Question:
If $x=a(\theta-\sin \theta)$ and, $y=a(1+\cos \theta)$, find $\frac{d y}{d x}$ at $\theta=\frac{\pi}{3}$.
Solution:
We have, $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$
$\Rightarrow \frac{d x}{d \theta}=\frac{d}{d \theta}[a(\theta-\sin \theta)]$ and $\frac{d y}{d \theta}=\frac{d}{d \theta}[a(1+\cos \theta)]$
$\Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta)$ and $\frac{d y}{d \theta}=a(-\sin \theta)$
$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-a \sin \theta}{a(1-\cos \theta)}$
Now, $\left[\frac{d y}{d x}\right]_{\theta=\frac{\pi}{3}}=-\frac{\sin \frac{\pi}{3}}{1-\cos \frac{\pi}{3}}=-\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$