The volume of a sphere is decreasing at a constant rate of 225 cubic centimeters per minute. At the instant when the volume of the sphere is 15 cubic centimeters.

Question

What is the rate of change of the surface area of the sphere?

Answer 1

This is related rate problem involving the a sphere's volume and surface area...

V_sphere = 4πR³/3

The rate of change of the volume with respect to time is dV/dt.

dV/dt = 4π/3 * 3R² * dR/dt = 4πR² * dR/dt

We are told the rate of volume decrease is 225 cm³/min = (dV/dt) when the radius is 15 cm.

225 = 4π(15)² * dR/dt --> dR/dt = 225/[ 4π*(15)² ] --> 1/(4π) = dR/dt

Surface Area of sphere --> S_sphere = 4πR² cm²

Following the same logic as above, we want to calculate the rate of change of the surface area by using the derivative...

dS/dt = 8πR* dR/dt cm²/min

But we know dR/dt from the rate of change of the volume...

dS/dt = 8π * (15) * 1/(4π) = 2 * 15 = 30 cm²/min