19. "Describe the motion of a particle with position (x,y) as t varies in the given interval" ?

x=5+2\cos\color(maroon)(cancel(t))\pit, y=3+2\sin\pit, 1\let\le2

*thanks to Ultrilliam for pointing it out,
x=5+2\color(red)(\cos\pi\t)

2 Answers
May 17, 2018

See explanation

Explanation:

\cos\pit=(x-5)/2 and \sin\pit=(y-3)/2
\therefore((x-5)/2)^2+((y-3)/2)^2=\cos^2\pit+\sin^2\pit=1

for t=1, x=3 and y=3. That makes for (3,3)
for t=2, x=7 and y=3. That makes for (7,3)

Based on Wolfram Alpha, the particle moves counterclockwise in a circular manner from (3,3) to (7,3).

SladerSlader

May 17, 2018

In cartesian: circle with centre (5,3), radius 2

Period T = 2.

Explanation:

Set it up for the Pytharorean identity like this:

  • x=5+2\cos\ \pi t implies cos pi t = (x-5)/2

  • y=3+2\sin\pi t implies sin pi t = (y - 3)/2

Pytharorean identity

  • sin^2 alpha + cos^2 alpha = 1 implies

  • ((x-5)/2)^2 + ((y - 3)/2)^2 = 1

implies (x-5)^2 + (y - 3)^2 = 2^2

Circular motion with centre (5,3), radius 2

Remember that cos omega t or sin omega t implies motion with constant angular frequency omega = 2 pi \ f = (2 pi)/T.

So you can also describe the periodicity of the motion as T = 2.

Finally, the interval: 1\let\le2, which amounts to half a period.

  • ((x(1)),(y(1))) = ((3),(3))

  • ((x(2)),(y(2))) = ((7),(3))

From the graph, you can see that it's making the CCW journey along the bottom half of the circle in that time interval

graph{ (x-5)^2 + (y - 3)^2 = 2^2 [-10, 10, -5, 5]}