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The Side of a Square is Increasing at the Rate of 0.2 Cm/Sec. Find the Rate of Increase of the Perimeter of the Square.

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Question

The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of the perimeter of the square.

Sum
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Solution

\[\text { Let x be the side and P be the perimeter of the square at any timet. Then },\]
\[P = 4x\]
\[ \Rightarrow \frac{dP}{dt} = 4\frac{dx}{dt}\]
\[ \Rightarrow \frac{dP}{dt} = 4 \times 0 . 2 \left[ \because\frac{dx}{dt}= 0.2 cm/sec \right]\]
\[ \Rightarrow \frac{dP}{dt} = 0 . 8 cm/\sec\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.2 [Page 19]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.2 | Q 3 | Page 19

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