PLEASE HELP Related Rates Calculus

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 V = (4/3)pi*r^3

Answer 1

The derivative with respect to t looks like this:

dV/dt = 4πr²dr/dt. The fact that helium is being pumped in at the given rate means we know dV/dt.

The question is what is the value of r after 2 minutes?

The value of V after 2 minutes (120 seconds) is 5 (120) = 600 ft³, i.e. 5ft³ added each second.

Substituting that value for V (in the original formula allows you to determine the value of r at that time.

r = ((600(3/4))/π)^1/3.

^{Now substitute the determined values for dV/dt and r into the derivative and solve for dr/dt.}

^{This all assumes that the balloon had zero helium at time t = 0.}