RELATED RATES


A "related rates" problem is a problem which involves at least two changing quantities and asks you to figure out the rate at which one is changing given sufficient information on all of the others. For example, as two vehicles drive in different directions we should be able to deduce the speed at which they are separating if we know the individual speeds and directions.

There are several key steps in setting up and solving related rates problems. They are

  1. Draw a picture or otherwise make a mathematical model of the situation.

  2. Label all quantities which can change as variables.

  3. Identify the underlying independent variable in the problem. This is usually time but it need not be.

  4. Identify in terms of the variables and their derivatives what is being asked and what is given.

  5. Use the mathematical model to write down a relation among the variables.

  6. Differentiate this relation with respect to the underlying independent variable, usually making heavy use of the chain rule.

  7. Solve for the quantity wanted.

  8. Go back over your work and write up a presentable solution.

Example:

One vehicle starts driving north and one vehicle starts driving west from an intersection (as indicated by the arrows above). At the time the first vehicle is .3 miles north the intersection it is traveling at 20 mph. Simultaneously the second vehicle is .4 miles west of the intersection and is traveling at 25 mph. At that instant, how fast are the two vehicles separating?

Solution:

The mathematical model of this situation is a right triangle whose legs are increasing with time. If we call the two legs x and y and the hypoteneuse z, then all three quantities are changing. The underlying independent variable in the problem is time which we will call t. Note that x, y, and z are really functions x(t), y(t), and z(t). We are asked to find dz/dt at a particular time t0. We are given the values of x and y at this time as well as the values of dx/dt and dy/dt at this time. The relation among the variable x, y, and z, using the Pythagorean theorem, is that

x2 + y2 = z2.

If we differentiate this with respect to t we obtain

2 x dx/dt + 2 y dy/dt = 2 z dz/dt.

We note that we were given four of the six quantities (x, y, dx/dt, and dy/dt) and want dz/dt. Missing is only the quantity z, which we can compute using

(.3)2 + (.4)2 = z2.

The result is that z = .5, and hence

2 (.3) (20) + 2 (.4) (25) = 2 (.5) dz/dt.

After doing the arithemetic,

dz/dt = 32 mph.

Answer:At the instant in question, the distance between the vehicles is increasing at 32 miles per hour.


The Practice area contains six related rates problems for you to solve. Solutions are given to Problems 1 and 2 -- just click on the button.

Practice With Related Rates


/Stage9/Lesson/relatedRates.html

COVER CQ DIRECTORY HUB CQ RESOURCES

© CalculusQuestTM
Version 1
All rights reserved---1996
William A. Bogley
Robby Robson