If $x = 3 \sin t - \sin 3t$, $y = 3 \cos t - \cos 3t$, then find $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$.

$-\frac{1}{\sqrt{3}}$

The correct answer is Option (4) → $-\frac{1}{\sqrt{3}}$ ##

Given, $x = 3 \sin t - \sin 3t$ and $y = 3 \cos t - \cos 3t$

Taking derivative of both equations w.r.t. $t$, we get

$\frac{dx}{dt} = 3 \cdot \frac{d}{dt} \sin t - \frac{d}{dt} \sin 3t$

$= 3 \cos t - \cos 3t \cdot \frac{d}{dt} 3t = 3 \cos t - 3 \cos 3t \quad \dots (i)$

and $\frac{dy}{dt} = 3 \cdot \frac{d}{dt} \cos t - \frac{d}{dt} \cos 3t$

$= -3 \sin t + \sin 3t \cdot \frac{d}{dt} 3t$

$∴\frac{dy}{dt} = 3 \sin 3t - 3 \sin t \quad \dots (ii)$

On dividing Eq. (ii) by Eq. (i), we get

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3(\sin 3t - \sin t)}{3(\cos t - \cos 3t)}$

Now, $\left( \frac{dy}{dx} \right)_{t = \pi/3} = \frac{\sin \frac{3\pi}{3} - \sin \frac{\pi}{3}}{\cos \frac{\pi}{3} - \cos \frac{3\pi}{3}} = \frac{0 - \sqrt{3}/2}{\frac{1}{2} - (-1)}$

$\left[ ∵\sin \pi = 0, \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \cos \pi = -1 \text{ and } \cos \frac{\pi}{3} = \frac{1}{2} \right]$

$= \frac{-\sqrt{3}/2}{3/2} = \frac{-\sqrt{3}}{3} = \frac{-1}{\sqrt{3}}$