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(C) Let $x$ be the side of the square. The area $A$ is given by $A = x^2$. Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 2x \frac{d

Vedclass · Accepted Answer

$0.1 	ext{ cm/sec}$

Vedclass · Answer

(C) Let $x$ be the side of the square. The area $A$ is given by $A = x^2$. Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 2x \frac{dx}{dt}$. Given $\frac{dA}{dt} = 0.5 	ext{ cm}^2/	ext{sec}$ and $x = 10 	ext{ cm}$. Substituting these values: $0.5 = 2(10) \frac{dx}{dt} \Rightarrow 0.5 = 20 \frac{dx}{dt}$. Thus,$\frac{dx}{dt} = \frac{0.5}{20} = 0.025 	ext{ cm/sec}$. The perimeter $P$ of the square is $P = 4x$. Differentiating with respect to $t$,we get $\frac{dP}{dt} = 4 \frac{dx}{dt}$. Substituting $\frac{dx}{dt} = 0.025$,we get $\frac{dP}{dt} = 4(0.025) = 0.1 	ext{ cm/sec}$.

The area of a square increases at the rate of $0.5 \text{ cm}^2/\text{sec}$. The rate at which its perimeter is increasing when the side of the square is $10 \text{ cm}$ long,is

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