Rate of change maths que.?

Question

The vol of a balloon is increasing at a rate of 10m^3/s. Find the rate at which its surface area is increasing at the instant when the radius is 5m. Ans is 4m^2/s.

Answer 1

Volume of a balloon is v = (4/3)πr³ .

The rate of change in volume is 10 m³/s .

Radius of balloon is 5 m.

Rate of change in volume dv/dt = (4/3) (3)πr² [dr/dt] 

⇒ 10 = 4 πr² [dr/dt]  ⇒ dr/dt = (10/4πr²) 

⇒ dr/dt = (10/4π (5)²) ⇒ dr/dt = (1/10π)

⇒ dr/dt = 0.0318

Surface area of balloon is s = 4 πr² .

Rate of change in surface area is ds/dt = 4(2) πr dr/dt

⇒ds/dt = 8 π(5) dr/dt ⇒ds/dt = 40π (0.0318)

⇒ds/dt = 4 .

So the rate of change in surface area is 4 m²/s .