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Question
If x = a(1 − cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{1}{a}\text { at } \theta = \frac{\pi}{2}\] ?
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Solution
Here,
\[x = a\left( 1 - \cos\theta \right) \text { and y } = a\left( \theta + \sin\theta \right)\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d x}{d \theta} = a \sin\theta \text { and }\frac{d y}{d \theta} = a + a \cos\theta\]
\[ \therefore \frac{d y}{d x} = \frac{a + a\cos\theta}{a\sin\theta} = \frac{1 + \cos\theta}{\sin\theta}\]
\[\text {Differentiating w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \left\{ \frac{- \sin^2 \theta - \left( 1 + \cos\theta \right)\cos\theta}{\sin^2 \theta} \right\}\frac{d \theta}{d x}\]
\[ = \left( - 1 - \frac{\left( 1 + cos\theta \right)cos\theta}{\sin^2 \theta} \right)\frac{1}{a \sin\theta}\]
\[ \therefore \frac{d^2 y}{d x^2} at \theta = \frac{\pi}{2} \]
\[ \Rightarrow \left[ \frac{d^2 y}{d x^2} \right]_\theta = \frac{\pi}{2} = \frac{1}{a}\left( - 1 - \frac{0}{1} \right) = \frac{- 1}{a}\]
Hence proved.
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