Answer
a) The rate at which the volume changes when $r=2$ is $16\pi (ft^3/ft)$.
b) The amount of increase the volume experiences is approximately $11.092(ft^3)$.
Work Step by Step
$$V(r)=\frac{4}{3}\pi r^3$$
a) Find $V'(r)$: $$V'(r)=\Big(\frac{4}{3}\pi r^3\Big)'=\frac{4}{3}\pi(3r^2)=4\pi r^2(ft^3/ft)$$
So the rate at which the volume changes when $r=2ft$ is $$V'(2)=4\pi\times2^2=16\pi(ft^3/ft)$$
b) We calculate the volume for $r=2ft$ and $r=2.2ft$.
$$V(2)=\frac{4}{3}\pi\times2^3=\frac{32\pi}{3}(ft^3)$$
$$V(2.2)=\frac{4}{3}\pi\times(2.2)^3=\frac{42.592\pi}{3}(ft^3)$$
The amount of increase the volume experiences is $$V(2.2)-V(2)=\frac{42.592\pi}{3}-\frac{32\pi}{3}=\frac{10.592\pi}{3}\approx11.092(ft^3)$$
