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

The change in the volume of the sphere as a function of time is equal to dV/dt which here is assumed to be constant. Let's call this quantity k (k=80cm^3/s)

We now write dV/dt=k as follows

dV/dt = (dV/dr)*(dr/dt)=k

Since we know that V=(4/3)π r³ , dV/dr = 4π r² and therefore we can write 4π r² *dr/dt=k.

This equation can now be solved for dr/dt=k/(4π r²).

We can plug in the values k=80cm^3/s r=15cm we obtain dr/dt=0.028 cm/s

In the same way we can evaluate the rate of change of the surface with respect to time as dS/dt=(dS/dr)*(dr/dt)

Since S=4π r² then dS/dr=8 π r . Therefore, we have dS/dt= 8 π r *dr/dt = 8 π *15 *0.028 = 10.55 cm²/s

Best,

Davide