Question

How do you find parametric equations for the path of a particle that moves along the `x^2 + (y-3)^2 = 16` Three times around counterclockwise, starting at (4,3) 0 ≤ t ≤ 6pi?

Answer 1

`x = 4 cos t`
`y = 3 + 4 sin t`

Explanation:

`x^2 + (y-3)^2 = 16` is a circle centre (0,3), radius 4

for circular motion, polar co-ordinates are the way to go.

we need however to tweak the usual Origin based system, ie `x = r cos t, y = r sin t` where r = 4, because the circle is centred on (0,3)

[to be clear we will use t as the parameter instead of `varphi`, as the question demands, but it's just another letter so we do that trivially]

so our start point is

`x = 4 cos t`
`y = 3 + 4 sin t`

now for t = 0 this form already starts at (4,3) because that is `(4cos 0, 3 + 4sin 0 )` so we are good to go on that count

it has to rotate 3 times counterclockwise in `0 le t le 6 pi`.

the usual period of the simplest polar parameterisation is `2 pi` for one full revolution reflecting the underlying sine and cosine functions .

so in that `3 times 2 pi` period, it will rotate 3 times.

plus in polar the usual `varphi` is measured counterclockwise with the x axis representing `varphi = 0` - see diagram above.