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If x = a (1 – cos3θ), y = a sin3 θ, Prove that \(\frac{d^2y}{dx^2}=\frac{32}{27}\) at \(\theta=\frac{\pi}{6}.\)

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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 (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 :

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