Question

If x = a(θ – sin θ), y = a(1 – cos θ) find `(d^2y)/(dx^2)`.

Answer 1

Here `dy/dx = (dy/(dθ))/(dx/(dθ))`

Finding `dy/(dθ)` and `dx/(dθ)` separately:

Calculating `dy/(dθ)`:

y = a(1 – cos θ)

`dy/(dθ) = (d(a (1 - cos θ)))/(dθ)`

= `a((d(1 - cos θ))/(dθ))`

= a (0 − (−sin θ))

= a (sin θ)

= a sin θ

Calculating `dx/(dθ)`:

x = a(θ – sin θ)

`dx/(dθ) = (d(a θ - a sin θ))/(dθ)`

= `(d(a θ))/(dθ) - (d(a sin θ))/(dθ)`

= a − a cos θ

= a (1 − cos θ)

Therefore, `dy/dx = (dy/(dθ))/(dx/(dθ))`

= `(a sin θ)/(a(1 - cos θ))`

= `sin θ/(1 - cos θ)`

= `(2 sin (θ/2) cos (θ/2))/(2 sin^2 (θ/2))`

= `(cos (θ/2))/(sin (θ/2))`

= `cot (θ/2)`

Finding `(d^2y)/(dx^2)`:

`dy/dx = cot (θ/2)`

Differentiating again:

`(d^2y)/(dx^2) = (d(cot (θ/2)))/dx`

= `(d(cot (θ/2)))/(dθ) xx (dθ)/dx`

= `- "cosec"^2 θ/2 xx 1/2 xx (dθ)/dx`

= `- 1/2 "cosec"^2 θ/2 xx 1/(dx/(dθ))`

= `-1/2 "cosec"^2 θ/2 xx 1/(a(1 - cos θ))`

Putting `1 - cos θ = 2 sin^2 (θ/2)`

= `-1/2 "cosec"^2 θ/2 xx 1/(a xx 2 sin^2 (θ/2))`

= `-1/(4a) xx "cosec"^2 θ/2 xx 1/(sin^2 (θ/2))`

= `-1/(4a) xx "cosec"^2 θ/2 xx "cosec"^2 (θ/2)`

= `-1/(4a) xx "cosec"^4 θ/2`