Answer
\[\begin{align}
& \mathbf{a}.\frac{db}{da}=\frac{b+3a}{3b+a} \\
& \mathbf{b}.\text{ }\frac{7}{9} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& V=\frac{{{\pi }^{2}}\left( b+a \right){{\left( b-a \right)}^{2}}}{4} \\
& \mathbf{a}\mathbf{.}\text{ For }V=64{{\pi }^{2}} \\
& \frac{{{\pi }^{2}}\left( b+a \right){{\left( b-a \right)}^{2}}}{4}=64{{\pi }^{2}} \\
& \left( b+a \right){{\left( b-a \right)}^{2}}=256 \\
& \text{Differentiate both sides with respect to }a \\
& \frac{d}{da}\left[ \left( b+a \right){{\left( b-a \right)}^{2}} \right]=\frac{d}{da}\left[ 256 \right] \\
& \left( b+a \right)\frac{d}{da}\left[ {{\left( b-a \right)}^{2}} \right]+{{\left( b-a \right)}^{2}}\frac{d}{da}\left[ \left( b+a \right) \right]=\frac{d}{da}\left[ 256 \right] \\
& 2\left( b+a \right)\left( b-a \right)\left( \frac{db}{da}-1 \right)+{{\left( b-a \right)}^{2}}\left( \frac{db}{da}+1 \right)=0 \\
& 2\left( b+a \right)\left( \frac{db}{da}-1 \right)+\left( b-a \right)\left( \frac{db}{da}+1 \right)=0 \\
& \text{Solve for }\frac{db}{da} \\
& 2\left( b+a \right)\frac{db}{da}-2\left( b+a \right)+\left( b-a \right)\frac{db}{da}+b-a=0 \\
& 2\left( b+a \right)\frac{db}{da}+\left( b-a \right)\frac{db}{da}=2\left( b+a \right)-b+a \\
& \left[ 2\left( b+a \right)+\left( b-a \right) \right]\frac{db}{da}=2\left( b+a \right)-b+a \\
& \frac{db}{da}=\frac{2\left( b+a \right)-b+a}{2\left( b+a \right)+\left( b-a \right)} \\
& \frac{db}{da}=\frac{2b+2a-b+a}{2b+2a+b-a} \\
& \frac{db}{da}=\frac{b+3a}{3b+a} \\
& \\
& \mathbf{b}.\text{ Evaluate when }a=6\text{ and }b=10 \\
& \frac{db}{da}=\frac{10+3\left( 6 \right)}{3\left( 10 \right)+6} \\
& \frac{db}{da}=\frac{7}{9} \\
\end{align}\]
