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? note: the volume of a sphere is given by .

Answer 1

The volume of a sphere is given by
V = (4π/3)r^3
so the radius is given in terms of volume as
r = (3V/(4π))^(1/3)

Here, we have V = 5t, where t is time in seconds, and V is measured in cubic feet. The the radius as a function of time is
r(t) = (3·5t/(4π))^(1/3)
and its derivative with respect to time is
r'(t) = (1/3)(15/(4π))^(1/3)·t^(-2/3)

At t=120 seconds, the rate of increase of the radius is
r'(120) = (1/3)(15/(4π))^(1/3)/(120^(2/3)) ≈ 0.014534 ft/second