Answer
a. $16\pi ft^3/ft$
b. $10.05ft^3$.
Work Step by Step
a. . Given the equation $V=\frac{4}{3}\pi r^3$, at $r=2$ ft, we have $\frac{dV}{dr}=3\times\frac{4}{3}\pi r^2=4\pi r^2|_{r=2}=16\pi ft^3/ft$
b. With $\frac{dV}{dr}=16\pi ft^3/ft$ and $\Delta r=2.2-2=0.2ft$, we have $\Delta V=\frac{dV}{dr}\Delta r=16\pi(0.2)=3.2\pi\approx10.05ft^3$.
We can also calculate the volume change directly as $\Delta V=V(2.2)-V(2)=\frac{4}{3}\pi (2.2^3-2^3)\approx11.09ft^3$
and the difference is caused by the linear approximation used in the first calculation.
