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If $x=a(\theta - \sin \theta) \quad ; y= a(1+\cos \theta) $ find $\large\frac{d^2y}{dx^2}$ at $\theta =\large\frac{\pi}{2}$

$\begin{array}{1 1} 1 \\ \pi \\ 0 \\ 2\end{array} $

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Solution :
$y=a(1+ \cos \theta)$
$x=a(\theta - \sin \theta)$
$\large\frac{dy}{d \theta} $$=a(- \sin \theta)$
$\large\frac{dx}{d \theta} $$=a(1 - \cos \theta)$
$\large\frac{dy}{dx} =\large \frac{-a \sin \theta}{a(1- \cos \theta)}=\frac{-\sin \theta}{(1-\cos \theta)}$
$\qquad= \large\frac{-2 \sin \theta/2 \cos \theta/2}{2 \sin ^2 \theta/2}$
$\qquad= \large\frac{- \cos \theta/2}{\sin \theta/2}$$=-\cot \theta/2$
$\large\frac{dy}{dx} $$=-\cot \large\frac{\theta}{2}$
$\large\frac{d^2y}{dx^2} = \large\frac{-1}{2} $$ cosec^2 \large\frac{\theta}{2}$
$\bigg( \large\frac{d^2y}{dx^2} \bigg)_{\theta = \Large\frac{\pi}{2}} =\large\frac{1}{2} $$cosec ^2 \bigg(\large\frac{\pi}{4}\bigg)$
$\qquad= \large\frac{1}{2}.$$2=1$
answered Feb 16 by meena.p
 

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