The sides of an equilateral triangle are increasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$. The rate…

The sides of an equilateral triangle are increasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$. The rate at which the area increases, when side is $10 \mathrm{~cm}$, is
  1. $\frac{\sqrt{3}}{10} \mathrm{~cm}^2 / \mathrm{sec}$
  2. $\frac{10}{\sqrt{3}} \mathrm{~cm}^2 / \mathrm{sec}$
  3. $\sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}$
  4. $10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}$

Solution

$\begin{aligned} & A=\frac{\sqrt{3}}{4} S^2 \\ & \Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \cdot 2 S \frac{\mathrm{d} s}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \times 2 \times 10 \times 2 \mathrm{~cm}^2 / \mathrm{sec} \\ & \Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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