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Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  6

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Answer: 1 \mathrm{~cm} / \mathrm{sec}

Hint: Here, we use basic concept of rate of change of bodies and quantities.

Given: The side of an equilateral triangle is increasing \frac{1}{3} \mathrm{~cm} / \mathrm{sec}

Solution:  Let a be the side of an equilateral triangle is

So,        \frac{d a}{d t}=\frac{1}{3} \mathrm{~cm} / \mathrm{sec}

We know the formula about perimeter of equilateral triangle

        \begin{aligned} &p=3 a \\\\ &\frac{d p}{d t}=3 \frac{d a}{d t} \end{aligned}    (differentiate with respect to time (t))

        \begin{aligned} &\frac{d p}{d t}=3 \times \frac{1}{3}\left(\frac{d a}{d t}=\frac{1}{3}\right) \\\\ &\frac{d p}{d t}=1 \mathrm{~cm} / \mathrm{sec} \end{aligned}

So, the rate change of equilateral triangle’s perimeter is 1cm/sec


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