A spherical balloon is being inflated at a rate of 8pi cm^3/s At what rate is the surface area of the balloon increasing at the instant that the radius of the balloon is 1cm?

Question

A spherical balloon is being inflated at a rate of 8pi cm^3/s At what rate is the surface area of the balloon increasing at the instant that the radius of the balloon is 1cm?

Answer 1

While a couple different approaches could work here, it is probably easiest to use the given information to calculate the instantaneous rate of change of radius with respect to time, dr/dt, then use that to calculate the instantaneous rate of change of surface area with respect to time, dA/dt.

We are given the rate of change of volume, dV/dt, as well as info about the instant in question (r = 1), so we want to set up the equation that gives volume as a function of radius, then differentiate with respect to time:

V = 4/3πr³

dV/dt = 4πr²dr/dt

8 = 4π·dr/dt

dr/dt = 2/π cm/sec

SA = 4πr²

dA/dt = 8πr·dr/dt

dA/dt = 16 cm²/sec