The radius \( r \) of a sphere is increasing at a rate of 2 inches per minute. (a) Find the rate of change of the volume ( \( \left.\ln \ln ^{3} / m i n\right) \) when \( r=a \) inches and when \( r=34 \) inches. \( r=8 \mathrm{in} \) \( \square \) \( 1 n^{3} / \) min \( r=34 \) in \( \square \) in \( 3 / \) min (b) Explain why the rate of change of the volume of the sphere is not constant even though \( \frac{d r}{d t} \) is constant. If \( \frac{d r}{d t} \) is constant, \( \frac{d V}{d t} \) is equal to \( r^{2} \). If \( \frac{d r}{d t} \) is constant, \( \frac{d V}{d r} \) is equal to \( t^{2} \). If \( \frac{d r}{d t} \) is constant, \( \frac{d v}{d r} \) is proportional to \( t^{2} \). If \( \frac{d r}{d t} \) is constant, \( \frac{d V}{d t} \) is proportional to \( r^{2} \).

To solve for the rate of change of the volume when \( r = 8 \) inches and \( r = 34 \) inches, we start with the formula for the volume of a sphere, \( V = \frac{4}{3} \pi r^3 \). Using the chain rule, we derive that \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \). Plugging in \( \frac{dr}{dt} = 2 \) inches/min, we find that at \( r = 8 \): \( \frac{dV}{dt} \) is \( 4 \pi (8)^2 \cdot 2 \approx 256\pi \), which is about \( 804.25 \) cubic inches per minute. For \( r = 34 \): \( \frac{dV}{dt} \) is \( 4 \pi (34)^2 \cdot 2 = 8,112\pi \), or approximately \( 25,483.46 \) cubic inches per minute. As for the rate of change of the volume not being constant, it's essential to note that while \( \frac{dr}{dt} \) remains steady at 2 inches/min, the volume's growth is contingent on the square of the radius, which continually alters as the radius increases. Thus, as the sphere expands, the function's output, \( \frac{dV}{dt} \), escalates significantly despite the input change (the increase in radius) being constant. It's like taking a ride on a roller coaster—the speed may be consistent, but the elevation leads to thrilling changes in how quickly you descend!