Let \(B \equiv(0,3)\) and \(C \equiv(4,0)\). The point \(A\) is moving on the line \(y=2 x\) at the rate of 2 units/second. The area of \(\triangle \mathrm{ABC}\) is increasing at the rate of

\(\text {Let } \mathrm{A}=(\mathrm{h}, 2 \mathrm{~h}) \)
\(\mathrm{OA}=\sqrt{\mathrm{h}^2+4 \mathrm{~h}^2}=\sqrt{5} \mathrm{~h} \)
\( \therefore \frac{\mathrm{d}(\mathrm{OA})}{\mathrm{dt}}=\sqrt{5} \frac{\mathrm{dh}}{\mathrm{dt}} \)
\( \Rightarrow 2=\sqrt{5} \frac{\mathrm{dh}}{\mathrm{dt}} \)
\(\Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{2}{\sqrt{5}}\)
\(\alpha= A (\triangle ABC )=\frac{1}{2}\left|\begin{array}{ccc} h & 2 h & 1 \\ 0 & 3 & 1 \\ 4 & 0 & 1\end{array}\right|\)
\(=\frac{1}{2}(3 h+8 h-12)\)
\(=\frac{11 h-12}{2}\)
\(\therefore \frac{ d \alpha}{ dt }=\frac{11}{2} \cdot \frac{ dh }{ dt }\)
\(=\frac{11}{2} \cdot \frac{2}{\sqrt{5}}\)
\(=\frac{11}{\sqrt{5}}(\text { units })^2 / sec\)