Question

Find `(d^2y)/(dx^2)` of the following : x = a cos θ, y = b sin θ at θ = `π/4`.

Answer 1

x = a cos θ, y = b sin θ

Differentiating x and y w.r.t. θ, we get,

`(dx)/(dθ) = (d)/(dθ) (a cos θ)`

`(dx)/(dθ) = a (d)/(dθ) (cos θ)`

`(dx)/(dθ) = a(– sin θ)`

`(dx)/(dθ) = –  a sin θ`                          ...(1)

and

`(dy)/(dθ) = (d)/(dθ) (b sin θ)`

`(dy)/(dθ) = b (d)/(dθ) (sinθ)`
`(dy)/(dθ) = b cos θ`

∴ `dy/dx = (((dy)/(dθ)))/(((dx)/(dθ)))`

`dy/dx = (b cos θ)/(– a sin θ)`

`dy/dx = (– b/a)cotθ`

and

`(d^2y)/(dx^2) = d/dx [(– b/a)cotθ]`

`(d^2y)/(dx^2) = – b/a. (d)/(dθ) (cot θ) × (dθ)/(dx)`

`(d^2y)/(dx^2) = (– b/a)(– cosec^2θ) × (1)/(((dx)/(dθ)))`

`(d^2y)/(dx^2) = (b/a) cosec^2θ × (1)/(– asinθ)`      ..[By (1)]

`(d^2y)/(dx^2) = (– b/a^2) cosec^3θ`

∴ `((d^2y)/(dx^2))_("at" θ = pi/(4)) = (– b/a^2) cosec^3 pi/(4)`

= `(– b)/(a^2) × (sqrt(2))^3`

= `– (2sqrt(2)b)/(a^2)`