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  • If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = cos theta - cos 2 theta , y =sin theta -sin 2 theta

Q5    If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta

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Given equations are
x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta
Now, differentiate both w.r.t  \theta
We get,
\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}
Therefore, answer is \frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}
 

Posted by

Gautam harsolia

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