Example CD-7.1 The Fast Lighthouse

A lighthouse is 1 mile off a straight shore with its beacon revolving 3 times per minute. How fast does the beam of light sweep down the beach at the points that are 2 miles from the point closest to the lighthouse?

The "chain" in this exercise is that the angle of the beacon depends on time and the distance to the point of contact down the beach depends on the angle. We will assume that the light revolves clockwise in the diagram Figure CD-7.1. We want to know the rate of change of distance with respect to time. We introduce the variables shown in the diagram Figure CD-7.1:

Figure CD-7.1: A lighthouse beam sweeps the shore

The link between the angle and the distance is

The link between the time and the angle can be expressed as an explicit function, but what is important is to know the rate of change,

The question in this problem is

SOLUTION: Distance as an explicit function of time is

This formula tells us the speed of the light's motion in terms of . Notice that is not the independent variable t, but instead is the link variable or output from the first function in the chain. We could use the first function to express the speed in terms of t, but this is neither necessary nor desirable (unless we want to know when the light gets to the point 2 miles down the beach.) In fact, it is not even necessary to know the derivative at any other time; it need not be constant as long as we know that .

The specific position when D=2 is shown in Figure CD-7.2.

Figure CD-7.2: The angle when D=2

The Pythagorean Theorem applied to Figure CD-7.2 tells us that the hypotenuse of the right triangle shown is . SOH-CAH-TOA tells us that

Finally, when D=2

Example CD-7.2 The Lightspeed Lighthouse

The beam of light in the previous example moves down the beach faster and faster, but real light can only move at the speed of light. The Scientific Project on the Lightspeed Lighthouse examines the example taking into account that the light must travel from the lighthouse to the shore at the speed of light.

Example CD-7.3 Implicit Solution of Exercise CD-7.1

After you solve Exercise CD-7.1, you should compare your solution to the following implicit method. We begin with the equation

The total differential of the equation is

When (a,b,c)=(1,-2,1), the only root of the original equation is x=1. In this case, the implicit differential breaks down

Figure CD-7.3: x²+bx+1=0

Of course, this degeneracy is only a hint of the trouble in the original implicit equation. If we change b to -1.99, there are no real solutions to the original problem. The equation "branches" here between the positive discriminant and negative one, and . The vertical tangent we just computed is the tangent touching these two branches shown in Figure CD-7.3.

Differentiating the explicit function x[b] is more complicated, as you saw in Exercise CD-7.1:

Example CD-7.4 Implicit Solution of the Balloon Exercise CD-7.5.1

In Exercise CD-7.5.1, you calculate the rate of change of surface area of a balloon that is being blown up so that its volume is increasing at the rate of 2 cubic inches per second. We want to know how the rate of change of volume effects the rate of change of surface area. In the exercise, we ask you to solve explicitly for surface area as a function of volume. The implicit solution of that problem does not require that we find the explicit function.

Here is the implicit solution of the problem:

We solve the volume differential for dr, and substitute into the surface differential, , obtaining

Example CD-7.5 The Fast Ladder

A ladder of length L rests against a vertical wall. The bottom of the ladder is pulled out horizontally at a constant rate r (m/sec). What is the vertical speed of the tip that rests against the wall? We introduce the variables shown in Figure CD-7.4:

Figure CD-7.4: Ladder and wall

The fact that the length of the ladder is fixed together with the Pythagorean Theorem gives us the implicit linkage between x and y:

The total differential of the implicit equation is

Example CD-7.6 A Limiting Case

As we pull the base of the ladder toward the point x=L away from the wall, what happens to the speed with which the other tip moves down the wall? We can see from the diagram, Figure CD-7.5, that when , . The formula for the vertical speed tends to minus infinity.

Humm, this calculation is a little mysterious. Here is another way to look at it: Solve the equation x²+y²=L² for and substitute into the expression for vertical speed,

1. A child is blowing up a balloon by adding air at the rate of 2 cubic inches per second. Well before it bursts, its radius is 6 inches. Assume that the balloon is a perfect sphere and find the surface area S as an explicit function of volume V. Use your explicit function to say how fast the surface is stretching at this point.

   HINTS: Give variables for surface area, volume, time, and whatever else you need. Solve for S in terms of V. The "chain" in this exercise is that volume depends on time (through the child's efforts), V=V[t] and surface area depends on volume, S=S[V]. We want to know how area changes with time, , where S=S[V[t]]. (See review Exercise CD-28.3.2 to express surface area as an explicit function of volume.)

2. The Explicit Fast Ladder
   Solve the equation x²+y²=L² for . Use the fact that x=x[t] is a function of time together with the Chain Rule to show that
   
   Verify that this model predicts that as x approaches L, the speed of the tip resting on the wall tends to infinity.

   The result of the previous exercise is wrong for a real ladder. The tip of such a ladder cannot go faster than the speed of light. There is a physical condition that this simple mathematical model does not take into account. The real falling ladder is explored in the Scientific Projects Chapter on Mechanics. It uses Galileo's Law of Gravity from Chapter 9.

3. Related Rates Drill
   
   • 1. Each edge of a cube is expanding at the rate of 3 inches per second. How fast is the volume expanding at the point where the volume equals 64 cubic inches?
   • 2. Each edge of a cube is expanding at the rate of 3 inches per second. How fast is the surface area expanding at the point where the volume equals 64 cubic inches?
   • 3. A 6-foot man walks away from an 8-foot lamp at the rate of 5 feet per second. How fast is his shadow growing at the point where he is 7 feet from the lamp?
   • 4. A snowball melts at a rate proportional to its surface area, that is, it loses volume per unit time in this proportion. Say the constant of proportionality is k. What is , the rate of change of radius with respect to time?
   • 5. Your snow-cone has melted filling the conical holder with sticky liquid. The liquid is dripping from the point at the bottom at the rate of 1 liter per hour. If the cone is 7 centimeters high and 5 centimers in diameter at the top, what is the rate of change of the height of the liquid when half of it has leaked out? (Note: The volume of a cone of height h and base radius r is .)