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If x and y are connected parametrically by the equations given in Exercise x=a(\theta-\sin \theta), y=a(1+\cos \theta) without eliminating the parameter, find dy/dx.
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.6 | Maths | NCERT
If x and y are connected parametrically by the equations given in Exercise x=a(\theta-\sin \theta), y=a(1+\cos \theta) without eliminating the parameter, find dy/dx.

Solution:

The provided functions are x=a(\theta-\sin \theta) and y=a(1+\cos \theta)

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

=-\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}

=-\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}

=-\cot \frac{\theta}{2}

i

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