Question

If $x=a(\theta -\sin \theta )$ and $y=a(1-\cos \theta )$, find $\frac{d^{2}y}{d{x}^2}$ at $\theta =\pi$.

Answer 1

$x=a(\theta -\sin \theta )$

$\Rightarrow y=a(1-\cos \theta )$

$\Rightarrow \frac{dy}{d\theta }=a\sin \theta$

$\frac{dx}{d\theta }=a(1-\cos \theta )$

$\frac{dy}{dx}=\frac{a\sin \theta }{a(1-\cos \theta )}=\cot \left(\frac{\theta }{2}\right)$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\cot \left(\frac{\theta }{2}\right)\right)=\left(\frac{-1}{2}{\csc }^2\frac{\theta }{2}\right)\times \frac{d\theta }{dx}$

$=\left(\frac{-1}{2}{\csc }^2\frac{\theta }{2}\right)\times \frac{1}{a(1-\cos \theta )}$

at $\theta =\pi$

$=\left(\frac{-1}{2}{\csc }^2\frac{\pi }{2}\right)\times \frac{1}{a(1-\cos \left(\pi \right))}$

$=\frac{-1}{2}{.1}^2\times \frac{1}{a(1-(-1))}$

$=\frac{-1}{4a}$

Hence, the answer is $\frac{-1}{4a}.$