides gives
$dV/dt=4\pi r^2\cdot dr/dt$.  We now substitute the values we know at the
instant in question: $7=4\pi 4^2\cdot dr/dt$.  Solving for $dr/dt$, we
obtain $dr/dt=7/(64\pi)=0.0348$ cm/sec.

Now the second question.

The formula for the area is $A=4\pi r^2$. Taking $d/dt$ of both sides, we
get $dA/dt = 4\pi 2r dr/dt= 8\pi r dr/dt $.

Now that we have worked out the answer to $dr/dt$, we can easily deduce
that 
$$dA/dt = 8 \pi (4) \frac{7}{64\pi} = 7/2 = 3.5.$$

{\bf Note:} If only the second question was asked, then we would have first
written down the two well known formulas:
$$V= 4 \pi/3 r^3 ~,~ A = 4 \pi r^2.$$
and tried to find a direct relation between $V$ and $A$. The relation
is $36\pi v^2 = a^3$. Then we would try to solve this for $A$ and take
derivatives, or take $d/dt$ of both sides and finally arrive at the
same result. 

The point is that it is not necessary to eliminate $r$ right away. It is
far easier to take $d/dt$ of the two relations first and get equations 
$$dV/dt = 4 \pi r^2 dr/dt ~,~ dA/dt = 8 \pi r dr/dt \mbox{~so~}2dV/dt = r dA/dt 
\mbox{~or~} dA/dt = (2/r) dV/dt.$$

Our final answer is now clear!

\end{example}


% Example 3
\begin{example}

A plane is flying at 500 mph at an altitude of 3 miles in a direction away
from where you're standing (i.e., the point on the ground directly beneath
the plane is moving away from you).  How fast is the plane's distance from
you increasing at the moment when the plane is flying over a point on the
ground 4 miles from you?  \index{related rates: airplane problem}

\smallskip
\begin{rightindent}{4.0 truein}
To see what's going on, we first draw a triangle whose \index{hypotenuse} hypotenuse is the
line from you to the plane (see figure at right).  The vertical leg of the
triangle is a constant --- $3$ mi --- which does not change in the course of
the problem.  The horizontal leg $x$ is the variable whose rate we know ---
its rate $\dot{x}$ is the speed of the plane.  The hypotenuse $y$ is the
variable whose rate we want.  The equation relating $x$ and $y$ is the
Pythagorean theorem: $x^2+3^2=y^2$.  We next take $d/dt$ of both sides,
getting $2x\cdot \dot{x}=2y\cdot\dot{y}$.  At the instant in question 
\end{rightindent}
\hfill
%--- figure MA --------------------------------------------------
\begin{psfigure}{1.45in}{1.45in}{figMA.ps}
\puttext{0, -40.}{\Small$x$}
\puttext{-6.67, 6.67}{\Small$y$}
\puttext{46.7, 0}{\Small$3$}
\end{psfigure}
%------------------------------
we're told that $x=4$
and $\dot{x}=500$.  What about $y$? From the \index{Pythagorean theorem} Pythagorean theorem, we have:
$y=\sqrt{4^2+3^2}=5$ at the instant in question.  Thus, we obtain: $2\cdot
4\cdot 500=2\cdot 5\cdot\dot{y}$, and solving for $\dot{y}$ gives the value
400 mph. That is, the plane is receding from you at 400 mph.  (Until you
think about this for a while, it might seem counterintuitive that, although
the side $y$ is greater than the side $x$, its rate of increase is less.)

\end{example}



% Example 4
\begin{example}
\index{related rates: water in cone problem}
Water is poured into a conical container at the rate of $10$ cm${}^3$/sec.
The cone points directly down (see side view at right), and it has a height
of $30$ cm and a base radius of $10$ cm.  How fast is the water level rising
when the water is $4$ cm deep (at its deepest point)? 

