If $x = a (θ - \sin θ), y = (1 + \cos θ), find \frac{d^{2} y}{{dx}^{2}}$

Solution

$x = a (θ - \sin θ), y = a (1 + \cos θ) Differentiating x and y w . r . t . θ, \frac{dx}{dθ} = a (1 - \cos θ) . . . . . . . . . (i) \frac{dy}{dθ} = - a \sin θ . . . . . . . . . . (ii) Dividing (2) by (1), \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} = \frac{- a \sin θ}{a (1 - \cos θ)}$

$\Rightarrow \frac{dy}{dx} = \frac{- sinθ}{1 - \cos θ} \Rightarrow \frac{dy}{dx} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} \Rightarrow \frac{dy}{dx} = \frac{- \cos \frac{θ}{2}}{\sin \frac{θ}{2}} \Rightarrow \frac{dy}{dx} = - \cot \frac{θ}{2}$

Differentiating w.r.t. x,

$\frac{d}{dx} (\frac{dy}{dx}) = \frac{d}{dθ} (\frac{dy}{dx}) x \frac{dθ}{dx} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dθ} (\frac{dy}{dx}) x \frac{dθ}{dx} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dθ} (- \cot \frac{θ}{2}) x \frac{dθ}{dx} . . . . [From equation (iii)] \frac{d^{2} y}{{dx}^{2}} = - (- {cosec}^{2} \frac{θ}{2} x \frac{1}{2}) x \frac{dθ}{dx} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} x \frac{1}{(\frac{dx}{dθ})}$

$= \frac{1}{2} {cosec}^{2} \frac{θ}{2} x \frac{1}{a (1 - \cos θ)} . . . . . . . . [From equation (i)] = \frac{{cosec}^{2} \frac{θ}{2}}{2 a (1 - \cos θ)} = \frac{{cosec}^{2} \frac{θ}{2}}{2 a (2 \sin^{2} \frac{θ}{2})} = \frac{1}{4 a} x {cosec}^{4} \frac{θ}{2}$

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Continuity And Differentiability

If $x = a (θ - \sin θ), y = (1 + \cos θ), find \frac{d^{2} y}{{dx}^{2}}$

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Continuity And Differentiability

If x = a θ - sin θ , y = 1 + cos θ , find d2ydx2

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If $x = a (θ - \sin θ), y = (1 + \cos θ), find \frac{d^{2} y}{{dx}^{2}}$