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 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:
Apply chain rule to determine \(\frac{d}{d\theta}\frac{1}{(1-cos\theta)}\)