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Please solve RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  5 maths textbook solution

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Answer: 3.14 \mathrm{~cm} / \mathrm{sec}

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

Given: The radius of a circle is increasing at the rate of 0.5cm/sec

Solution:   Let r be the radius of a circle


        \frac{d r}{d t}=0.5 \mathrm{~cm} / \mathrm{sec}

We know the formula about circumference of circle

        \begin{aligned} &\mathrm{C}=2 \pi \mathrm{r} \\ &\frac{d c}{d t}=2 \pi \frac{d r}{d t} \end{aligned}   (differentiate with respect to t)

                \begin{aligned} &=2 \pi \times 0.5\left(\frac{d r}{d t}=0.5\right) \\ &=\pi \\ &=3.14 \mathrm{~cm} / \mathrm{sec} \end{aligned}

So, the rate of increase about circle’s circumference is 3.14 cm/sec .

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