The formula for the volume of a cone is given below. Find the rate of change (in in \( { }^{3} / m i n \) ) of the volume for each of the rad given below if \( d r / d t \) is 3 inches per minute and \( h=12 r \). \[ V=\left(\frac{1}{3}\right) \pi r^{2} h \] (a) \( r=3 \) in \[ v^{\prime}=\square \mathrm{in}^{3} / \mathrm{min} \] (b) \( r=23 \) in \[ v^{\prime}=\square \ln ^{3} / \mathrm{min} \]

To find the rate of change of the volume as the radius changes, we start with the formula for the volume of a cone. Given that \( h = 12r \), we can substitute this into the volume formula: \[ V = \frac{1}{3} \pi r^2 (12r) = 4\pi r^3 \] Next, we need to differentiate \( V \) with respect to \( t \) using the chain rule: \[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \] Calculating \( \frac{dV}{dr} \): \[ \frac{dV}{dr} = 12\pi r^2 \] Now we substitute \( \frac{dr}{dt} = 3 \) in/min into our expression for \( \frac{dV}{dt} \): \[ \frac{dV}{dt} = 12\pi r^2 \cdot 3 = 36\pi r^2 \] Now we can calculate \( \frac{dV}{dt} \) for each radius: (a) For \( r = 3 \) in: \[ \frac{dV}{dt} = 36\pi (3^2) = 36\pi \cdot 9 = 324\pi \approx 1017.88 \text{ in}^3/\text{min} \] (b) For \( r = 23 \) in: \[ \frac{dV}{dt} = 36\pi (23^2) = 36\pi \cdot 529 = 19164\pi \approx 60274.63 \text{ in}^3/\text{min} \] Thus, the final outputs are: (a) \( v^{\prime} \approx 1017.88 \, \text{in}^3/\text{min} \) (b) \( v^{\prime} \approx 60274.63 \, \text{in}^3/\text{min} \)