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  • Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 6

Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 6

Answers (1)

Answer:

            \frac{dy}{dx}=1  At  \theta =\frac{\pi}{2}

Hint:

            Use differentiation formula

Given:

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

Solution:

x=a(1-\cos \theta) \\

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

\begin{aligned} & &=a\left[0-\frac{d \cos \theta}{d \theta}\right] \end{aligned}                                                                    \left(\frac{d(\operatorname{CONSTANT})}{d \theta}=0\right)

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

\frac{d x}{d \theta}=a \sin \theta \\

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

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

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

\frac{d x}{d \theta}  At  \quad \theta=\frac{\pi}{2} \\

=a\left(1+\cos \frac{\pi}{2}\right) \\

=a(1+0) \\

\begin{aligned} &=a \end{aligned}

\frac{d x}{d \theta}   At  \quad \theta=\frac{\pi}{2} \\

=a \sin \frac{\pi}{2} \\

\begin{aligned} &=a \end{aligned}

\frac{d y}{d \theta}   At  \quad \theta=\frac{\pi}{2} \

\begin{aligned} &\ &\frac{\left(\frac{d y}{d \theta}\right) a t \theta=\frac{\pi}{2}}{\left(\frac{d x}{d \theta}\right) \operatorname{at} \theta=\frac{\pi}{2}}=\frac{a}{a}=1 \end{aligned}

 

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