Parametric equations for particle motion

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The motion of the particle described by the parametric equations x=2+cos(t) and y=3+sin(t) for t in the interval [0, 2π] traces a circular path. By rewriting the equations and applying the Pythagorean identity, it can be shown that the trajectory satisfies the equation of a circle: (x - 2)² + (y - 3)² = 1. This confirms that the particle moves in a circle with a radius of 1, centered at the point (2, 3). To visualize this motion, plotting various points for t will illustrate the circular path clearly. The discussion emphasizes the importance of graphing to understand the motion accurately.
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Describe the motion of the particle with position (x,y) as t varies over the given interval.

x=2+cost y=3+sint

where t is greater than or equal to 0 and less than or equal to 2 pi


i've tried to eliminate t and came up with

y=3+sin(arccos(x-2))

i don't know if this is even right, but if it is, I'm not too sure how to describe the motion. picking random points came up with a very strange graph. i am concerned because all of the other questions in this section had graphs which were circles or ellipses.

any advice?
 
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All you have to do is describe the motion? Just plug in many different times and then plot it on graph paper. If the jump between two times is so great that you can not tell what happens between two points, then take another point with at time between the two other times. If you have a plotting program you can use that too. Some graphing calculators will do this too. However all you need is graphing paper and calculator with sin and cos.
 
Describe the motion of the particle with position (x,y) as t varies over the given interval.

x=2+cost y=3+sint

where t is greater than or equal to 0 and less than or equal to 2 pi
Rewrite the equations:
(x - 2) = cos(t)
(y - 3) = sin(t)
Square both sides:
(x - 2)^2 = cos^2(t)
(y - 3)^2 = sin^2(t)
Since both equations are true simultaneously, add them, and use {sin^2(t) + cos^2(t) = 1}:
(x - 2)^2 + (y - 3)^2 = cos^2(t) + sin^2(t)
(x - 2)^2 + (y - 3)^2 = 1

Look familiar??
Path is Circle of Radius 1 Centered at (x=2, y=3)


~~
 
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