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  • Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 9

Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 9

Answers (1)

Answer:

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

Hint:

            Use product rule

Given:

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

Solution:

x=a(\cos \theta+\theta \sin \theta) \\

\frac{d x}{d \theta}=a \frac{d}{d \theta}(\cos \theta+\theta \sin \theta)

\begin{aligned} &\\ &=a\left[\frac{d \cos \theta}{d \theta}+\frac{d}{d \theta}(\theta \sin \theta)\right] \end{aligned}

=a\left[-\sin \theta+\left(\theta \frac{d \sin \theta}{d \theta}+\sin \theta \cdot \frac{d \theta}{d \theta}\right)\right]                                                [Using product rule]

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

=a(\theta \cos \theta) \\

\begin{aligned} & &\frac{d x}{d \theta}=a \theta \cos \theta \end{aligned}                         (1)

y=a(\sin \theta-\theta \cos \theta) \\

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

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

\begin{aligned} & &=a\left[\cos \theta-\left(\theta \cdot \frac{d \cos \theta}{d \theta}+\cos \theta \cdot \frac{d \theta}{d \theta}\right)\right] \end{aligned}                                               [Using product rule]

=a[\cos \theta-(\theta(-\sin \theta)+\cos \theta)] \\                                            \left[\begin{array}{l} \because \frac{d \cos \theta}{d \theta}=-\sin \theta \\ \frac{d x}{d x}=1 \end{array}\right]

=a[\cos \theta+\theta \sin \theta-\cos \theta] \\

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

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

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

\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta \\

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

 

Posted by

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