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# If $x=a\sin 2t(1+\cos 2t)\;and\;y=b\cos 2t(1-\cos 2t),show\;that\bigg(\large \frac{dy}{dx}\bigg)_{at\;t=\frac{\pi}{4}}=\frac{-b}{a}$

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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}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}} Step 1: x=a\sin 2t(1+\cos 2t) and y=b\cos 2t(1-\cos 2t) Consider x=a\sin 2t+a\sin 2t\cos 2t But \sin 2t\cos 2t=\large\frac{1}{2}$$\sin 4t$
Therefore $x=a\sin 2t+\large\frac{a}{2}$$\sin 4t Differentiating this w.r.t t we get, \large\frac{dx}{dt}$$=2a\cos 2t+\large\frac{a}{2}$$\times 4\cos 4t \quad\;=2a\cos 2t+2a\cos 4t \quad\;=2a(\cos 2t+\cos 4t) Step 2: Consider y=b\cos 2t-b\cos^22t But \cos^22t=\large\frac{1+\cos 4t}{2} Therefore y=b\cos 2t-b\big(\large\frac{1+\cos 4t}{2}\big) \qquad\;\;\;\;\;\;\;\;=b\cos 2t-\large\frac{b}{2}-\large\frac{b}{2}$$\cos 4t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}$$=-2b\sin 2t-\big(\large\frac{-b}{2}$$\times 4\sin 4t)$
$\qquad=-2b\sin 2t+2b\sin 4t$
$\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}} \quad=\large\frac{2b(\sin 4t-\sin 2t)}{2a(\cos 2t+\cos 4t)} \quad=\large\frac{b(\sin 4t-\sin t)}{a(\cos 2t+\cos 4t)} Step 3: Now \bigg(\Large\frac{dy}{dx}\bigg)_{at\;t=\large\frac{\pi}{4}}=\large\frac{b}{a}\bigg[\large\frac{\sin 4\large\frac{\pi}{4}-\sin2.\Large\frac{\pi}{4}}{\cos 2.\Large\frac{\pi}{4}-\normalsize\cos4.\Large\frac{\pi}{4}}\bigg] \qquad\qquad\qquad\qquad\;\;=\large\frac{b}{a}\bigg[\large\frac{\sin \pi-\sin\Large\frac{\pi}{2}}{\cos\Large\frac{\pi}{2}-\cos \pi}\bigg] But \sin\pi=\cos\large\frac{\pi}{2}$$=0$