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.

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