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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\cos\theta-\cos2\theta,\text{y}=\sin\theta-\sin2\theta$
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Answer

The given equations are $\text{x}=\cos\theta-\cos2\theta\text{ and y}=\sin\theta-\sin2\theta$
Then, $\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\cos\theta-\cos2\theta)=\frac{\text{d}}{\text{d}\theta}(\cos\theta)-\frac{\text{d}}{\text{d}\theta}(\cos2\theta)$
$=-\sin\theta-(-2\sin2\theta)=2\sin2\theta-\sin\theta$
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\sin-\sin2\theta)=\frac{\text{d}}{\text{d}\theta}(\sin\theta)-\frac{\text{d}}{\text{d}\theta}(\sin2\theta)$
$=\cos\theta-2\cos2\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{\cos\theta-2\cos2\theta}{2\sin2\theta-\sin\theta}$

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