Assume that the radius r of a sphere is expanding at a rate of 60 cm/min. The volume of a sphere is V = 4 3 πr3 and its surface area is 4πr2. Determine the rate of change of volume when r = 17 cm. 1156π cm3/min

Answer: dV/dt = 69,360πcm^3/min

Therefore, the rate of change of volume when r = 17 cm is = 69,360πcm^3/min

Step-by-step explanation:

The volume of a sphere is given by;

V = (4/3) πr^3

dV/dt = dV/dr * dr/dt

dr/dt = 60cm/min (given)

dV/dr = d/dr (4/3 πr^3) = 4πr^2

r = 17cm

So,

dV/dt = 4πr^2 (dr/dt)

Substituting the values into the equation

dV/dt = 4π (17^2) (60)

dV/dt = 69,360πcm^3/min

Therefore, the rate of change of volume when r = 17 cm is = 69,360πcm^3/min