# Solve this

Question:

Find $\frac{d y}{d x}$, when

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

Solution:

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

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

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

$\Rightarrow \frac{d x}{d \theta}=a[-\sin \theta+\theta \cos \theta]$ and $\frac{d y}{d \theta}=a[\cos \theta+\theta \sin \theta-\cos \theta]$

$\Rightarrow \frac{d x}{d \theta}=a \theta \cos \theta$ and $\frac{d y}{d \theta}=a \theta \sin \theta$

$\Rightarrow \frac{d x}{d \theta}=a \theta \cos \theta$ and $\frac{d y}{d \theta}=a \theta \sin \theta$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$