Question

Find the length of normal to the curve $r=a(\theta +\sin \theta ),y=a(1-\cos \theta )at\theta =\frac{\pi }{2}$.

Answer 1

Given $r=a(\theta +\sin \theta )\therefore \frac{dr}{d\theta }=a(1+\cos \theta )$

and, $y=a(1-\cos \theta )$ $\therefore \frac{dy}{d\theta }=a\sin \theta$

$\therefore \frac{dy}{dr}=\frac{dy}{d\theta }.\frac{d\theta }{dr}=\frac{a\sin \theta }{a(1+\cos \theta )}=\frac{\sin \theta }{1+\cos \theta }$

$\therefore {\left[\frac{dy}{dr}\right]}_{\theta =\pi /2}=\frac{\sin \left(\frac{\pi }{2}\right)}{1+\cos \left(\frac{\pi }{2}\right)}=\frac{1}{1}=1$

[$\frac{dy}{dr}$ is the equation of the normal to the given curve]

$\therefore$ The length of the normal to the curve at $\theta =\frac{\pi }{2}$ is $1$