#### Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 6

$\frac{dy}{dx}=1$  At  $\theta =\frac{\pi}{2}$

Hint:

Use differentiation formula

Given:

$x=a(1-\cos \theta) \\$ , \begin{aligned} &y=a(\theta+\sin \theta) \end{aligned}

Solution:

$x=a(1-\cos \theta) \\$

$\frac{d x}{d \theta}=a \frac{d(1-\cos \theta)}{d \theta} \\$

\begin{aligned} & &=a\left[0-\frac{d \cos \theta}{d \theta}\right] \end{aligned}                                                                    $\left(\frac{d(\operatorname{CONSTANT})}{d \theta}=0\right)$

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

$\frac{d x}{d \theta}=a \sin \theta \\$

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

$\frac{d y}{d \theta}=a\left(\frac{d \theta}{d \theta}+\frac{d \sin \theta}{d \theta}\right) \\$

\begin{aligned} & &=a(1+\cos \theta) \end{aligned}

$\frac{d x}{d \theta}$  At  $\quad \theta=\frac{\pi}{2} \\$

$=a\left(1+\cos \frac{\pi}{2}\right) \\$

$=a(1+0) \\$

\begin{aligned} &=a \end{aligned}

$\frac{d x}{d \theta}$   At  $\quad \theta=\frac{\pi}{2} \\$

$=a \sin \frac{\pi}{2} \\$

\begin{aligned} &=a \end{aligned}

$\frac{d y}{d \theta}$   At  $\quad \theta=\frac{\pi}{2} \$

\begin{aligned} &\ &\frac{\left(\frac{d y}{d \theta}\right) a t \theta=\frac{\pi}{2}}{\left(\frac{d x}{d \theta}\right) \operatorname{at} \theta=\frac{\pi}{2}}=\frac{a}{a}=1 \end{aligned}