The diagonal of a square is changing at the rate of $0.5 \mathrm{~cm} / \mathrm{sec}$. Then the rate of change of area when the area is $400 \mathrm{~cm}^2$ is equal to

Question

Aicharya · Accepted Answer

$\begin{array}{ll} & \frac{\mathrm{d} x}{\mathrm{dt}}=0.5 \mathrm{~cm} / \mathrm{sec} \ 	herefore \quad & 	ext { Area }=\frac{x^2}{2} \ 	herefore \quad & \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{2 x}{2} \cdot \frac{\mathrm{d} x}{\mathrm{dt}}=x \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{1}{2} x \end{array}$ $\begin{aligned} 	herefore \quad\left[\frac{\mathrm{dA}}{\mathrm{dt}}ight]_{\mathrm{A}=400} & =\frac{1}{2} \sqrt{800} \quad \ldots\left[\begin{array}{l}\because \mathrm{A}=400 \mathrm{~cm}^2 \ x=\sqrt{800} \mathrm{~cm}\end{array}ight] \ & =10 \sqrt{2} \mathrm{~cm}^2 / \mathrm{sec}\end{aligned}$

The diagonal of a square is changing at the rate of $0.5 \mathrm{~cm} / \mathrm{sec}$. Then the rate of…

Solution