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Question
If \[x = a \left( \theta + \sin \theta \right), y = a \left( 1 + \cos \theta \right), \text{ find} \frac{dy}{dx}\] ?
Solution
\[\text{ We have, x} = a\left( \theta + \sin\theta \right) \text{ and y }= a\left( 1 + \cos\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = a\left\{ \frac{d}{d\theta}\left( \theta \right) + \frac{d}{d\theta}\left( \sin\theta \right) \right\} \text{ and }\frac{dy}{d\theta} = a\left( 0 - \sin\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = a\left( 1 + \cos\theta \right) \text{ and } \frac{dy}{d\theta} = - a\sin\theta \]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- a\sin\theta}{a\left( 1 + \cos\theta \right)} = \frac{- 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = - \tan\frac{\theta}{2} \]
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