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#).

Slader

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]}