If *x* = *a* (1 + cos θ), *y* = *a*(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = \frac{- 1}{a}at \theta = \frac{\pi}{2}\]

#### Solution

Here,

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

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

\[\frac{d x}{d \theta} = - a\sin\theta \text{ and } \frac{d y}{d \theta} = a + a \cos\theta\]

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

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

\[\frac{d^2 y}{d x^2} = \frac{d}{d\theta}\left\{ \frac{d y}{d x} \right\}\frac{d\theta}{dx}\]

\[\frac{d^2 y}{d x^2} = - \left\{ \frac{- \sin^2 \theta - \cos\theta - \cos^2 \theta}{\sin^2 \theta} \right\}\frac{d\theta}{dx}\]

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

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

\[\text{ At } \theta = \frac{\pi}{2}: \frac{d^2 y}{d x^2} = \frac{- \left( 1 + \cos\frac{\pi}{2} \right)}{a \left( \sin\frac{\pi}{2} \right)^3} = \frac{- 1}{a}\]