#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 27

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

Hint:

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

Given:

\begin{aligned} &x=3 \sin t-\sin 3 t \\ &y=3 \cos t-\cos 3 t \end{aligned}

Solution:

$x=3 \sin t-\sin 3 t$

\begin{aligned} &\\ &\frac{d x}{d t}=3 \frac{d(\sin t)}{d t}-\frac{d(\sin 3 t)}{d t} \end{aligned}

$\frac{d x}{d t}=3 \cos t-3 \cos 3 t \\$                                                                             $\left[\because \frac{d \sin x}{d x}=\cos x\right]$

\begin{aligned} & &\frac{d x}{d t}=3(\cos t-\cos 3 t) \end{aligned}                                                                                                                                  (1)

$\therefore y=3 \cos t-\cos 3 t \\$

\begin{aligned} & &\frac{d y}{d t}=\frac{d(3 \cos t)}{d t}-\frac{d(\cos 3 t)}{d t} \end{aligned}                                                               $\left[\because \frac{d \cos x}{d x}=-\sin x\right]$

$=-3 \sin t-(-3 \sin 3 t) \\$

$\frac{d y}{d t}=-3 \sin t+3 \sin 3 t \\$

$\frac{d y}{d t}=3(\sin 3 t-\sin t) \\$                                                                                                                 (2)

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

Putting the value of  $\frac{d x}{d t} \text { and } \frac{d y}{d t}$   from the equation (1) and (2) respectively

$\frac{d y}{d x}=\frac{3(\sin 3 t-\sin t)}{3(\cos t-\cos 3 t)} \\$

\begin{aligned} & &\frac{d y}{d x}=\frac{\sin 3 t-\sin t}{\cos t-\cos 3 t} \end{aligned}

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

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

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

\begin{aligned} & &=\frac{0-\frac{\sqrt{3}}{2}}{\frac{1}{2}-(-1)} \end{aligned}

$=\frac{\frac{-\sqrt{3}}{2}}{\frac{3}{2}}=\frac{-\sqrt{3} \times 2}{3 \times 2}=\frac{-1}{\sqrt{3}}$

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