Helium is pumped into a spherical balloon at a constant rate of 4 cubic feet per second. How fast is the radius increasing after 3 minutes? At what time (if any) is the radius increasing at a rate of 100 feet per second?

Step 1 ：We are given that the volume of the balloon is increasing at a constant rate of 4 cubic feet per second, so \(\frac{dV}{dt} = 4\).

Step 2 ：The volume of a sphere is given by the formula \(V = \frac{4}{3} \pi r^3\).

Step 3 ：We can use the chain rule to differentiate the volume with respect to time, which gives us \(\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}\).

Step 4 ：Solving this equation for \(\frac{dr}{dt}\) gives us \(\frac{dr}{dt} = \frac{1}{\pi r^2}\).

Step 5 ：We are asked to find how fast the radius is increasing after 3 minutes (or 180 seconds), which is \(\frac{dr}{dt}\) when \(t = 180\).

Step 6 ：Substituting \(t = 180\) into the equation gives us \(\frac{dr}{dt} = \frac{1}{\pi r^2} = 0.015\) feet per second.

Step 7 ：We are also asked to find at what time the radius is increasing at a rate of 100 feet per second. We can set \(\frac{dr}{dt} = 100\) and solve for \(t\).

Step 8 ：However, because the balloon is being filled at a constant rate of 4 cubic feet per second, the radius will never increase at a rate of 100 feet per second.