A square plate is contracting at the uniform | vedclass

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(D) Let $A$, $P$, and $X$ be the area, perimeter, and side length of the square at time $t$ seconds, respectively. Then, $A = X^2$ and $P = 4X$. From

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$\frac{2}{5} \,cm / sec$.

Vedclass · Answer

(D) Let $A$, $P$, and $X$ be the area, perimeter, and side length of the square at time $t$ seconds, respectively. Then, $A = X^2$ and $P = 4X$. From these, we have $P = 4 \sqrt{A}$. Differentiating both sides with respect to $t$, we get: $\frac{dP}{dt} = 4 \cdot \frac{1}{2 \sqrt{A}} \cdot \frac{dA}{dt} = \frac{2}{\sqrt{A}} \cdot \frac{dA}{dt}$. Since $A = X^2$, we have $\sqrt{A} = X$. Thus, $\frac{dP}{dt} = \frac{2}{X} \cdot \frac{dA}{dt}$. Given that the area is contracting at a rate of $4 \,cm^2 / sec$, we have $\frac{dA}{dt} = -4 \,cm^2 / sec$. At $X = 20 \,cm$: $\frac{dP}{dt} = \frac{2}{20} 	imes (-4) = -\frac{8}{20} = -\frac{2}{5} \,cm / sec$. The negative sign indicates that the perimeter is decreasing at a rate of $\frac{2}{5} \,cm / sec$.

$A$ square plate is contracting at the uniform rate $4 \,cm^2 / sec$. The rate at which the perimeter is decreasing, when the side of the square is $20 \,cm$, is

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