related rates (2 formulas)

Question

The volume of a sphere is increasing at a constant rate of 139 cubic feet per second. At the instant when the volume of the sphere is 15 cubic feet, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation V=34​πr3 and the surface area can be found with S=4πr2. Round your answer to three decimal places (if necessary).

Answer 1

V = (4/3)πr³ ft³

S= 4πr² ft²

r = ((3/4)V/π)^1/3

dV/dt = 4πr²(dr/dt) = 139 ft³/s

dr/dt = (dV/dt)/(4πr²)

dS/dt = 2πr (dr/dt) = 2πr(dV/dt)/(4πr²) = (dV/dt)/(2r) = 139/(2r)

when V= 15 ft³

r = ((3/4)15/π)^1/3 ≈ 1.5299 ft

dS/dt = 139/30.598 = 45.428 ft²/s