A rod of 10 feet long moves with its ends A and B always on the axes of x and y respectively. If A is 8 ft from the origin and is moving away at the rate of 2 ft/sec. At what rate the area formed by AB and the axes changing ?

$$\begin{gathered} \mathrm{x}=8 \frac{\mathrm{dx}}{\mathrm{dt}=2} \\ {\mathrm{x}}^2+{\mathrm{y}}^2=100 \Rightarrow {\mathrm{y}}^2=100-64 \\ =36 \\ \mathrm{y}=6 \\ 2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dt}} =0 \\ \Rightarrow 8\times 2+6\frac{\mathrm{dy}}{\mathrm{dt}} =0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\raisebox{1ex}{$-8$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ \mathrm{area} \mathrm{A}=\frac{1}{2} \mathrm{xy} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{1}{2}\left(\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y}.\frac{\mathrm{dx}}{\mathrm{dt}}\right) \\ =\frac{1}{2}\left(8\times \frac{-8}{3}+6\times 2\right) \\ =\frac{1}{2}\left(\frac{-64+36}{3}\right)=\frac{1}{2}\left(\frac{-28}{3}\right)=\frac{-14}{3} \end{gathered}$$