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?

1 Answer
Jun 27, 2016

#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.

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.