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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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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}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
$x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$
Consider $x=a\sin 2t+a\sin 2t\cos 2t$
But $\sin 2t\cos 2t=\large\frac{1}{2}$$\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$
$\sin\large\frac{\pi}{2}$$=1$ and $\cos \pi=-1$
Therefore $\large\frac{dy}{dx}=\bigg(\frac{0-1}{0+1}\bigg)\frac{b}{a}$
$\qquad\qquad=-1\big(\large\frac{b}{a}\big)$
Therefore $\large\frac{dy}{dx}=\large\frac{-b}{a}$
answered Jul 1, 2013 by sreemathi.v
 

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