A spherical balloon is inflating with helium at a rate of $128 \pi \frac{\mathrm{ft}^{3}}{\mathrm{~min}}$. How fast is the balloon's radius increasing at the instant the radius is $4 \mathrm{ft}$ ? Write an equation relating the volume of a sphere, $\mathrm{V}$, and the radius of the sphere, $r$. (Type an exact answer, using $\pi$ as needed.)

Step 1 ：Given that the volume of a sphere is \(V = \frac{4}{3}\pi r^3\), we need to find the rate of change of the radius with respect to time, given the rate of change of the volume with respect to time. This is a problem of related rates, and we can solve it by differentiating the volume equation with respect to time.

Step 2 ：The differentiation of the volume with respect to time did not yield the expected result. This is because the differentiation was done with respect to 't', but 't' does not appear in the volume equation. Instead, we should differentiate the volume with respect to 'r', and then multiply by 'dr/dt' to get 'dV/dt'. This will give us an equation that relates 'dV/dt' and 'dr/dt', which we can then solve for 'dr/dt' given that 'dV/dt' is \(128 \pi \frac{\mathrm{ft}^{3}}{\mathrm{~min}}\).

Step 3 ：By differentiating the volume with respect to 'r', we get \(dV/dr = 4.0\pi r^2\).

Step 4 ：Substituting the given value of 'dV/dt' into the equation, we get \(dr/dt = 32.0/r^2\).

Step 5 ：Substituting the given radius of 4 ft into the equation, we find that the rate at which the balloon's radius is increasing at the instant the radius is 4 ft is \(\boxed{2 \frac{\mathrm{ft}}{\mathrm{~min}}}\).