The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of increase of its surface area, when the radius is 2 cm.

Solution:

Let ‘V’ be the volume, ‘A’ be the surface area and ‘r’ be the radius of the sphere.

Given, dV/dt = 3 cm²

Rate of increase of surface area of the sphere when r = 2 cm is dA/dt.

We have volume of sphere, V = 4/3 ℼr³

Differentiate w.r.t ‘t’

dV/dt = 4/3 × ℼ × 3 × r² dr/dt

3 = 4ℼ (2)² dr/dt

dr/dt = 3 /16ℼ cm/sec

Surface area of sphere, A = 4ℼr²

Differentiate w.r.t ‘t’

dA/dt = 4ℼ × (2r) dr/dt

dA/dt = 4ℼ × (2×2) × (3/16ℼ)

dA/dt = 3 cm²/sec

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The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of increase of its surface area, when the radius is 2 cm.

Summary:

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. The rate of increase of its surface area, when the radius is 2 cm, is 3 cm²/sec.