If the volume of a spherical ballon is increasing at 4 pi m^3/ min when its radius is 2m, how fast is its surface are increasing at the same time
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
using the given:
4π = 4π(4) dr/dt
dr/dt = 4π/(4π(4)) = 1/4
SA = 4πr^2
d(SA) = 8πr dr/dt
= 8π(2)(1/4)
= 4π m^2/min