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If $x$ and $y$ are connected parametrically by the equations given in $ x = \cos \theta -\ cos 2\theta, y = \sin \theta -\ sin 2\theta $ without eliminating the parameter, Find $\large\frac{dy}{dx}$

$\begin{array}{1 1} -\large\frac{\cos\theta-2\cos \theta}{\sin\theta-2\sin \theta} \\ \large\frac{\cos\theta-2\cos 2\theta}{\sin\theta-2\sin 2\theta} \\ -\large\frac{\cos\theta-2\cos 2\theta}{\sin\theta-2\sin 2\theta} \\ -\large\frac{\cos\theta-2\cos 3\theta}{\sin\theta-2\sin 3\theta} \end{array} $

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  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
Step 1:
$x=\cos\theta-\cos 2\theta$
Differentiating with respect to $\theta$
$\large\frac{dx}{d\theta}$$=-\sin\theta-(-\sin 2\theta).2$
$\quad\;=-\sin\theta+\sin 2\theta.2$
$\quad\;=-\sin\theta+2\sin 2\theta$
$f'(\theta)=-\sin\theta+2\sin 2\theta$
Step 2:
$y=\sin\theta-\sin 2\theta$
$\large\frac{dy}{d\theta}$$=\cos\theta-(\cos 2\theta).2$
$\quad\;=\cos\theta-2\cos 2\theta$
$g'(\theta)=\cos\theta-2\cos 2\theta$
Step 3:
$\large\frac{dy}{dx}=\frac{g'(\theta)}{f'(\theta)}$
$\quad\;=\large\frac{\Large\frac{dy}{d\theta}}{\Large\frac{dx}{d\theta}}$
$\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
$\quad\;=\large\frac{\cos\theta-2\cos 2\theta}{-\sin\theta+2\sin 2\theta}$
$\quad\;=-\large\frac{\cos\theta-2\cos 2\theta}{\sin\theta-2\sin 2\theta}$
answered May 10, 2013 by sreemathi.v
edited May 10, 2013 by sreemathi.v
 

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