Answer
$16 \pi~ft^3/ft$
Work Step by Step
Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as
$f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$
We use the above equation to find the rate of change of volume with respect to the radius.
Consider $V(r)=\dfrac{4}{3} \pi r^3$. Then at $r=2$, we have:
$V'(2) =\lim\limits_{r \to 3} \dfrac{V(r)-V(2)}{r-2} \\=\lim\limits_{r \to 2} \dfrac{\dfrac{4}{3} \pi r^3 -\dfrac{32\pi}{3}}{r-2}\\=\lim\limits_{r \to 2} \dfrac{ \dfrac{4}{3}\pi (r^3-8) }{r-3}\\=\lim\limits_{r \to 2} \dfrac{ \dfrac{4}{3}\pi (r-2)(r^2+2r+4) }{r-2}\\=\lim\limits_{r \to 2} \dfrac{4}{3}\pi \times (r^2+2r+4) \\=\dfrac{48}{3}\pi \\=16 \pi~ft^3/ft$