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Question
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Solution
Applying parametric differentiation \[\frac{dx}{d\theta}\] =2a − 2acos2 \[\theta\] \[\frac{dy}{d\theta}\] = 0 + 2asin2 \[\theta\] \[\frac{dy}{dx}\] = \[\frac{dy}{d\theta} \times \frac{d\theta}{dx} = \frac{\sin2\theta}{1 - \cos2\theta}\] Now putting the value of \[\theta\] = \[\frac{\pi}{3}\]
\[\frac{dy}{dx}_\theta = \frac{\pi}{3} = \frac{\sin2\left( \frac{\pi}{3} \right)}{1 - \cos2\left( \frac{\pi}{3} \right)}\]
\[ = \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\]
\[ = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{1}{\sqrt{3}}\]
So,
\[\frac{dy}{dx}\] \[\frac{1}{\sqrt{3}}\] at \[\theta = \frac{\pi}{3}\] .
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