Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$

Solution:

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

Differentiating it with respect to $\theta$,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta) \ldots \ldots(1)$

And,

$y=a(1-\cos \theta)$

Differentiating it with respect to $\theta$,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)$

$\frac{d y}{d \theta}=a \sin \theta$        .....(2)

Using equation (1) and (2),

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}$

$=\frac{a \sin \theta}{a(1-\cos \theta)}$

$=\frac{\frac{2 \sin \theta(\cos \theta)}{2}}{\frac{2 \sin ^{2} \theta}{2}}$,

\{since, $\left.1-\cos \theta=\frac{2 \sin ^{2} \theta}{2}\right\}$

$=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\tan \theta}{2}$

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