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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta),\text{y}=\text{ae}^\theta(\sin\theta+\cos\theta)$
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Answer

We have, $\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta)$ and $\text{y}=\text{ae}^{\theta}(\sin\theta+\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta-\cos\theta)+(\sin\theta-\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta+\cos\theta)+(\sin\theta+\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta+\sin\theta)+(\sin\theta-\cos\theta)(\text{e}^\theta)\Big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta-\sin\theta)+(\sin\theta+\cos\theta)(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\sin\theta)\big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\cos\theta)\big]$
$\therefore\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}(2\text{e}^\theta\cos\theta)}{\text{a}(2\text{e}^\theta\sin\theta)}=\cot\theta$

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