The radius r of a sphere is increasing at a rate of 8 inches per minute. (a) Find the rate of change of the volume when r = 8 inches. in.3/min (b) Find the rate of change of the volume when r = 38 inches. in.3/min

a) 2421.743 in³/ min , b) 54437.52 in³/ min

Explanation: The formulae for the volume of a sphere is given below as

V = (4π/3) ×r³ where V =volume of sphere and r = radius of sphere.

By taking the time derivative of the formulae, we have the rate of change of volume with time (we do so by using implicit diffrenciation) , hence we have that

dV/dt = (4π/3) × 3r²×dr/dt

Where dV/dt = rate of change of volume and dr/dt= rate of change of radius.

A)

r = 8 inches, dr/dt = 3 in/min

By substituting this into the formulae, we have that

dV/dt = (4π/3) × 3r²×dr/dt

dV/dt = (4π/3) × 3(8)² × 3

dV/dt = (4π/3) × 576

dV/dt = 2421.743 in³/ min

B)

r = 38 inches, dr/dt = 3 in/min

By substituting this into the formulae, we have that

dV/dt = (4π/3) × 3r²×dr/dt

dV/dt = (4π/3) × 3(38)² × 3

dV/dt = (4π/3) × 12996

dV/dt = 54437.52 in³/ min

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.