# if x = 2cosθ and y= 6sinθ find dy/dx in term of θ This is Parametric equation

### 1 Answer | Add Yours

You need to form the polar equation and then you need to convert the polar form in cartesian form.

Hence, differentiating `y = 6 sin theta` both sides yields:

`(dy)/(dx) = (dy)/(d theta)*(d theta)/(dx)`

`(dy)/(dx) = 6 cos theta *(d theta)/(dx)`

You need to differentiate both sides `x = 2cos theta` such that:

`dx = -2sin theta d theta`

You need to divide by `dx` both sides such that:

`1 =-2sin theta *(d theta)/(dx)`

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

You need to substitute `-1/(2sin theta)` for `(d theta)/(dx)` in `(dy)/(dx) = 6 cos theta *(d theta)/(dx)` such that:

`(dy)/(dx) = 6 cos theta *(-1/(2sin theta))`

`(dy)/(dx) = -3 cot theta`

**Hence, evaluating `(dy)/(dx)` in terms of theta yields `(dy)/(dx) = -3 cot theta.` **