Section
7.4 Related Rates
Several exercises in Exercise
Set 6.3.9 asked you to use the chain rule to answer questions about
situations that were modeled with composite functions. Exercise 6.3.9.7
(Mischievous Monkey) is repeated below as Figure 7.4.1. A 10foot ladder
leans against a wall; the ladder's bottom slides away from the wall at a
rate of 1.3 ft/sec after the monkey kicks it.
The question was, "At what rate does the ladder's height against the wall
change with respect to the number of seconds elapsed since the monkey kicked
the bottom?"
Figure 7.4.1. A ladder falls after a
mischievous monkey kicks its bottom.
In Section 6.3.9 you were directed to express y, the distance
between the floor and the ladder's top, as a function of x, the
distance between the wall and the ladder's foot. You then used the chain
rule to derive a rate of change function $r_y(t)$ for y(t).
Another way to approach this question is to model the situation using the
Pythagorean Theorem, letting y and x each vary with
time. Then:
$$\sqrt{x^2+y^2}=10,$$
which, not surprisingly, says that the ladder's length is always 10 feet.
Since x and y are both greater than or equal to 0, we
will make life easier when finding rate of change functions by
writing$$x^2+y^2=100,$$which says that for all values of x and y
from 0 to 10, the sum of the square of the ladder's distance from the wall
and the square of the ladder's height on the wall is always 100.
Because $x^2+y^2$ is always 100, the sum does not change as values of x
and y vary. Therefore, the rate of change of the sum will be the
rate of change of 100.
$$\begin{align}\frac{d}{dt}\left(x^2+y^2\right)&=\frac{d}{dt}{100}\\[1ex]&=0.\end{align}$$
But, by the chain rule,
$$\frac{d}{dt}\left(x^2+y^2\right)=2x\frac{dx}{dt}+2y\frac{dy}{dt}.$$
We know that $\dfrac{dx}{dt}=1.3$ because the ladder's foot is moving at 1.3
ft/sec, so
$$\begin{align}\frac{d}{dt}\left(x^2+y^2\right)&=2x(1.3)+2y\frac{dy}{dt}\\[1ex]&=0.\end{align}$$
Solving for $\dfrac{dy}{dt}$ in $2x(1.3)+2y\dfrac{dy}{dt}=0$, we get
$$\frac{dy}{dt}=\frac{1.3x}{y}.$$
In other words, the rate at which the distance between the floor and the
ladder's top changes with respect to time, at any moment in time, depends on
the how far the ladder's top and bottom are from the point at which the wall
meets the floor.
You might wonder about expressing $\dfrac{dy}{dt}$ in terms of x
and y. Where is time? Actually, time is hidden within x
and y. They are both functions of time. We would be more explicit
about the relationships we are modeling were we to
write,$$\frac{dy}{dt}=\frac{1.3x(t)}{y(t)}.$$We can then relate the
functions x and y as $$y(t)=\sqrt{100x(t)^2},0 \le t
\le T, \text{ where $T$ is the number of seconds elapsed at which the ladder
hits the ground.}$$
So, $\dfrac{dy}{dt}$, or equivalently $y'(t)$, can be expressed entirely in
terms of $x(t)$ by
$$y'(t)=\frac{dy}{dt}=\frac{1.3x(t)}{\sqrt{100x(t)^2}},0 \le t \le T,
\text{ where $T$ is the number of seconds at which the ladder hits the
ground and }x(t)=1.3t.$$
We can say even more about how changes in y (height above floor)
and x (distance from wall) are related.
We know that $dy$, an incremental change in $y$, is $dy=y'(t)dt$.
We also know that $dx$, an incremental change in $x$, is
$dx=x'(t)dt$. We can then relate $dy$ and $dx$ by
relating the incremental changes in $y$ and $x$ during
the same incremental change in $t$:
$$\begin{align}
\frac{dy}{dx}&=\frac{y'(t)dt}{x'(t)dt}\\[1ex]
&=\frac{y'(t)}{x'(t)}\\[1ex]
&=\dfrac{\left(\dfrac{1.3x(t)}{\sqrt{100x(t)^2}}\right)}{1.3}\\[1ex]
&=\dfrac{x(t)}{\sqrt{100x(t)^2}}\\[1ex]
&=\dfrac{1.3x}{\sqrt{100(1.3x)^2}}\end{align}$$
Summary
We were given a situation that involved two rates of change—the
