Helium is injected into a balloon sphere at a constant rate of 80 (cm^3)/s...

Question

Helium is injected into a balloon sphere at a constant rate of 80 (cm^3)/s. Find the rate of change of its radius and how fast the surface of the balloon change at the instant when the radius is 15 centimeters?

Answer 1

Volume of sphere = 4/3πr³.

Given dV/dt = 80 cm³/s = 4πr²dr/dt

dr/dt = 80/(4πr²)

When r = 15:

dr/dt = 20/(π225) cm/s

Surface Area of sphere, S = 4πr²

dS/dt = 8πr dr/dt

When r = 15 and dr/dt = 20/(π225)

dS/dt = 160(15)π/(π/225) = 2400/225 = 10.67 cm/s

Answer 2

V = 4/3πr³; dV/dt = 4πr²dr/dt; dr/dt = (dV/dt)/(4πr²) = 80cm³/s/(4π··15²cm²) = 0.0283 cm/s;

Surface area A = 4πr² change with rate dA/dt = 8πr·dr/dt = 8π·15cm·0.0283cm/s = 10.67 cm²/s