A spherical balloon is inflating with helium at a rate of \( 72 \pi \mathrm{ft}^{3} / \mathrm{min} \). How fast is the balloon's radus increasing at the instant the radius is 3 ft ? How fast is the surface area increasing? The balloon's radus is increasing at a rate of \( \square \mathrm{ft} / \mathrm{min} \) at the instant the radus is 3 ft . (Simplify your answer.)

To determine how fast the balloon's radius is increasing, we start with the volume formula for a sphere: \[ V = \frac{4}{3} \pi r^3. \] Given that the volume is changing at a rate of \( \frac{dV}{dt} = 72 \pi \mathrm{ft}^{3}/\mathrm{min} \), we can apply the chain rule: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}. \] Now, plug in \( r = 3 \) ft: \[ 72 \pi = 4 \pi (3^2) \frac{dr}{dt}. \] This simplifies to: \[ 72 \pi = 36 \pi \frac{dr}{dt}, \] which leads us to find: \[ \frac{dr}{dt} = \frac{72 \pi}{36 \pi} = 2 \, \mathrm{ft/min}. \] So, the radius is increasing at a rate of 2 ft/min when the radius is 3 ft. Next, to find how fast the surface area is increasing, we use the surface area formula: \[ S = 4 \pi r^2. \] Taking the derivative with respect to time gives: \[ \frac{dS}{dt} = 8 \pi r \frac{dr}{dt}. \] Substituting \( r = 3 \) ft and \( \frac{dr}{dt} = 2 \, \mathrm{ft/min} \): \[ \frac{dS}{dt} = 8 \pi (3) (2) = 48 \pi \, \mathrm{ft}^2/\mathrm{min}. \] In conclusion, the balloon's radius is increasing at a rate of \( 2 \, \mathrm{ft/min} \) and the surface area is increasing at \( 48 \pi \, \mathrm{ft}^2/\mathrm{min} \).