The side of a square is increasing at the rat | vedclass

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Question

(D) Let the side of the square be $a$. The perimeter $P$ of a square is given by $P = 4a$.To find the rate of increase of the perimeter with respect t

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) Let the side of the square be $a$. The perimeter $P$ of a square is given by $P = 4a$.To find the rate of increase of the perimeter with respect to time $t$,we differentiate $P$ with respect to $t$:$\frac{dP}{dt} = \frac{d}{dt}(4a) = 4 \left( \frac{da}{dt} ight)$.Given that the rate of increase of the side is $\frac{da}{dt} = 0.2\,cm/s$.Substituting this value into the equation:$\frac{dP}{dt} = 4 	imes 0.2 = 0.8\,cm/s$.Therefore,the rate of increase of the perimeter is $0.8\,cm/s$.

The side of a square is increasing at the rate of $0.2\,cm/s$. The rate of increase of perimeter with respect to time is $...........\,cm/s$.

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