If the volume of a spherical ball is increasing at the rate of 4π cc/sec, then the rate of increase of its radius (in cm/sec), when the volume is 288π cc, is

Solution:

Given, volume of a spherical ball is increasing at the rate of 4π cc/sec

Volume of the spherical ball = 288π cc

We have to find the rate of increase of radius.

Volume of sphere = (4/3)πr³

288π = (4/3)πr³

72(3) = r³

r³ = 216

Taking cube root,

r = 6cm

On differentiating,

dV/dt = d[(4/3)πr³]/dt

dV/dt = (4/3)π(3r²)dr/dt

4π = 4πr²(dr/dt)

dr/dt = 1/r²

dr/dt = 1/(6)²

dr/dt = 1/36 cm/sec

Therefore, the rate of increase of radius is 1/36 cm/sec

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If the volume of a spherical ball is increasing at the rate of 4π cc/sec, then the rate of increase of its radius (in cm/sec), when the volume is 288π cc, is

Summary:

If the volume of a spherical ball is increasing at the rate of 4π cc/sec, then the rate of increase of its radius (in cm/sec), when the volume is 288π cc, is 1/36 cm/sec.