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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 $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta),\text{y}=\text{a}(\sin\theta-\theta\cos\theta)$
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Answer

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

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