If $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ), find \frac{dy}{dx} when θ = \frac{π}{3}$

Solution

$y = a [1 - \cos 2 θ], \frac{dy}{dθ} = a (0 - 2 \sin 2 θ) ∴ \frac{dy}{dθ} = - 2 a \sin 2 θ x = a (2 θ - \sin 2 θ), \frac{dy}{dθ} = a (2 - 2 \cos 2 θ) \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}} = \frac{- 2 a \sin 2 θ}{2 a [1 - \cos 2 θ]} (∵ \frac{dx}{dθ} \neq 0) \Rightarrow \frac{- 2 \sin θ cosθ}{2 \sin^{2} θ} = - \cot θ ∴ \frac{dy}{dx} = - \cot (\frac{π}{3}) = - \frac{1}{\sqrt{3}}$

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Differential Equations

If $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ), find \frac{dy}{dx} when θ = \frac{π}{3}$

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Differential Equations

If x =a (2θ - sin 2θ) and y = a(1- cos 2θ), finddydx when θ =π3

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ), find \frac{dy}{dx} when θ = \frac{π}{3}$