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  • Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 23 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 23 Maths Textbook Solution.

Answers (1)

Answer:

            \frac{d y}{d x}\; \: \mathrm{At}\; \; \left(\theta=\frac{\pi}{3}\right)=-\sqrt{3}

Hint:

            Use chain rule and   \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}

Given:

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

Solution:

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

\frac{d x}{d \theta}=a\left[\frac{d \theta}{d \theta}-\frac{d \sin \theta}{d \theta}\right] \\                                                            \left[\frac{d \sin \theta}{d \theta}=\cos \theta\right]

\frac{d x}{d \theta}=a(1-\cos \theta) \\                                                        (1)                                              

\begin{aligned} & &y=a(1+\cos \theta) \end{aligned}                                                                      

\frac{d y}{d \theta}=a\left(\frac{d 1}{d \theta}+\frac{d \cos \theta}{d \theta}\right) \\

=a(0+(-\sin \theta)) \\                                                                                     \left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]

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

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

Put the values of  \frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}  from the equation (2) and (1) respectively

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

\frac{d y}{d x}=\frac{-\sin \theta}{1-\cos \theta} \\

\begin{aligned} &\text { At } \theta=\frac{\pi}{3} \end{aligned}

\frac{d y}{d x}=-\left(\frac{\sin \frac{\pi}{3}}{1-\cos \frac{\pi}{3}}\right)

=-\left(\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}\right)                                                                       \left[\begin{array}{rl} \because \sin \frac{\pi}{3} & =\frac{\sqrt{3}}{2} \\\\ \cos \frac{\pi}{3} & =\frac{1}{2} \end{array}\right]                                                                       

=-\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) \\                                                   

\begin{aligned} & &\frac{d y}{d x} a t \theta=\frac{\pi}{3}=-\sqrt{3} \end{aligned}

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