If x = 3sint – sin 3t, y = 3cost – cos 3t, find

If = 3sin– sin 3t= 3cos– cos 3t, find

$\frac{d y}{d x}$ at $t=\frac{\pi}{3}$



= 3sin– sin 3t= 3cos– cos 3t

Now, differentiating both the parametric functions w.r.t t, we have

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

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

So, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3(-\sin t+\sin 3 t)}{3(\cos t-\cos 3 t)}=\frac{-\sin t+\sin 3 t}{\cos t-\cos 3 t}$

Putting, $t=\frac{\pi}{3}$

$\frac{d y}{d x}=\frac{-\sin \frac{\pi}{3}+\sin 3\left(\frac{\pi}{3}\right)}{\cos \frac{\pi}{3}-\cos 3\left(\frac{\pi}{3}\right)}$

$=\frac{-\frac{\sqrt{3}}{2}+\sin \pi}{\frac{1}{2}-\cos \pi}=\frac{-\frac{\sqrt{3}}{2}+0}{\frac{1}{2}-(-1)}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}+1}=\frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\frac{-1}{\sqrt{3}}$

Thus, $\frac{d y}{d x}=\frac{-1}{\sqrt{3}}$.


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