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Explain solution RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  4 maths

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Answer: 0.4 \mathrm{~cm} / \mathrm{sec}

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

Given: The side of square is increasing at the rate of 0.1cm/sec

Solution:  Let a be side of square     

 So, \frac{d a}{d t}=0.1 \mathrm{~cm} / \mathrm{s}

 We know about the formula of square perimeter (p)


\begin{gathered} \mathrm{p}=4 \mathrm{a} \\ \frac{d p}{d t}=4 \frac{d a}{d t} \end{gathered}     (taking differentiate of a with respect to time)

      \begin{aligned} &=4 \times 0.1 \quad\left(\frac{d a}{d t}=0.1\right) \\ &=0.4 \end{aligned}

So, the rate of increase of square’s perimeter is 0.4cm/sec

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