Example 1: Air is being pumped into a spherical balloon such that its radius increases at a rate of .75 in/min. Find the rate of change of its volume when the radius is 5 inches.
The volume ( V) of a sphere with radius r is
Differentiating with respect to t, you find that
The rate of change of the radius dr/dt = .75 in/min because the radius is increasing with respect to time.
At r = 5 inches, you find that
hence, the volume is increasing at a rate of 75π cu in/min when the radius has a length of 5 inches.
Example 2: A car is traveling north toward an intersection at a rate of 60 mph while a truck is traveling east away from the intersection at a rate of 50 mph. Find the rate of change of the distance between the car and truck when the car is 3 miles south of the intersection and the truck is 4 miles east of the intersection.
- Let x = distance traveled by the truck
- y = distance traveled by the car
- z = distance between the car and truck
The distances are related by the Pythagorean Theorem: x 2 + y 2 = z 2(Figure 1) .
The rate of change of the truck is dx/dt = 50 mph because it is traveling away from the intersection, while the rate of change of the car is dy/dt = −60 mph because it is traveling toward the intersection. Differentiating with respect to time, you find that
hence, the distance between the car and the truck is increasing at a rate of 4 mph at the time in question.
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