Assume that the radius \( r \) of a sphere is expanding at a rate of \( 15 \mathrm{in} . / \mathrm{min} \). The volume of a sphere is \( V=\frac{4}{3} \pi r^{3} \). Determine the rate at which the volume is changing with respect to time when \( r=13 \mathrm{in} \). The volume is changing at a rate of \( \square \mathrm{in}^{3} / \mathrm{min} \).

To determine the rate at which the volume \( V \) of the sphere is changing with respect to time, we will use the formula for the volume of the sphere and apply the chain rule. Given the volume formula: \[ V = \frac{4}{3} \pi r^3 \] we differentiate \( V \) with respect to \( t \): \[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \] Calculating \( \frac{dV}{dr} \): \[ \frac{dV}{dr} = 4 \pi r^2 \] Now, substituting in our known values: \( r = 13 \) in and \( \frac{dr}{dt} = 15 \) in/min. \[ \frac{dV}{dr} \Big|_{r=13} = 4 \pi (13)^2 = 4 \pi (169) = 676 \pi \] Now substituting into the equation for \( \frac{dV}{dt} \): \[ \frac{dV}{dt} = (676 \pi) \cdot (15) = 10140 \pi \, \mathrm{in}^3/\mathrm{min} \] Numerically calculating \( 10140 \pi \) to provide a specific answer: \[ \frac{dV}{dt} \approx 31825.45 \, \mathrm{in}^3/\mathrm{min} \] So, the volume is changing at a rate of approximately \[ \boxed{31825.45 \, \mathrm{in}^3/\mathrm{min}} \]