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Solution for The Side of a Square is Increasing at the Rate of 0.1 Cm/Sec. Find the Rate of Increase of Its Perimeter ? - CBSE (Commerce) Class 12 - Mathematics

ConceptRate of Change of Bodies Or Quantities

Question

The side of a square is increasing at the rate of 0.1 cm/sec. Find the rate of increase of its perimeter ?

Solution

$\text { Let x be the side and P be the perimeter of the square at any timet. Then },$
$P = 4x$
$\Rightarrow \frac{dP}{dt} = 4\frac{dx}{dt}$
$\Rightarrow \frac{dP}{dt} = 4 \times 0 . 1 \left[ \because\frac{dx}{dt}=0.1 cm/sec \right]$
$\Rightarrow \frac{dP}{dt} = 0 . 4 cm/\sec$

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Solution for question: The Side of a Square is Increasing at the Rate of 0.1 Cm/Sec. Find the Rate of Increase of Its Perimeter ? concept: Rate of Change of Bodies Or Quantities. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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