## 5.4 Related Rates

(This topic is also in Section 5.4 in Applied Calculus or Section 13.4 of Finite Mathematics and Applied Calculus)

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## Related Rates

If Q is a quantity that is varying with time, we know that the derivative measures how fast Q is increasing or decreasing. Specifically, if we let t stand for time, then we know the following.

Rate of Change of a Quantity
Rate of change of Q= dQdt

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

W=
 1t
- 4t2 ,
where t is time in seconds. At time t = 10 seconds, the amount of fuel in the launcher is:

 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
=2r
=2r
 drdt
=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 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)

 Find the area the rate of change of area the radius the rate of change of radius given that the area the rate of change of area the radius the rate of change of radius = 12, at the instant when the area the rate of change of area the radius the rate of change of radius = 10.

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

The derived equation is

=2r
 drdt

Substituting the above values and solving for the unknown gives:
 drdt
=
 35
cm/sec.
=240cm2/sec.
 drdt
=
 512
cm/sec.
=120cm2/sec.

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

 Solving a Related Rates Problem Step 1: Identify the changing quantities, possibly with the aid of a sketch. Step 2: Write down an equation that relates the changing quantities. Step 3: Differentiate both sides of the equation with respect to t. 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 . . . ." Last Step: Substitute the given values in the derived equation you obtained above, and solve for the required quantity.