1. **State the problem:** A spherical balloon is inflated at a rate of $\frac{dV}{dt} = 200\pi$ ft³/min. We want to find how fast the radius $r$ is increasing, i.e., $\frac{dr}{dt}$, when $r = 5$ ft. 2. **Formula:** The volume of a sphere is given by $$V = \frac{4}{3} \pi r^3$$ 3. **Differentiate both sides with respect to time $t$:** $$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$ 4. **Plug in known values:** $$200\pi = 4 \pi (5)^2 \frac{dr}{dt}$$ 5. **Simplify:** $$200\pi = 4 \pi \times 25 \times \frac{dr}{dt} = 100\pi \frac{dr}{dt}$$ 6. **Divide both sides by $100\pi$ to solve for $\frac{dr}{dt}$:** $$\frac{\cancel{200\pi}}{\cancel{100\pi}} = \frac{\cancel{100\pi} \frac{dr}{dt}}{\cancel{100\pi}} \implies 2 = \frac{dr}{dt}$$ 7. **Answer:** The radius is increasing at a rate of $2$ ft/min when the radius is 5 ft.