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

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  • 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 $\sin\pi=0,\sin\large\frac{\pi}{3}=\frac{\sqrt 3}{2}$
$\cos \large\frac{\pi}{3}=\frac{1}{2},$$\cos \pi=-1$
Therefore $\large\frac{dy}{dx}=\frac{0-\large\frac{\sqrt 3}{2}}{\large\frac{1}{2}+1}$
$\qquad\qquad\;\;\;=-\large\frac{\Large\frac{\sqrt 3}{2}}{\Large\frac{3}{2}}$
$\qquad\qquad\;\;\;=\large\frac{-1}{\sqrt 3}$
answered Jul 1, 2013 by sreemathi.v

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