A square plate is contracting at the uniform | vedclass

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(B) Let $A$, $P$, and $X$ be the area, perimeter, and length of the side of the square respectively at time $t$ seconds.Given that the area is contrac

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $A$, $P$, and $X$ be the area, perimeter, and length of the side of the square respectively at time $t$ seconds.Given that the area is contracting at a rate of $\frac{dA}{dt} = -3 \,cm^2 / sec$.The area of the square is $A = X^2$ and the perimeter is $P = 4X$.From $A = X^2$, we have $X = \sqrt{A}$.Substituting this into the perimeter formula: $P = 4\sqrt{A}$.Differentiating with respect to $t$:$\frac{dP}{dt} = 4 \cdot \frac{1}{2\sqrt{A}} \cdot \frac{dA}{dt} = \frac{2}{X} \cdot \frac{dA}{dt}$.Given $X = 15 \,cm$ and $\frac{dA}{dt} = -3 \,cm^2 / sec$ (since it is contracting):$\frac{dP}{dt} = \frac{2}{15} \cdot (-3) = -\frac{6}{15} = -\frac{2}{5} \,cm / sec$.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 $3 \,cm^2 / sec$. The rate at which the perimeter is decreasing, when the side of the square is $15 \,cm$, is

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