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,
y = a (θ + sin θ) ……equation 1
x = a (1+ cos θ) ……equation 2
to prove : \(\frac{d^2y}{dx^2}=\frac{-1}{a}\) at \(\theta=\frac{\pi}{2}\).
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:
[using equation 4]
As we have to find \(\frac{d^2y}{dx^2}=-\frac{1}{a}\) at \(\theta=\frac{\pi}{2}\)
∴ put θ = π/2 in above equation: