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

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Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :\(\frac{dy}{dx}\)\(=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

Given,

x = a (θ + sin θ) ……equation 1

y = a (1+ cos θ) ……equation 2

to prove :\(\frac{d^2y}{dx^2}=-\frac{a}{y^2}\) 

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

 Let’s find \(\frac{d^2y}{dx^2}\)

As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\) 

So, lets first find dy/dx using parametric form and differentiate it again.

Similarly,

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

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