If x = 3sint – sin 3t, y = 3cost – cos 3t, find
$\frac{d y}{d x}$ at $t=\frac{\pi}{3}$
Given,
x = 3sint – sin 3t, y = 3cost – 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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