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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 = theta - sin theta y = a 1 + cos theta

 Q6   If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx  x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )

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Given equations are
x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )
Now, differentiate both w.r.t  \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)
Similarly,
\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})
Therefore, the answer is \frac{dy}{dx}=-\cot \frac{\theta}{2}

Posted by

Gautam harsolia

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