Solve this


Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when


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

$\frac{d x}{d t}=3 \cos t-3 \cos 3 t$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{-3 \sin \mathrm{t}+3 \sin 3 \mathrm{t}}{3 \cos \mathrm{t}-3 \cos 3 \mathrm{t}}$

When $t=\frac{\pi}{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-3 \sin \left(\frac{\pi}{3}\right)+3 \sin 3\left(\frac{\pi}{3}\right)}{3 \cos \left(\frac{\pi}{3}\right)-3 \cos 3\left(\frac{\pi}{3}\right)}$

$\frac{d y}{d x}=\frac{-3 \times \frac{\sqrt{3}}{2}+0}{\frac{3}{2}-3(-1)}=\frac{1}{\sqrt{3}}$

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