Question

If x = a sin³ θ, y = b cos³ θ, then find `(d^2y)/dx^2` at θ = `(pi)/4`.

Answer 1

Given, x = a sin³ θ and y = a cos³ θ

`(dx)/(dθ) = 3a sin^2 θ cos θ`

and `(dy)/(dθ) = -3a cos^2 θ sin θ`

∴ `(dy)/(dx) = (-3a cos^2 θ sin θ)/(3a sin^2 θ cos θ)`

= −cot θ

and `(d^2y)/(dx^2) = d/(dθ) (-cot θ) (dθ)/(dx)`

= cosec² θ `1/(3a sin^2 θ cos θ)`

= `1/(3a sin^4 θ cos θ)`

Now, `(d^2y)/(dx^2)|_(at  x = (pi)/4) = 1/(3.a(1/sqrt2)^4 . 1/sqrt2)`

= `1/(3.a 1/4 . 1/sqrt2)`

= `(4sqrt2)/(3a)`