air is being pumped into a spherical balloon at a rate of 8 cubic feet per minute.

Question

find the rate at which the surface area of the balloon is changing at the instant the radius is 3 feet

Answer 1

Volume flow rate, dV/dt = 8 cubic feet per minute

Radius, r = 3 feet

Volume of the spherical balloon, V = (4/3)(πr³ )

Take derivative both sides with respect to t,

dV/dt = (4/3)3(πr² )dr/dt

8 = (4/3)3(π(3² )dr/dt

dr/dt = 1/36π

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Find the surface area rate.

S = 4πr²

Take derivative both sides with respect to t,

dS/dt = 8πrdr/dt

          = 8π(3)(1/36π)

          = 2/3  square feet/min

So, rate of surface area is 2/3 square feet/min