Air is being pumped into a spherical balloon so that its volume increases at a rate of 50cm3/s. How fast is the surface area of the balloon increasing when its radius is 11cm? Recall that a ball of radius r has volume V=43πr3 and surface area S=4πr2.

Answer 1

dS/dt = 8πr dr/dt
dV/dt = 4πr^2 dr/dt
so, dV/dt = r/2 dS/dt
dS/dt = dV/dt * 2/r

The given data thus mean that

dS/dt = 500*2/11 cm^2/min