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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 $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
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Answer

The given equations are $\text{x}=\text{a}\sec\theta\text{ and y}=\text{b}\tan\theta$
Then, $\frac{\text{dx}}{\text{d}\theta}= \text{a}.\frac{\text{d}}{\text{d}\theta}(\sec\theta)=\text{a}\sec\theta\tan\theta$
$\frac{\text{dy}}{\text{d}\theta}=\text{b}\frac{\text{d}}{\text{d}\theta}(\tan\theta)=\text{b}\sec^2\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}}{\text{a}}\sec\theta\cot\theta$ $=\frac{\text{b}\cos\theta}{\text{a}\cos\theta\sin\theta}=\frac{\text{b}}{\text{a}}\times\frac{1}{\sin\theta}=\frac{\text{b}}{\text{a}}\ \text{cose}\theta$

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