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  • Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 14

Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 14

Answers (1)

Answer:

            \frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)

Hint:

            Use chain rule

Given:

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

Solution:

=2 \cos \theta-\cos 2 \theta \\

\frac{d x}{d \theta}=\frac{d(2 \cos \theta)}{d \theta}-\frac{d(\cos 2 \theta)}{d \theta} \\

\begin{aligned} &x &=2(-\sin \theta) \cdot \frac{d \cos 2 \theta}{d(2 \theta)} \times \frac{d(2 \theta)}{d \theta} \end{aligned}                                                                      [Using chain rule]

=-\sin \theta-(-\sin 2 \theta) \times 2 \\

=-2 \sin \theta+2 \sin 2 \theta \\

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

y=2 \sin \theta-\sin 2 \theta \\

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

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

=2 \cos \theta-[\cos 2 \theta \times 2] \\

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

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

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

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

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

=\frac{-2 \sin \left(\frac{\theta+2 \theta}{2}\right) \cdot \sin \left(\frac{\theta-2 \theta}{2}\right)}{2 \cos \left(\frac{2 \theta+\theta}{2}\right) \cdot \sin \left(\frac{2 \theta-\theta}{2}\right)}                                     \left[\begin{array}{l} \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \\ \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \end{array}\right]

=\frac{-\sin \left(\frac{3 \theta}{2}\right) \cdot \sin \left(\frac{-\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \cdot \sin \left(\frac{\theta}{2}\right)} \\

\begin{aligned} & &=\frac{-\sin \left(\frac{3 \theta}{2}\right) \cdot\left(-\sin \frac{\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \cdot \sin \frac{\theta}{2}} \end{aligned}

=\frac{\sin \left(\frac{3 \theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right)} \\

\begin{aligned} & &\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right) \end{aligned}                                                                                                \left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]

 

Posted by

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