Air is pumped into a spherical balloon at a constant rate of 80 cm^3/s.

Question

What is the rate of change of the radius and how fast does the surface of the balloon change at the instant when the radius is 15cm?

Hello! I've solved this problem and I would just like to double-check if my answer is correct.

ANSWER: The surface of the radius changes at a rate of 32/45 cm^2/s.

Answer 1

V = 4/3πr³ , dV/dt = + 80 , r = 15 , d(SA) / dt = ?

dV/dt = 4πr² dr/dt

80 = 4π(225) dr/dt

dr/dt = 4/(45π)

SA = 4πr²

d(SA)/dt = 8πr dr/dt

d(SA)/dt = 8π(15) ⋅ 4/(45π) = 32/3 in²/sec

Is it possible you didn't include an r in your equation for d(SA)/dt?