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### Answers (1)

Answer:

$\frac{d y}{d x}=\tan \theta$

Hint:

Use product rule

Given:

\begin{aligned} &x=a(\cos \theta+\theta \sin \theta) \\ &y=a(\sin \theta-\theta \cos \theta) \end{aligned}

Solution:

$x=a(\cos \theta+\theta \sin \theta) \\$

$\frac{d x}{d \theta}=a \frac{d}{d \theta}(\cos \theta+\theta \sin \theta)$

\begin{aligned} &\\ &=a\left[\frac{d \cos \theta}{d \theta}+\frac{d}{d \theta}(\theta \sin \theta)\right] \end{aligned}

$=a\left[-\sin \theta+\left(\theta \frac{d \sin \theta}{d \theta}+\sin \theta \cdot \frac{d \theta}{d \theta}\right)\right]$                                                [Using product rule]

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

$=a(\theta \cos \theta) \\$

\begin{aligned} & &\frac{d x}{d \theta}=a \theta \cos \theta \end{aligned}                         (1)

$y=a(\sin \theta-\theta \cos \theta) \\$

$\frac{d y}{d \theta}=a \frac{d}{d \theta}(\sin \theta-\theta \cos \theta) \\$

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

\begin{aligned} & &=a\left[\cos \theta-\left(\theta \cdot \frac{d \cos \theta}{d \theta}+\cos \theta \cdot \frac{d \theta}{d \theta}\right)\right] \end{aligned}                                               [Using product rule]

$=a[\cos \theta-(\theta(-\sin \theta)+\cos \theta)] \\$                                            $\left[\begin{array}{l} \because \frac{d \cos \theta}{d \theta}=-\sin \theta \\ \frac{d x}{d x}=1 \end{array}\right]$

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

\begin{aligned} & &\frac{d y}{d \theta}=a \theta \sin \theta \end{aligned}                                                                                                 (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Put the value of   $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$    from equations (2) and (1) respectively

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

\begin{aligned} & &\frac{d y}{d x}=\tan \theta \end{aligned}

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