Question

How do you find the horizontal and vertical tangents to `x = Cos(3t)` and `y = 2sin(t)`?

Answer 1

Recall that

`(dy/(dt))/(dx/(dt)) = dy/dx`

Therefore

`dy/dx = (2cost)/(-3sin(3t))`

Horizontal tangents occur when the derivative equals `0`.

`0 = 2cost -> t = pi/2 + pin`

Vertical tangents occur when the derivative is undefined.

`-3sin(3t) =0 -> 3t = pin -> t = pi/3n`

Hopefully this helps!

Answer 2

Horizontal Tangents occur when `t = pi/2, (3pi)/2`

Vertical Tangents occur when `t = 0, pi/3, (2pi)/3, pi, (4pi)/3, (5pi)/3`

Explanation:

We have the following parametric equation:

`x = cos(3t)` ... [A]
`y =2sin(t)` ... [B]

The entire curve is mapped out for `tin[0,2pi]`.

Horizontal tangents occurs when `dy/dt=0`, and vertical tangents occur when `dx/dt=0`, This arrises from the chain rule, as:

`dy/dx = (dy//dt) /(dx//dt)`

Horizontal Tangents:

Differentiating [B] wrt `t` we get:
`dy/dt = 2cost`

`dy/dt = 0 => 2cost = 0`
`:. cost=0 => t = pi/2, (3pi)/2 \ " for " \ t in [0,2pi]`

Vertical Tangents:

Differentiating [A] wrt `t` we get:
`dx/dt = -3sin(3t)`

`dx/dt = 0 => -3sin(3t) = 0`
`:. sin(3t)=0 => 3t = 0, pi, 2pi, 3pi, 4pi, 5pi, 6pi \ " for " \ 3t in [0,6pi]`
`:. t = 0, pi/3, (2pi)/3, pi, (4pi)/3, (5pi)/3, 2pi \ " for " \ t in [0,2pi]`

The last value of `t` also corresponds to `t=0`, so can omit this value