If x and y are connected parametrically by the equation,
Question:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.

$x=a(\theta-\sin \theta), y=a(1+\cos \theta)$

Solution:

The given equations are $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$

Then, $\frac{d x}{d \theta}=a\left[\frac{d}{d \theta}(\theta)-\frac{d}{d \theta}(\sin \theta)\right]=a(1-\cos \theta)$

$\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(1)+\frac{d}{d \theta}(\cos \theta)\right]=a[0+(-\sin \theta)]=-a \sin \theta$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}$

 

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