A spherical balloon is inflated and its volume increases at a rate of \( 23 \mathrm{in}^{3} / \mathrm{min} \). What is the rate of change of its radius when the radius is 9 in.? The balloon's radius is increasing at a rate of (Round to three decimal places as needed.)

To find the rate of change of the radius when the radius is 9 inches, we start with the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3. \] We know that the volume \( V \) is changing over time, so we can differentiate both sides with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}. \] Given that \( \frac{dV}{dt} = 23 \, \text{in}^3/\text{min} \) and \( r = 9 \, \text{in} \), we can plug in these values: \[ 23 = 4 \pi (9^2) \frac{dr}{dt}. \] Calculating \( 9^2 \): \[ 9^2 = 81, \] so our equation becomes: \[ 23 = 4 \pi (81) \frac{dr}{dt}. \] Now, simplify \( 4 \pi (81) \): \[ 4 \pi (81) = 324 \pi. \] Now we have: \[ 23 = 324 \pi \frac{dr}{dt}. \] To solve for \( \frac{dr}{dt} \), divide both sides by \( 324 \pi \): \[ \frac{dr}{dt} = \frac{23}{324 \pi}. \] Now calculating \( \pi \): \[ \frac{dr}{dt} \approx \frac{23}{324 \times 3.14159} \approx \frac{23}{1017.876} \approx 0.0226 \, \text{in/min}. \] Thus, rounding to three decimal places, the balloon's radius is increasing at a rate of approximately: \[ \frac{dr}{dt} \approx 0.023 \, \text{in/min}. \]