#### Question

If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find `dy/dx` when `theta = pi/3`

#### Solution

Applying parametric differentiation

`dx/(d theta) = 2a - 2acos2theta`

`dy/(d theta) = 0 + 2asin 2theta`

`dy/dx = dy/(d theta) xx (d theta)/dx = (sin 2 theta)/(1-cos 2 theta)`

Now putting the value of `theta = pi/3`

`dy/dx|_(theta = pi/3) = (sin 2(pi/3))/(1-cos2(pi/3))`

`= (sqrt3/2)/(1+ 1/2)`

`= (sqrt3/2)/(3/2) = 1/sqrt3`

So `dy/dx is 1/sqrt3` at `theta = pi/3`

Is there an error in this question or solution?

Solution If X = A (2θ – Sin 2θ) And Y = A (1 – Cos 2θ), Find `Dy/Dx` When `Theta = Pi/3` Concept: Derivatives of Functions in Parametric Forms.