Answer
$16 \pi \text{ cubic feet per 1 foot change in radius}$
Work Step by Step
Let us say that the tangent Line contains the points $(x, c)$. The slope the tangent Line to the graph of $f(x)$ at $(x, c)$ can be written as $f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c} ...(1)$
In order to find the rate of change of volume with respect to radius, we will use equation (1).
Plug in the given data to obtain:
$$V'(2) =\lim\limits_{r \to 3} \dfrac{V(r)-V(2)}{r-2} \\=\lim\limits_{r \to 2} \dfrac{4/3 r^3 - 32/3 \pi }{(r-2)}\\=\lim\limits_{r \to 2} \dfrac{ \dfrac{4 \pi}{3} (r^3-8) }{(r-2)}\\=\lim\limits_{r \to 2} \dfrac{ \dfrac{4 \pi}{3} [(r-2)(r^2+2r+4) ]}{r-2}\\=\lim\limits_{r \to 2} \dfrac{4}{3}\pi \times (r^2+2r+4) \\=\dfrac{48}{3}\pi \\=16 \pi \text{ cubic feet per foot}$$
