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Question
If x = a (θ + sin θ), y = a (1 + cos θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{a}{y^2}\] ?
Solution
Here,
\[x = a\left( \theta + \sin\theta \right) \text{ and y } = a\left( 1 + \cos\theta \right)\]
\[\text { Differentiating w . r . t . }\theta, \text { we get }\]
\[\frac{d x}{d \theta} = a + a\cos\theta \text { and } \frac{d y}{d \theta} = - a \sin\theta\]
\[ \therefore \frac{d y}{d x} = \frac{- a \sin\theta}{a + a \cos\theta} = \frac{- \sin\theta}{1 + \cos\theta}\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = - \left\{ \frac{\left( 1 + \cos\theta \right)\cos\theta + \sin^2 \theta}{\left( 1 + cos\theta \right)^2} \right\}\frac{d\theta}{dx}\]
\[ = \frac{- \cos\theta - \cos^2 \theta - \sin^2 \theta}{\left( 1 + \cos\theta \right)^2} \times \frac{1}{a + a\cos\theta}\]
\[ = \frac{- \left( 1 + \cos\theta \right)}{a \left( 1 + \cos\theta \right)^3}\]
\[ = \frac{- 1}{a \left( 1 + \cos\theta \right)^2}\]
\[ = \frac{- a}{y^2} \left[ \because y = a\left( 1 + \cos\theta \right) \right]\]
Hence proved.
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