Unit 3 - Objective 3 - Related Rates


Example 1

The derivative also represents rate of change. Do the following for related rate problems:


Try it!

The voltage of a certain thermocouple as a function of temperature is given by

E = 2.800T + 0.006T²

If the temperature is increasing at the rate of 0.5° C/min, how fast is the voltage increasing when T = 100°C?

Given: E = 2.8T + 0.006T²

Find: when T=100° C

Solution: E = 2.8T + 0.006T²

use implicit differentiation

= 2.8 + 2 (0.006) T

Now substitute what is given

= 2.8 (0.5) + 2 (0.006) (100) (0.5) = 1.4 + 0.6 = 2



Example 2

To solve related rate problems (after translating "given" and "to find"):

  1. Write an equation which relates the variables whose derivatives are given and those whose derivatives are asked for.
  2. Differentiate each term of that equation implicitly with respect to t (time).
  3. Substitute in given values.
  4. Solve for expression asked for.


Try it!

Problem 1:

A spherical balloon is being blown up at a constant rate of 2 ft³/min. Find the rate at which the radius is increasing when the radius is 3 feet.

Rate means derivative, so translation becomes:
Given: = 2 ft³/min

(ft³ means volume)

Find: = ? when r = 3 ft

First we need an equation relating v and r (see inside front cover of text):
1. Equation for volume of sphere:
2. Differentiate equation (with respect to t)
3. Substitute in given
4. Solve for dr/dt


Problem 2:

A 15 foot ladder is sliding down the side of a vertical building. If the top of the ladder is moving at the rate of .25 ft/min, how fast is the foot of the ladder moving when it is 9 feet from the building?

Given: dx/dt = -0.25 ft/min (negative because the ladder is moving down the wall)
Find: dy/dt when y = 9

x² + y² = 15²

differentiate (with respect to t)

2x + 2y = 0

You need to find the value for x before you substitute into this equation:

x² + 9² = 15²
x² + 81 = 225
x² = 144
x = 12

(only positive values apply)

Now since:


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