A spherical snowball is melting in such a way that its radius is decreasing at a rate of $0.2 \mathrm{~cm} / \mathrm{min}$. At what rate is the volume of the snowball decreasing when the radius is $17 \mathrm{~cm}$. (Note the answer is a positive number). $\frac{\mathrm{cm}^{3}}{\min }$

Step 1 ：The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\).

Step 2 ：We are given that the radius is decreasing at a rate of \(0.2 \, \mathrm{cm} / \mathrm{min}\), which we can denote as \(\frac{dr}{dt} = -0.2 \, \mathrm{cm} / \mathrm{min}\).

Step 3 ：We are asked to find the rate at which the volume is decreasing when the radius is \(17 \, \mathrm{cm}\), which we can denote as \(\frac{dV}{dt}\) when \(r = 17 \, \mathrm{cm}\).

Step 4 ：We can find \(\frac{dV}{dt}\) by differentiating the volume formula with respect to time \(t\) and then substituting the given values.

Step 5 ：The derivative of the volume with respect to time is \(-0.8\pi r^2\).

Step 6 ：We can substitute \(r = 17 \, \mathrm{cm}\) into this equation to find the rate at which the volume is decreasing.

Step 7 ：\(\frac{dV}{dt}\) at \(r = 17\) is \(-231.2\pi\).

Step 8 ：Final Answer: The rate at which the volume of the snowball is decreasing when the radius is \(17 \, \mathrm{cm}\) is \(\boxed{231.2\pi \, \mathrm{cm}^{3}/\mathrm{min}}\).