Question

If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a (θ – sin θ), y = a (1 + cos θ)

Answer 1

Here x = a (θ – sin θ)  ...(1)

y = a (1 + cos θ)  ...(2)

Differentiating (1) and (2) w.r.t. θ, we get

`dx/(dθ)` = a [1 – cos θ]

`dy/(dθ)` = a [–sin θ]

= –a sin θ

`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)`

= `-cot  θ/2`