Question

Assume that the radius of a sphere is expanding at a rate of 80 cm/min. The volume of a sphere is 4/3 pi r^3 and its surface area is 4 pi r^2. Determine the rate at which the surface area is changing with respect to time when r=40 cm.

Look for dA/dt= ??? cm^2/min

Answer 1

Let radius of a sphere r and the surface area S.

dr/dt = 80 cm/min

And the surface area is

S = 4πr^2

Differentiate the equation as respect to t, then

dS/dt = 8πr × dr/dt

For r=40cm

dS/dt = 8π×40cm×80cm/min=2560π cm^2/min