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Question
If \[x = a \left( \theta - \sin \theta \right) and, y = a \left( 1 + \cos \theta \right), \text { find } \frac{dy}{dx} \text{ at }\theta = \frac{\pi}{3} \] ?
Solution
\[\text { We have, x} = a\left( \theta - \sin\theta \right) \text { and y } = a\left( 1 + \cos\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = \frac{d}{d\theta}\left[ a\left( \theta - \sin\theta \right) \right] \text { and } \frac{dy}{d\theta} = \frac{d}{d\theta}\left[ a\left( 1 + \cos\theta \right) \right]\]
\[ \Rightarrow \frac{dx}{d\theta} = a\left( 1 - \cos\theta \right) \text { and } \frac{dy}{d\theta} = a\left( - \sin\theta \right)\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- a \sin\theta}{a\left( 1 - \cos\theta \right)}\]
\[\text { Now, } \left[ \frac{dy}{dx} \right]_\theta = \frac{\pi}{3} = - \frac{\sin\frac{\pi}{3}}{1 - \cos\frac{\pi}{3}} = - \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = - \sqrt{3}\]
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