Question

A spherical party balloon is being inflated with helium pumped in at a rate of 11 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 2 ft? HINT [See Example 1.] (The volume of a sphere of radius r is
V = 4/3𝜋r^3


Round your answer to two decimal places.)


____ ft/min

Answer 1

The CHANGE in volume is the surface area times the change in radius.
V = (4/3) pi r^3
dV/dr = 4 pi r^2 = surface area of of sphere. (here 16 pi = 50.2 ft^2)
dV/dt = 4 pi r^2 dr/dt = 50.2 dr/dt
so
dr/dt = dV/dt / (4 pi r^2)
= 11 /(4 pi *4) = 11/ (16 pi ) = 0.219 ft/minute