Question

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

Answer 1

$x=a(\theta -\sin \theta )$ & $y=a(1+\cos \theta )$
Now, $\frac{dx}{d\theta }$ $=a(1-\cos \theta )$ & $\frac{dy}{d\theta }$ $=a(-\sin \theta )$.
$\therefore \frac{dy}{dx}=-\frac{\sin \theta }{(1-\cos \theta )}$
$=-\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$
$=-\frac{2\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}}$
$=-\cot \frac{\theta }{2}$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{1}{2}cose{c}^2\left(\frac{\theta }{2}\right)\frac{d\theta }{dx}$
$=\frac{1}{2}cose{c}^2\left(\frac{\theta }{2}\right)\frac{1}{2a{\sin }^2\frac{\theta }{2}}$
$=\frac{1}{4a{\sin }^4\frac{\theta }{2}}$
$=\frac{1}{4a{(1-\frac{y}{2a})}^2}$ $[Since,y=2a{\cos }^2\frac{\theta }{2}\Rightarrow {\sin }^2\frac{\theta }{2}=(1-\frac{y}{2a})]$