\smallskip

\noindent\begin{rightindent}{3.6in}
In this problem, the water forms a conical shape within the big cone whose
height and base radius --- and hence also its volume --- are all increasing
as water is poured into the container.  This means that we actually have
three things varying with time: the water level $h$ (the height of the cone
of water), the radius $r$ of the circular top surface of water (the base
radius of the cone of water), and the volume of water $V$.  And we have the
relation $V=\frac{1}{ 3}\pi r^2h$.  We know $dV/dt$, and we want $dh/dt$.  At
first something seems to be wrong: we have a third variable $r$ whose rate
we don't know.  
\end{rightindent}
\hfill
%--- figure MB --------------------------------------------------
\begin{psfigure}{1.85in}{1.9in}{figMB.ps}
\puttext{-3,-20}{\Tiny$h$}
\puttext{6,-3}{\Tiny$r$}
\puttext{15, 43}{\Tiny$10$}
\puttext{40, 0}{\Tiny$30$}
\end{psfigure}
%------------------------------
But the dimensions of the cone of water must have the same
proportions as those of the container.  In other words, $r$ and $h$ fit
into the above diagram of similar triangles, from which it follows that
$\frac{r}{ h}=\frac{10}{ 30}$, i.e., $r=h/3$.  So we can eliminate $r$ from the
problem entirely: $V=\frac{1}{ 3}\pi(h/3)^2h=\frac{\pi}{ 27}h^3$.  We now take
$d/dt$ of both sides and then plug in $h=4$ and $dV/dt=10$, obtaining
$10=\frac{\pi}{ 27}3\cdot 4^2\cdot\frac{dh}{ dt}$.  Thus, $dh/dt=90/16\pi=1.79$
cm/sec.

{\bf Note:}
Again we can take the lazy approach and write the equations:
$$V=\frac{1}{ 3}\pi r^2h \mbox{~and~} r=h/3 $$
that we noticed above. Instead of eliminating the variable $r$ we could have
just taken separate derivatives first and then eliminated $dr/dt$.
Of course, here the second equation is so easy to solve (it is already solved for $r$), we did eliminate $r$ right away in our solution. 

In general, the lazy method is better, unless the elimination is very easy.

\end{example}

% Example 5
\begin{example}
\index{related rates: swing problem}
A swing consists of a board at the end of a $10$ ft long rope.  Think of the
board as a point $P$ at the end of the rope, and let $Q$ be the point of
attachment at the other end.  Suppose that the swing is directly below $Q$
at time $t=0$, and is being pushed by someone who walks at the speed of $6$
ft/sec from left to right.  Find (a) how fast the swing is rising after $1$
sec; (b) the angular speed of the rope in deg/sec after $1$ sec.

\smallskip
\begin{rightindent}{3.5in}
In doing this problem, we must start out by asking: What is the geometric
quantity whose rate of change we know, and what is the geometric quantity
whose rate of change we're being asked about?  Note that the person pushing
the swing is moving {\bf horizontally} at a rate we know.  In other words,
the {\bf horizontal} coordinate of $P$ is increasing at $6$ ft/sec.  In the
$xy$-plane let us make the convenient choice of putting the origin at the
location of $P$ at time $t=0$, i.e., a distance $10$ directly below the point
$Q$ of attachment.  Then the rate we know is $dx/dt$, and in part 
\end{rightindent}
\hfill
%--- figure MC --------------------------------------------------
\begin{psfigure}{1.95in}{2.0in}{figMC.ps}
%\putcrosshair{0,0}
\puttext{-6.67, 46.7}{\Small$Q$}
\puttext{5,16}{\Small$\theta$}
\puttext{46.7, -40.}{\Small$P$}
\puttext{20., -26.7}{\Small$x$}
\puttext{-6.67, -36.7}{\Small$y$}
\puttext{33.3, 0.}{\Small$10$}
\end{psfigure}
%------------------------------
(a) the rate we want is $dy/dt$ (the rate at which $P$ is rising).  In part
(b) the rate we want is $\dot{\theta}=d\theta/dt$, where $\theta$ stands for
the angle in radians through which the swing has swung from the vertical.
(Actually, since we want our answer in deg/sec, at the end we must convert
$d\theta/dt$ from rad/sec by multiplying by $180/\pi$.)

