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 $\frac{d y}{d x}$.

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

Solution:

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

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

$=a[-\sin \theta+\theta \cos \theta+\sin \theta]=a \theta \cos \theta$

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

$=a[\cos \theta+\theta \sin \theta-\cos \theta]$

$=a \theta \sin \theta$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$

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