The volume of a spherical balloon is increasing at a constant rate of 3 in3/s. At what rate is the radius of the balloon increasing at the instant its radius is 2 inches?

0.0597in/s

Step-by-step explanation:

The volume of the spherical balloon V= 4/3πr³

r is the radius of the balloon

The rate at which the volume is changing dV/dt is expressed as;\

dV/dt = dV/dr*dr/dt

dV/dr = 3(4/3πr²)

dV/dr = 4πr²

dV/dr = 4π(2)²

dV/dr = 16π

dV/dt = 3 in³/s

Substitute inti the formula

3 = 16πdr/dt

dr/dt = 3/16π

dr/dt = 3/50.24

dr/dt = 0.0597in/s

Hence the balloon is increasing at the rate of 0.0597in/s

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step-by-step explanation: