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

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

Answers (1)

Answer:

            \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}

 

                                                                          

 

 

 

Posted by

infoexpert27

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