rate of change of y with respect to time and the rate of change
of x with respect to time. By modeling the situation in terms of
x and y each changing as a function of time, and by
using the chain rule, we were able to determine the rate of change of y
with respect to time because we were given an actual value for the rate of
change of x with respect to time.
We were able to do even more.
Once we had the rates of change of both x and y with
respect to time, we were then able to relate changes in x to
changes in y. We did this by thinking of changes in both x
and y as happening during the same increment in time, allowing
us to derive the rate of change of y with respect to x.
Note that we could just as easily have derived the rate of change of x
with respect to y.
Situations that involve two rates of change that are related are called,
quite naturally, related rates situations.
Example 2: A decorative pool
The animation in Figure 7.4.2 shows an architect's vision for a decorative
pool that she will propose for the Montreal Trade Center. The pool's water
will be contained within two circular walls that expand and contract. The
walls will be 0.5 meters high.
When the radius of the outer wall is x meters, the radius of the
inner wall will be $\sqrt x$ meters. The outer wall varies in radius from
$x=1.2$ meters to $x=2.5$ meters at a constant rate of 0.8 meters/min.
The architect's problem is that she must know at what rate to pump water
into and out of the pool at each moment in time so that the pool remains
completely filled as the walls' radii vary. In other words, she must
determine at what rate the pool's volume changes with respect to time so
that she can program a variablerate pump that will put water into the pool
and draw water from the pool so that the pool remains completely filled at
all moments in time.
Figure 7.4.2. An architect's vision for
a dynamic decorative pool.
Solution
Since the outer wall expands and contracts at 0.8 meters/min,
we can focus on the phase when the pool expands.
Let $x=x(t)$ be the radius of the outer wall in meters, $1.2
\le x \le 2.5$. Then the radius of the inner wall is $\sqrt {x(t)}$
meters. Moreoever, $x'(t)=0.8$.
The outer wall's radius changes at 0.8 meters/min, it starts at 1.2
meters, and it ends at 2.5 meters. It will therefore take
$\frac{2.51.2}{0.8}=1.625$ minutes to go from 1.2 meters to 2.5 meters,
so t varies from 0 to 1.625 minutes while the tank is
expanding. The outer wall's radius will be $x(t)=1.2+0.8t, 0 \le t \le
1.625$.
The pool's volume will be the difference between the volume of the right
cylinder having radius x meters and the volume of the right
cylinder having radius $\sqrt x$ meters. The volume of any right
cylinder is $\left(\text{Area of base} \cdot \text{Height}\right)$.
Therefore, the pool's volume at each moment in time that it expands is
given by the function V, defined as
$$\begin{align}
V(t)&=0.5\left(\text{Floor area enclosed by outer wall}\text{Floor
area enclosed by inner wall}\right)\\[1ex]
&=0.5\left( \pi \cdot
x(t)^2\pi\cdot\left(\sqrt{x(t)}\right)^2\right)\\[1ex]
&=0.5\pi\left(x(t)^2x(t)\right), 0 \le t \le 1.625\end{align}$$
The pool's rate of change of volume with respect to time at values of t
from 0 to 1.625 minutes will therefore be
$$\begin{align}
V'(t)&=0.5\pi\left(2x(t)x'(t)x'(t)\right)\\[1ex]
&=0.5\pi\left(2x(t)(0.8)0.8\right)\\[1ex]
&=0.5\pi(0.8)\left(2x(t)1\right)\\[1ex]
&=0.5 \pi (0.8)\left(2(1.2+0.8t)1\right)\\[1ex]
&=\pi (0.64t+0.56)\end{align}$$
So, t minutes after the outer wall begins to expand, the pump
must put water into the pool at the rate of $V'(t)=\pi(0.64t+0.56)$
$\mathrm{m^3/min}$ in order that the pool remain completely full at each
moment that it expands.
Exercise Set 7.4

