Question:
Find $\frac{d y}{d x}$, when
$x=a(1-\cos \theta)$ and $y=a(\theta+\sin \theta)$ at $\theta=\frac{\pi}{2}$
Solution:
We have, $x=a(1-\cos \theta)$ and $y=a(\theta+\sin \theta) \therefore \frac{d x}{d \theta}=\frac{d}{d \theta}[a(1-\cos \theta)]$
$=a(\sin \theta)$ and $\frac{d y}{d \theta}=\frac{d}{d \theta}[a(\theta+\sin \theta)]$
$=a(1+\cos \theta) \therefore\left[\frac{d y}{d x}\right]_{\theta=\frac{\pi}{2}}=\left[\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\right]_{\theta=\frac{\pi}{2}}$
$=\left[\frac{a(1+\cos \theta)}{a(\sin \theta)}\right]_{\theta=\frac{\pi}{2}}$
$=\frac{a(1+0)}{a}=1$
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