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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
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Verified Answer
The correct answer is:
$\frac{2}{5} \mathrm{~cm} / \mathrm{sec}$.
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
$\begin{aligned}
\frac{\mathrm{dP}}{\mathrm{dt}} & =4 \frac{1}{2 \sqrt{\mathrm{A}}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{\mathrm{X}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{20} \times 4...\left[\begin{array}{l}
\text { side }=20 \mathrm{~cm} \\
\frac{\mathrm{dA}}{\mathrm{dt}}=4 \mathrm{~cm}^2 / \mathrm{sec}
\end{array}\right] \\
& =\frac{2}{5} \mathrm{~cm} / \mathrm{sec}
\end{aligned}$
$\begin{aligned}
\mathrm{A} & =\mathrm{X}^2, \mathrm{P}=4 \mathrm{X} \\
\therefore \quad \mathrm{P} & =4 \sqrt{\mathrm{A}}
\end{aligned}$
Differentiating w.r.t. t, we get
$\begin{aligned}
\frac{\mathrm{dP}}{\mathrm{dt}} & =4 \frac{1}{2 \sqrt{\mathrm{A}}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{\mathrm{X}} \cdot \frac{\mathrm{dA}}{\mathrm{dt}} \\
& =\frac{2}{20} \times 4...\left[\begin{array}{l}
\text { side }=20 \mathrm{~cm} \\
\frac{\mathrm{dA}}{\mathrm{dt}}=4 \mathrm{~cm}^2 / \mathrm{sec}
\end{array}\right] \\
& =\frac{2}{5} \mathrm{~cm} / \mathrm{sec}
\end{aligned}$
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