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 (1 – cos3θ) ……equation 1
y = a sin3 θ, ……equation 2
to prove : \(\frac{d^2y}{dx^2}=\frac{32}{27}\) at \(\theta=\frac{\pi}{6}.\)
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.
\(\frac{dx}{d\theta}\)\(=\frac{d}{d\theta}a(1-cos^3\theta)=3acos^2\theta sin\theta\)........equation 3 [using chain rule]
Similarly,
Differentiating again w.r.t x :
[ using chain rule and \(\frac{d}{dx}tanx=sec^2x\)]
From equation 3:
Putting the value in equation 5 :