Application of Derivatives — Class 12 Maths Solution

Let two men start from the point $C$ with velocity $v$ each at the same time.

Also, $\angle BCA = {45^\circ }$

Since, $A$ and $B$ are moving with same velocity $v$, so they will cover same distance in same time.

Therefore, $\Delta ABC$ is an isosceles triangle with $AC = BC$.

Now, draw $CD \bot AB$.

Let at any instant $t$, the distance between them is $AB$.

Let us assume $AC = BC = x$ and $AB = y$

In $\Delta ACD$ and $\Delta DCB$,
$\angle CAD = \angle CBD$

$\angle CDA = \angle CDB = {90^\circ }$

Therefore,$\angle ACD = \angle DCB$

or $\angle ACD = \frac{1}{2} \times \angle ACB$

$\Rightarrow$ $\angle ACD = \frac{1}{2} \times {45^\circ }$

$\Rightarrow$ $\angle ACD = \frac{\pi }{8}$

Therefore,$\sin \frac{\pi }{8} = \frac{{AD}}{{AC}}$
$\Rightarrow$ $\sin \frac{\pi }{8} = \frac{{y/2}}{x}$

$\Rightarrow$ $\frac{y}{2} = x\sin \frac{\pi }{8}$

$\Rightarrow$ $y = 2x \cdot \sin \frac{\pi }{8}$

Now, differentiating both sides w.r.t. $t$, we get

$\frac{{dy}}{{dt}} = 2 \cdot \sin \frac{\pi }{8} \cdot \frac{{dx}}{{dt}}$
$= 2 \cdot \sin \frac{\pi }{8} \cdot v$

$= 2v \cdot \frac{{\sqrt {2 - \sqrt 2 } }}{2}$

$= \sqrt {2 - \sqrt 2 } v$ unit/s

which is the rate at which $A$ and $B$ are being separated.