Question:
Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when
If $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$, find $\frac{d y}{d x}$ at $\theta=\frac{\pi}{3}$
Solution:
Here,
$x=(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$
then,
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1-\cos \theta)$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{\mathrm{a}(-\sin \theta)}{\mathrm{a}(1-\cos \theta)}=\frac{(-\sin \theta)}{(1-\cos \theta)}$
At $x=\frac{\pi}{3}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}\left(-\sin \frac{\pi}{3}\right)}{\mathrm{a}\left(1-\cos \frac{\pi}{3}\right)}=\frac{\sqrt{3} / 2}{1-1 / 2}$
$=\sqrt{3}$