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# If $x=3\sin t-\sin 3t,y=3\cos t-\cos 3t,find \;\large \frac{dy}{dx}\;\normalsize at \;t=\large \frac{\pi}{3}$

Toolbox:
• To find $\large\frac{dy}{dx}$ in the case of parametric functions,if $x=\phi(t)$ and $y=\psi(t)$,then $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}} Step 1: x=3\sin t-\sin 3t and y=3\cos t-\cos 3t Consider x=3\sin t-\sin 3t Differentiating w.r.t t we get, \large\frac{dx}{dt}=$$3\cos t-3\cos 3t$
Step 2:
Consider $y=a\cos t-\cos 3t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}$$=-3\sin t-(-3\sin 3t) \quad\;=-3\sin t+3\sin 3t Step 3: Therefore \large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
$\qquad\qquad\;\;=\large\frac{-3(\sin t-\sin 3t)}{3(\cos t-\cos 3t)}$
$\qquad\qquad\;\;=\large\frac{\sin 3t-\sin t}{\cos t-\cos 3t}$
Step 4:
When $t=\large\frac{\pi}{3}$
$\large\frac{dy}{dx}=\frac{\sin 3.\Large\frac{\pi}{3}-\sin\Large\frac{\pi}{3}}{\cos\Large\frac{\pi}{3}-\cos 3.\Large\frac{\pi}{3}}$
But we know $\sin\pi=0,\sin\large\frac{\pi}{3}=\frac{\sqrt 3}{2}$