   Chapter 3.9, Problem 6E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 80 mm?

To determine

To find: The rate of change of the volume of the sphere where the radius of the sphere is increasing at a rate of 4 mm/s and the diameter of the sphere is 80 mm.

Explanation

Given:

The diameter of the sphere is 80 mm.

The rate of change of the radius of the sphere is 4 mm/s.

Formula used:

(1) Chain rule dydx=dydududx

(2) Volume of the sphere of radius r, V=43πr3.

Calculation:

Let V be the volume and r be the radius of the sphere respectively.

Therefore, the volume is, V=43πr3.

The volume of the sphere increases as the time t increases.

Therefore, the radius of the sphere also increases as the time t increases.

Thus, the volume V and the radius r are the functions of the time variable t.

Obtain the derivative dVdt.

Differentiate the volume V with respect to the time variable t.

dVdt=ddt(43πr3)=(43π)ddt(r3)

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