A square plate is contracting at the uniform rate \(4 \mathrm{~cm}^2 / \mathrm{sec}\), then the rate at which the perimeter is decreasing, when side of the square is \(20 \mathrm{~cm}\), is

Let \(\mathrm{A}, \mathrm{P}\) and \(\mathrm{X}\) be the area, perimeter and length of side of square respectively at time ' \(\mathrm{t}\) ' seconds. Then,
\(\begin{aligned}
\mathrm{A} & =\mathrm{X}^2, \mathrm{P}=4 \mathrm{X} \\
\therefore \quad \mathrm{P} & =4 \sqrt{\mathrm{A}}
\end{aligned}\)
Differentiating w.r.t. t, we get
\(\begin{aligned}
\frac{\mathrm{dP}}{\mathrm{dt}} & =4 \frac{1}{2 \sqrt{\mathrm{A}}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{\mathrm{X}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{20} \times 4...\left[\begin{array}{l}
\text { side }=20 \mathrm{~cm} \\
\frac{\mathrm{dA}}{\mathrm{dt}}=4 \mathrm{~cm}^2 / \mathrm{sec}
\end{array}\right] \\
& =\frac{2}{5} \mathrm{~cm} / \mathrm{sec}
\end{aligned}\)