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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$
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Answer

The given equations are $\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then, $\frac{\text{dx}}{\text{d}\theta}= \text{a}\Big[\frac{\text{d}}{\text{d}\theta}(\theta)-\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big]=\text{a}(1-\cos\theta)$
$\frac{\text{dy}}{\text{d}\theta}= \text{a}\Big[\frac{\text{d}}{\text{d}\theta}(1)+\frac{\text{d}}{\text{d}\theta}(\cos\theta)\Big]=\text{a}[0+(-\sin\theta)]=-\text{a}\sin\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}=\frac{-\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}=-\cot\frac{\theta}{2}$

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