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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
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Answer

We have, $\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
$\Rightarrow\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] $ and
$\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] $
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[-\sin\theta+\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)+\sin\theta\frac{\text{d}}{\text{d}\theta}(\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\cos\theta-\left\{\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)+\cos\theta\frac{\text{d}}{\text{d}\theta}(\theta)\right\}\Big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[-\sin\theta+\theta\cos\theta\big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[\cos\theta+\theta\sin\theta-\cos\theta\big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\theta\cos\theta$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}\theta\sin\theta}{\text{a}\theta\cos\theta}=\tan\theta$

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