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?

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Answer 1 · 2016-06-27T08:27:00

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

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

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

Wikipedia

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.

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