The radius of a spherical ball is increasing at a rate of 2 cm/min. At exactly what rate (in cm2/min) is the surface area of the ball increasing when the radius is 9 cm?

Question

The radius of a spherical ball is increasing at a rate of 2 cm/min. At exactly what rate (in cm²/min) is the surface area of the ball increasing when the radius is 9 cm?

Answer 1

S = surface area at time t, and r = radius at time t

Given: dr/dt = 2 cm/min

S = 4πr²

So, dS/dt = 8πr(dr/dt) = 8π(9cm)(2cm/min) = 144π cm² / min