Question

If x = cos θ and y = sin³θ, show that `y(d^2y)/(dx^2) + (dy/dx)^2` = 3sin²θ(5cos²θ – 1)

Answer 1

x = cos θ,

y = sin³θ

Differentiating w.r.t. θ

`dx/(dθ)` = – sinθ

and `dy/(dθ)` = 3sin²θ·cosθ   ....(1)

∴ `dy/dx = ((dy)/(dθ))/((dx)/(dθ))` 

= `(3sin^2θcosθ)/(-sinθ)`

`dy/dx` = – 3 sin θ cos θ   ....(2)

Differentiating (2) w.r.t x

`d/dx(dy/dx) = d/dx(-3sinθcosθ)`

∴ `(d^2y)/(dx^2) = d/(dθ)(-3sinθcosθ)*(dθ)/dx`

= `-3[sinθ(-sinθ) + cosθ*cosθ]*(-1/sinθ)`

= `3[(-sin^2θ + cos^2θ)/sinθ]`

∴ L.H.S. = `y(d^2y)/(dx^2) + (dy/dx)^2`

= `sin^3θ*3((cos^2θ - sin^2θ))/sinθ + (-3sinθcosθ)^2`

= 3sin²θ(cos²θ – sin²θ) + 9sin²θcos²θ

= 3sin²θ[cos²θ – sin²θ + 3cos²θ]

= 3sin²θ[4cos²θ – (1 – cos²θ)]

= 3sin²θ(5cos²θ – 1)

= R.H.S.