Question

If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?

Answer 1

 Here,

\[x = a\left( \theta - \sin\theta \right) \text { and y } = a\left( 1 + \cos\theta \right)\]

\[\text { Differentiating w . r . t . } \theta, \text { we get }\]

\[\frac{d x}{d \theta} = a - a\cos\theta, \frac{d y}{d \theta} = - a \sin\theta\]

\[ \Rightarrow \frac{d y}{d x} = \frac{- a \sin\theta}{a - a \cos\theta} = \frac{- \sin\theta}{1 - \cos\theta}\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{- \cos\theta + \cos^2 \theta + \sin^2 \theta}{\left( 1 - \cos\theta \right)^2} \times \frac{d\theta}{dx}\]

\[ = \frac{- \cos\theta + \cos^2 \theta + \sin^2 \theta}{\left( 1 - \cos\theta \right)^2} \times \frac{1}{a - a\cos\theta}\]

\[ = \frac{\left( 1 - \cos\theta \right)}{a \left( 1 - cos\theta \right)^3}\]

\[ = \frac{1}{a \left( 1 - \cos\theta \right)^2}\]