**Toolbox:**

- If $s=f(t)$ is the distance function, representing the distance 's' travelled by a particle in time t, then the velocity and acceleration functions are $v=\large\frac{ds}{dt}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}$$=f''(t)$
- It $y=f(x)$ and both depend on 't' then $\large\frac{dy}{dt}$$=f'(x).\large\frac{dx}{dt}$

Let OA,OB be the sides of $\Delta OAB$ where $OA=12\;m,OB=15\; m$ and $\angle AOB=\theta$ at time t

The length h of the third side AB at time t is given by

$s^2=OA^2+OB^2-2OA-OB\cos \theta$

$s^2=144+225-360\;\cos \theta$

$\quad= 369-360 \;\cos \theta$-----(i)

Differentiating (i) with respect to 't' we have

$2s \large\frac{ds}{dt}$$=+360 \sin \theta \large\frac{d\theta}{d t}$

$=>\large\frac{ds}{dt}=\large\frac{180}{s} $$\sin \theta \large\frac{d \theta}{dt}$-----(ii)

When $\theta=60 ^{\circ}=\large \frac{\pi ^c}{3}$

$\large\frac{d\theta}{dt}$$=2^{\circ}/min=\large\frac{\pi^c}{90}$$/min$

and $s^2=369-360 \times \large\frac{1}{2}$$=189$

ie $s=\sqrt {189}=3 \sqrt {21}$

At this instant substituting for $s, \large\frac{d \theta}{dt}$ in (ii)

$\large\frac{ds}{dt}=\large \frac{+180}{3 \sqrt {21}}\;$$ \sin \large\frac{\pi}{3} \times \frac{\pi}{90}$

$\qquad=\large\frac{+60}{\sqrt {21}} \times \frac{\sqrt 3}{2} \times \frac{\pi}{90}$

$\qquad=\large\frac{+ \sqrt 3 \pi}{3 \sqrt {21}}=\frac{+\pi}{\sqrt {63}}$$m/sec$

The third side is increasing at the rate of $\large\frac{+\pi}{\sqrt {63}}$$m/sec$