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

Answer: 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