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Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 2

Answers (1)


B. \quad 10\sqrt{3}\; cm^{2}/sec


The area of an equilateral triangle side is

A=\frac{\sqrt{3}}{4}a^{2} \quad \quad \quad \quad.....(i)


a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}


Differentiating (i) with respect to t

We get

\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a \frac{d a}{d t}

→ Substituting values of

        a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}

We get

\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a\times \frac{d a}{d t} \\ &\quad=\frac{\sqrt{3 }a}{2} \times \frac{d a}{d t} \\ &\text { When } a=10 \\ &\frac{d A}{d t}=\frac{10 \sqrt{3}}{2} \times 2=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}

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Gurleen Kaur

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