5.4 Related Rates

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Related Rates

The weight (in kg.) of rocket fuel in a rocket launcher is given by

W

=

1 t - 4 t2 ,

	decreasing at a rate of 0.002 kg. per second			increasing at a rate of 0.002 kg. per second
	increasing at a rate of 0.8 kg. per second			not changing at all

In a related rates problem, we are given the rate of change of certain quantities, and are required to find the rate of change of related quantities.

For the text, we have developed a simple, step-by-step approach to solving related rates problems, which we shall illustrate with an example, similar to Exercise 9 in Section 5.5 of Applied Calculus, or Section 12.5 inFinite Mathematics and Applied Calculus.

Example
The area of a circular doggie puddle is growing at a rate of 12 cm²/s. How fast is the radius growing at the instant when it equals 10 cm?

Step 1: Identify the changing quantities, possibly with the aid of a sketch.

Here, the changing quantities are:

	derivative of the area			radius of a disc
	time			area of a disc

Here is a little sketch of the puddle showing the changing quantities.

Note At this stage, we do not substitute values for the changing quantities. That comes at the end.

Step 2: Write down an equation that relates the changing quantities.

A formula that relates the area A and radius r is:

=

Enter both sides of the equation and press "Check." Use standard computer format; for example, write the equation r = A³+2 as

r = pi*A^3 + 2

(Spaces are optional. (Press here to see more examples.)

Step 3: Differentiate both sides of the equation with respect to t.

Taking (d/dt) of both sides yields

	0 = 2r			dA dt = 2r
	dA dt = 2r dr dt			dA dr = 2r

Step 4: Go through the whole problem and restate it in terms of the quantities and their rates of change. Rephrase all statements regarding changing quantities using the phrase "the rate of change of . . . ."

The original problem reads:

The area of a circular doggie puddle is growing at a rate of 12 cm²/s. How fast is the radius growing at the instant when it equals 10 cm?

A valid restatement of the problem is: (make the appropriate selections)

Last Step: Substitute the given values in the derived equation you obtained above, and solve for the required quantity.

The derived equation is

dA dt = 2r

dr dt

Substituting the above values and solving for the unknown gives:

	dr dt = 3 5 cm/sec.			dA dt = 240 cm2/sec.
	dr dt = 5 12 cm/sec.			dA dt = 120 cm2/sec.

Here is a summary of the steps we used in solving the related rates problem.

Here is a "ladder problem."
 

Joey is perched precariously the top of a 10-foot ladder leaning against the back wall of an apartment building (spying on an enemy of his) when it starts to slide down the wall at a rate of 4 ft per minute. Joey's accomplice, Lou, is standing on the ground 6 ft. away from the wall. How fast is the base of the ladder moving when it hits Lou?