Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 29

$\frac{d y}{d x} \text { At }\left(t=\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}}$

Hint:

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

Given:

\begin{aligned} &x=a(2 \theta-\sin 2 \theta) \\ &y=a(1-\cos 2 \theta) \end{aligned}

Solution:

$x=a(2 \theta-\sin 2 \theta) \\$

$\frac{d x}{d \theta}=a \frac{d(2 \theta-\sin 2 \theta)}{d \theta} \\$

\begin{aligned} & &=a \times\left[\frac{d(2 \theta)}{d \theta}-\frac{d(\sin 2 \theta)}{d \theta}\right] \end{aligned}

$=a \times[2-2 \cos 2 \theta]$                                                        $\quad\left[ \because \frac{d(\sin \theta)}{d \theta}=\cos \theta\right]$

$=a \times 2(1-\cos 2 \theta) \\$

\begin{aligned} & &\frac{d x}{d \theta}=2 a(1-\cos 2 \theta) \end{aligned}                                                                                                                 (1)

$y=a(1-\cos 2 \theta) \\$

$\frac{d y}{d \theta}=a \frac{d(1-\cos 2 \theta)}{d \theta} \\$

\begin{aligned} & &=a \times\left[\frac{d 1}{d \theta}-\frac{d \cos 2 \theta}{d \theta}\right] \end{aligned}

\begin{aligned} &=a[0+2 \sin 2 \theta] \\ & \end{aligned}                                                            $\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]$

$\frac{d y}{d \theta}=2 a \sin 2 \theta$                                                                                                                                                               (2)

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

So, put the value of   $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$   from equation (2) and (1) respectively

$\frac{d y}{d x}=\frac{2 a \sin 2 \theta}{2 a(1-\cos 2 \theta)} \\$

\begin{aligned} & &\frac{d y}{d x}=\frac{\sin 2 \theta}{1-\cos 2 \theta} \end{aligned}

$\text { At } \theta=\frac{\pi}{3} \\$

\begin{aligned} & &\frac{d y}{d x}=\frac{\sin \left(2 \times \frac{\pi}{3}\right)}{1-\cos \left(2 \times \frac{\pi}{3}\right)} \end{aligned}

$=\frac{\sin \frac{2 \pi}{3}}{1-\cos \frac{2 \pi}{3}} \\$

$=\frac{\sin \left(\pi-\frac{\pi}{3}\right)}{1-\cos \left(\pi-\frac{\pi}{3}\right)} \\$

\begin{aligned} & =& \frac{\sin \frac{\pi}{3}}{1-\left(-\cos \frac{\pi}{3}\right)} \end{aligned}                                                                              $\left[\begin{array}{c} \sin (\pi-\theta)=\sin \theta \\ \cos (\pi-\theta)=-\cos \theta \end{array}\right]$

$=\frac{\sin \frac{\pi}{3}}{1+\cos \frac{\pi}{3}} \\$

\begin{aligned} & &=\frac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}} \end{aligned}                                                                                               $\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, \cos \frac{\pi}{3}=\frac{1}{2}\right]$

$=\frac{\frac{\sqrt{3}}{2}}{\frac{(2+1)}{2}} \\$

$=\frac{\sqrt{3} \times 2}{3 \times 2} \\$

\begin{aligned} & &\frac{d y}{d x} \text { at }\left(\theta=\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}