Question

Find $\frac{d^{2}y}{d{x}^2}$ If $x=a\left(\theta -\sin \theta \right),y=a\left(1+\cos \theta \right)$

Answer 1

$x=a(\theta -\sin \theta ),y=a(1+\cos \theta )$

$\therefore \frac{dx}{d\theta }=\frac{d}{d\theta }[a\theta -a\sin \theta ]=a-a\cos \theta$

$\therefore \frac{dy}{d\theta }=\frac{d}{d\theta }[a+a\cos \theta ]=-a\sin \theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-a\sin \theta }{a(1-\cos \theta )}$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(dy/dx\right)=\frac{d}{dx}\left[\frac{dy/d\theta }{dx/d\theta }\right]$

$=\frac{d}{dx}\left[\frac{-a\sin \theta }{a(1-\cos \theta )}\right]$

$=\frac{-\sin \theta \times \frac{d}{d\theta }(1-\cos \theta )\frac{d\theta }{dx}-(1-\cos \theta )\frac{d}{d\theta }(-\sin \theta )\frac{d\theta }{dx}}{{(1-\cos \theta )}^2}$

$=[\frac{-\sin \theta \sin \theta }{dx/d\theta }\times \frac{(1-\cos \theta )\cos \theta }{dx/d\theta }]\frac{1}{{(1-\cos \theta )}^2}$

$=\frac{\cos \theta -({\cos }^2\theta +{\sin }^2\theta )}{a(1-\cos \theta ){(1-\cos \theta )}^2}$

$=\frac{\cos \theta -1}{a{(1-\cos \theta )}^3}$

$=\frac{-1}{a{(1-\cos \theta )}^2}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{-1}{a{(1-\cos \theta )}^2}$