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

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

Answers (1)

Answer:

            \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}}

 

 

Posted by

infoexpert27

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