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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
$\text{If}\ \text{x}=\text{a}(2\theta-\sin2\theta)\ \text{and}\ \text{y}=\text{a}(1-\cos2\theta),\ \text{find}\ \frac{\text{dy}}{\text{dx}}\ \text{when}\ \theta=\frac{\pi}{3}.$
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Answer

$\text{x}=\text{a}(2\theta-\sin2\theta),\ \ \ \text{y}=\text{a}(1-\cos2\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(2-2\cos2\theta),\ \ \ \frac{\text{dy}}{\text{d}\theta}=\text{a}(0+2\sin2\theta)$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}$
$=\frac{2\text{a}\ \sin2\theta}{2\text{a}(1-\cos2\theta)}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{3}}=\frac{\sin\frac{2\pi}{3}}{1-\cos\frac{2\pi}{3}}$
$=\frac{\sin\big(\pi-\frac{\pi}{3}\big)}{1-\cos\big(\pi-\frac{\pi}{3}\big)}$
$=\frac{\sin\frac{\pi}{3}}{1+\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}}$
$=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

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