Radius increasing sphere Calculus Problem

Question

Helium is pumped into a spherical balloon at a rate of 5 cubic feet per second. How fast is the radius increasing after 2 minutes?

Rate of change of radius (in feet per second) = 14400

Answer 1

Say the radius of the balloon, as a function of time, is r(t).

We know that the volume of the sphere is then V(t)=(4/3)π(r(t))³. Differentiating both sides with respect to t (and remembering the chain rule) yields

V'(t)=4π(r(t))²r'(t).

We are told that V'(t)=5. Therefore, assuming the initial volume of the balloon is 0, we have V(t)=5t. Plugging these into our formulas yields

5t=(4/3)π(r(t))³

5=4π(r(t))²r'(t)

Finally, we care about the moment when t=2 minutes=120 seconds. Plugging this in yields a system of equations

5·120=(4/3)π(r(120))³

5=4π(r(120))²r'(120)

Solving the first equation, we get that r(120)=(450/π)^1/3. Plugging this into the second equation yields

5=4π(450/π)^2/3r'(120)

and now solving for r'(120) is simple.