**Question:**

The diagonal of a square is changing at the rate of $\frac{1}{2} \mathrm{~cm} / \mathrm{sec}$. Then the rate of change of area, when the area is $400 \mathrm{~cm}^{2}$, is equal to_______________

**Solution:**

Let the side of the square be $x \mathrm{~cm}$ at any time $t$.

$\therefore$ Length of the diagonal of square, $I=\sqrt{2} \times$ Side of square $=\sqrt{2} x$

$l=\sqrt{2} x$

Differentiating both sides with respect to $t$, we get

$\frac{d l}{d t}=\sqrt{2} \times \frac{d x}{d t}$

It is given that, $\frac{d l}{d t}=\frac{1}{2} \mathrm{~cm} / \mathrm{sec}$

$\therefore \frac{1}{2}=\sqrt{2} \times \frac{d x}{d t}$

$\Rightarrow \frac{d x}{d t}=\frac{1}{2 \sqrt{2}} \mathrm{~cm} / \mathrm{sec}$

Now,

Area of the square, $A=x^{2}$

Differentiating both sides with respect to $t$, we get

$\frac{d A}{d t}=2 x \frac{d x}{d t}$ .....(1)

When $A=400 \mathrm{~cm}^{2}$

$x^{2}=400 \mathrm{~cm}^{2}=(20 \mathrm{~cm})^{2}$

$\Rightarrow x=20 \mathrm{~cm}$

Putting $x=20 \mathrm{~cm}$ and $\frac{d x}{d t}=\frac{1}{2 \sqrt{2}} \mathrm{~cm} / \mathrm{sec}$ in $(1)$, we get

$\frac{d A}{d t}=2 \times 20 \times \frac{1}{2 \sqrt{2}}=10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$

Thus, the rate of change of area of the square is $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$

The diagonal of a square is changing at the rate of $\frac{1}{2} \mathrm{~cm} / \mathrm{sec}$. Then the rate of change of area, when the area is $400 \mathrm{~cm}^{2}$, is equal to $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$