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 10-foot 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 vary 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 vary 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 varies 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 $$y(t)=\sqrt{100-x(t)^2},0 \le t \le T, \text{ where $t$ is the number of seconds elapsed at which the ladder hits the ground.}$$
So, $\dfrac{dy}{dt}$, or equivalently $y'(t)$, can be expressed entirely in terms of $x(t)$ by $$y'(t)=\frac{dy}{dt}=\frac{-1.3x(t)}{\sqrt{100-x(t)^2}},0 \le t \le T, \text{ where $t$ is the number of seconds at which the ladder hits the ground and }x(t)=1.3t.$$
We can say even more about how variations in y (height above floor) and x (distance from wall) are related.
We know that $dy$, an incremental variation in y, is $dy=y'(t)dt$. We also know that $dx$, an incremental variation in x, is $dx=x'(t)dt$.
We can then relate $dy$ and $dx$by relating the incremental variations in y and x during the same incremental variation in t: $$\begin{align}
\frac{dy}{dx}&=\frac{y'(t)dt}{x'(t)dt}\\[1ex]
&=\frac{y'(t)}{x'(t)}\\[1ex]
&=\dfrac{\left(\dfrac{-1.3x(t)}{\sqrt{100-x(t)^2}}\right)}{1.3}\\[1ex]
&=\dfrac{-x(t)}{\sqrt{100-x(t)^2}}\\[1ex]
&=\dfrac{-1.3x}{\sqrt{100-(1.3x)^2}}\end{align}$$
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 varying 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 variations in x to variations in y. We did this by thinking of variations 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 varies with respect to time so that she can program a variable-rate 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 varies at 0.8 meters/min, it starts at 1.2 meters, and it ends at 2.5 meters. It will therefore take $\frac{2.5-1.2}{0.8}=1.625$ 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
$$\begin{align} V(t)&=0.5\left(\text{Floor area enclosed by outer wall}-\text{Floor area enclosed by inner wall}\right)\\[1ex] &=0.5\left( \pi \cdot x(t)^2-\pi\cdot\left(\sqrt{x(t)}\right)^2\right)\\[1ex]
&=0.5\pi\left(x(t)^2-x(t)\right), 0 \le t \le 1.625\end{align}$$ 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
The solution to Example 2 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 vary. 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 gravity-resistant 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 with his gun in 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 varying when the water's volume is 0.15 $\text{m}^3$?
