Find $\frac{d y}{d x}$, when
$x=a e^{\theta}(\sin \theta-\cos \theta), y=a e^{\theta}(\sin \theta+\cos \theta)$
We have, $x=a e^{\theta}(\sin \theta-\cos \theta)$ and $y=a e^{\theta}(\sin \theta+\cos \theta)$
$\Rightarrow \frac{d x}{d \theta}=a\left[e^{\theta} \frac{d}{d \theta}(\sin \theta-\cos \theta)+(\sin \theta-\cos \theta) \frac{d}{d \theta}\left(e^{\theta}\right)\right]$
and $\frac{d y}{d \theta}=a\left[e^{\theta} \frac{d}{d \theta}(\sin \theta+\cos \theta)+(\sin \theta+\cos \theta) \frac{d}{d \theta}\left(e^{\theta}\right)\right]$
$\Rightarrow \frac{d x}{d \theta}=a\left[e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) e^{\theta}\right]$
$\Rightarrow \frac{d x}{d \theta}=a\left[2 e^{\theta} \sin \theta\right]$ and $\frac{d y}{d \theta}=a\left[2 e^{\theta} \mathrm{cc}\right.$
$\therefore \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a\left(2 e^{\theta} \cos \theta\right)}{a\left(2 e^{\theta} \sin \theta\right)}=\cot \theta$
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