Related Rates

A key part of solving related rates problems is translating the word problem to mathematical terms.

Verbal StatementMathematical Model
A gear is revolving at the rate of 30 revolutions per minute ( 1 rev = 2pi rad).A = Angle of revolution dA/dt = 30(2pi) rad/min
Water is being pumped into a swimming pool at the rate of 12 cubic feet per minute.V = Volume of water in the pool dV/dt = 12 ft3/min
The velocity of a car after traveling 2 hours is 50 miles per hour.x = Distance traveled dx/dt = 50 when t = 2

Follow the steps below for solving related rates problems.

 1 Identify all given quantities and quantities that must be found. Make a sketch and label the quantities. 2 Write an equation involving the variables whose rates of change either are given or are to be found. 3 Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 4 Substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change.

Example:

Situation: You're sitting at your computer trying to remember all of the things about Calculus that you should know. You then see a red balloon inflating on your screen. You some how find out that air is being pumped into the round red ballon at 4.5 cubic inches per minute. You job is to find the rate of change of the radius when the radius is 2 inches.

Solution: Let V be the volume of the ballon and let r be its radius. Since the volume is increasing at the rate of 4.5 cubic inches per minute, you know that at time t the rate of change of the volume is dV/dt = 9/2.

Identify all quantities
Given rate = dV/dt = 9/2
Find dr/dt = when r = 2

Write an equation involving the variables
Volume of a Sphere V = 4/3pi r3

Differentiate with respect to time t
dV/dt = 4pi r2 dr/dt
dr/dt = 1/(4pi r2) (dV/dt)

Subtitute all values for the variables and their rates of change
dr/dt = (1/(16pi)) (9/2) which = .09 in./min.

If you're wondering where that red round balloon you saw is click here!