Answer
\[\begin{align}
& \mathbf{a}.\frac{dr}{dh}=\frac{h-2r}{h} \\
& \mathbf{b}.-3 \\
\end{align}\]
Work Step by Step
\[\begin{align}
& V=\frac{\pi {{h}^{2}}\left( 3r-h \right)}{3} \\
& \mathbf{a}\mathbf{.}\text{ For }V=\frac{5\pi }{3} \\
& \frac{\pi {{h}^{2}}\left( 3r-h \right)}{3}=\frac{5\pi }{3} \\
& {{h}^{2}}\left( 3r-h \right)=5 \\
& \text{Differentiate both sides with respect to }h \\
& \frac{d}{dh}\left[ {{h}^{2}}\left( 3r-h \right) \right]=\frac{d}{dh}\left[ 5 \right] \\
& {{h}^{2}}\frac{d}{dh}\left[ 3r-h \right]+\left( 3r-h \right)\frac{d}{dh}\left[ {{h}^{2}} \right]=\frac{d}{dh}\left[ 5 \right] \\
& {{h}^{2}}\left( 3\frac{dr}{dh}-1 \right)+\left( 3r-h \right)\left( 2h \right)=0 \\
& \text{Solve for }\frac{dr}{dh} \\
& 3{{h}^{2}}\frac{dr}{dh}-{{h}^{2}}+2h\left( 3r-h \right)=0 \\
& \frac{dr}{dh}=\frac{{{h}^{2}}-2h\left( 3r-h \right)}{3{{h}^{2}}} \\
& \frac{dr}{dh}=\frac{h-2\left( 3r-h \right)}{3h} \\
& \frac{dr}{dh}=\frac{h-6r+2h}{3h} \\
& \frac{dr}{dh}=\frac{3h-6r}{3h} \\
& \frac{dr}{dh}=\frac{h-2r}{h} \\
& \\
& \mathbf{b}.\text{ Evaluate when }r=2\text{ and }h=1 \\
& \frac{dr}{dh}=\frac{1-2\left( 2 \right)}{1} \\
& \frac{dr}{dh}=-3 \\
\end{align}\]
