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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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Toolbox:
• 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
edited May 10, 2013