Produce a complete solution to Example 2. The solution that was
developed in the example addressed only the phase where the outer wall
expands. Derive a rate of change function $r_V$ that will have water
pumped at the correct rate throughout 24 hours of operation. Take into
consideration that the outer wall will remain stationary for 0.5
minutes after expanding and for 0.5 minutes after contracting.

The diagram below shows a kite 100 ft above the ground moving
horizontally away from you at a rate of 8 feet per second. The kite’s
height does not change. The string between you and the kite plays out
as the kite moves away from you.
 Let the function L give the string's length as a
function of the number of seconds since the kite was over you
vertically. Use the Pythagorean Theorem to derive $r_L$, the
function that gives the rate of change of the string's length with
respect to time.
 At what rate is the string unraveling after 20 seconds have
elapsed since it was directly above you?

At what rate is the string unraveling with with respect to the
kite's horizontal distance between it and you?

Amalia, an officer of the Arizona Highway Patrol, wants to park her
car 150 feet from a highway so that motorists will not see it while
approaching. However, this creates a problem. Her radar gun registers
the speed at which a car is approaching the officer's location, which
in Amalia's case will not be the speed that the car is traveling along
the highway.
What Amalia needs is a function that she can put into her laptop so
that when she enters the car's speed according to the radar gun it
will give her the car's actual speed along the highway. Amalia will
aim her gun at a point 400 feet up the highway, measured from the
point on the highway that is aligned perpendicularly with her car.
Provide Amalia with the function that she needs.
 A red car (R) and a green car (G) are traveling on two roads that
intersect, making an angle of 1.35 radians. R is 100 km from the
intersection going 73 km/hr. G is at the intersection going 58 km/hr. In
how many hours will the cars be closest together?

A camera operator will film a rocket launch. The rocket accelerates
from liftoff at a rate of 45 $\mathrm{\dfrac{m/s}{s}}$. Let $\theta$
be the measure of the angle that the camera makes with the horizontal
direction and with the current position of the rocket's center. Define
a function that gives the rate at each moment in time since launch
that the camera operator must rotate his camera to keep it focused on
the rocket?
 Paintballs leave the barrel of a paintball gun at a speed of 90 meters
per second. James' gun shoots gravityresistant paint balls (they do not
fall to the ground). James mounted his gun on a stand that rotates at a
rate of 0.2 radians/sec. The gun is 50 meters from a straight,
infinitely long wall.
James started rotating his gun from an initial direction parallel to the
wall, rotating toward it. The gun fired balls as it turned,
shooting a ball every 0.01 seconds (his gun holds a lot of
paintballs).
 Let $\theta$ be the measure of the gun's angle from its initial
direction. Show that there is an interval $a\lt \theta\lt b$ where
from the perspective of an observer standing far behind the gun
(away from the wall), the paint balls are hitting the wall at
positions that move in the opposite direction of the gun's rotation.
 A lighthouse sits on an island that is 2 km from a straight,
infinitely long coast. Argue by analogy with Part (a) that as the
lighthouse's beam rotates, there is some part of the rotation in
which an observer from space would see the light beam hitting the
coast at positions that move in the opposite direction of the beam's
rotation. (Use c to represent the speed of light.)

The rate at which a spherical snowball melts with respect to time is
proportional to the snowball's surface area at moments in time.
Wilhelm made a spherical snowball having a radius of 10 cm.
 What is the rate of change of the snowball's volume with respect to
time at each moment in time?

What is the rate of change of the snowball's radius with respect to
time at each moment in time?
 A right parallelepiped (rectangular) tank filled with water has a base
area of 0.75 $\text{m}^2$ and a height of 0.25 m. The tank has a
circular drain of radius 2 cm in its bottom.
The rate at which the amount of water exits the tank depends on the the
water's height in the tank: $r_V(t)=\text{Area}_\text{hole}\cdot
\sqrt{19.6h(t)}$.

Define a function that gives the volume of water in the tank as a
function of the water's height.
 Define a function that gives the volume of water in the tank as a
function of the elapsed time since the drain is opened.

At what rate with respect to time is the tank draining when the
water's height is 0.12 meters?

At what rate with respect to time is the water's height changing
when the water's volume is 0.15 $\text{m}^3$?
 How long does it take the tank to drain?