A square plate is contracting at the uniform rate \(3 \mathrm{~cm}^2 / \mathrm{sec}\), then the rate at which the perimeter is decreasing, when the side of the square is 15 cm , is

Let \(\mathrm{A}, \mathrm{P}\) and X be the area, perimeter and length of side of square respectively at time ' \(t\) ' seconds. Then,
\(\begin{aligned}
& A=X^2, P=4 X \\
\therefore \quad & P=4 \sqrt{A}
\end{aligned}\)
Differentiating w.r.t. t, we get
\(\begin{aligned} \frac{\mathrm{dP}}{\mathrm{dt}} & =4 \frac{1}{2 \sqrt{\mathrm{~A}}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\ & =\frac{2}{\mathrm{X}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\ & =\frac{2}{15} \times 3 \\ & =\frac{2}{5} \mathrm{~cm} / \mathrm{sec}\end{aligned}\)
\(\cdots\begin{array}{l}\text { side }=15 \mathrm{~cm} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=3 \mathrm{~cm}^2 / \mathrm{sec}\end{array}\)