A spherical balloon is inflated so that its volume is increasing at the rate of 3.2 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?

Answer 1

V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when r = .6 , dV/dt = 3.2

3.2 = 4π(.6)^2 dr/dt
dr/dt = 3.2/(1.44π) = appr .707

so d(diameter)/dt = 1.415 ft/min