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…

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
  1. $\frac{11}{\sqrt{5}}$ (units $)^2 / \mathrm{sec}$
  2. $\frac{11}{5}$ (units) $)^2 / \mathrm{sec}$
  3. $\frac{43}{\sqrt{5}}$ (units $)^2 / \mathrm{sec}$
  4. $\frac{13}{5}$ (units) $)^2 / \mathrm{sec}$

Solution

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

Asked in: MHT CET 2023 (14 May Shift 1)

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