\smallskip

(a)~From the diagram above we see that we have a right triangle whose legs
are $x$ and $10-y$, and whose \index{hypotenuse} hypotenuse is 10.  Hence $x^2+(10-y)^2=100$.
Taking $d/dt$ of both sides --- and recalling that the derivative of the
square of a function of $t$ is twice the function being squared times the
time derivative of that function --- we obtain:
$2x\dot{x}+2(10-y)(0-\dot{y})=0$.  We now look at what we know after $1$ sec,
namely $x=6$ (because $x$ started at 0 and has been increasing at the rate
of $6$ ft/sec for $1$ sec), $y=2$ (because we get $10-y=8$ from the Pythagorean
theorem applied to the triangle with hypotenuse 10 and leg 6), and
$\dot{x}=6$.  Putting in these values leaves us with $2\cdot 6\cdot
6-2\cdot 8\dot{y}=0$, from which we can easily solve for $\dot{y}$, getting
$\dot{y}=4.5$ ft/sec.

\smallskip

(b)~Here our two variables are $x$ and $\theta$, so we want to use the same
right triangle as in part (a), but this time to relate $\theta$ to $x$.
Since the hypotenuse is constant (equal to $10$), the best way to do this is
to use the sine: $\sin\theta=x/10$.  Taking $d/dt$ of both sides, we obtain
$(\cos\theta)\dot{\theta}=0.1\dot{x}$.  At the instant in question ($t=1$
sec), when we have a right triangle with sides 6-8-10, the $\cos\theta$ in
this equation is $8/10$.  We also substitute $\dot{x}=6$ on the right.  We
get $0.8\dot{\theta}=0.6$, i.e., $\dot{\theta}=0.6/0.8=0.75$ rad/sec $=43$
deg/sec.

\end{example}

Sometimes there are \index{several variables} several variables that change with time, you know the
rates of all but one of them, and you want to know the remaining rate.  As
in the case when there are just two variables, take $d/dt$ of both sides of
the equation relating all of the variables, and then plug in all of the
known values and solve for the unknown rate.
In other words, our lazy method is the recommended method!

% Example 6
\begin{example}
\index{related rates: two roads crossing problem}
A road going South--North crosses a road going West--East at the point O.
Car A is driving North along the first road, and car B is driving East
along the second road.  At time $t$ car A is $a(t)$ to the North of O, and
car B is $b(t)$ to the East of O.  Suppose that at some instant you know
the values of $a(t)$, $b(t)$, the velocity $v_1(t)=\dot{a}(t)$ of car A,
and the velocity $v_2(t)=\dot{b}(t)$ of car B. Find a formula 

\medskip

\begin{rightindent}{3.0in}
\noindent
in terms of these known values for the rate at which
the distance between the two cars is increasing.

Let $c$ be the distance from car A to car B.  By the Pythagorean Theorem,
$c^2=a^2+b^2$.  We now take $d/dt$ of both sides of the Pythagorean
Theorem, obtaining $2c\dot{c}=2a\dot{a}+2b\dot{b}$.  Dividing by $2c$ gives
the following formula for the unknown rate:
\[
  \dot{c}=\frac{a\dot{a}+b\dot{b}}{ c}= \frac{av_1+bv_2}{\sqrt{a^2+b^2}}.
\]
\smallskip

\
\end{rightindent}
\hfill
%--- figure MD --------------------------------------------------
\begin{psfigure}{2.0in}{2.0in}{figMD.ps}
\puttext{23.3, -26.7}{\Small $B= (b(t),0)$}
\puttext{-43.3, 16.7}{\Small $A = (0,a(t))$}
\puttext{-20., -20.}{\Small $O$}
\puttext{0., 46.7}{\Small North}
\puttext{0., -46.7}{\Small South}
\puttext{-46.7, -10.}{\Small West}
\puttext{46.7, -10.}{\Small East}
\end{psfigure}
%------------------------------

\medskip

Notice how this problem differs from Example 12.3.  In both cases we took
$d/dt$ of the \index{Pythagorean Theorem} Pythagorean Theorem.  However, in Example 12.3 one of the sides
was a constant (the altitude of the plane), and so $d/dt$ of the square of
that side was simply zero.  In Example 12.6, on the other hand, all three
sides of the right triangle are variables.  As always in a story problem,
it's important to read the problem carefully enough to determine at the
start what are the variables and what are the constants.

\end{example}


