Equation of the path of the particle

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The particle's motion is described by the parametric equations x=8sin(t) and y=6cos(t), indicating a path that forms an ellipse. The discussion clarifies that the "path of the particle" refers to the curve connecting all points defined by these equations, rather than a straight line. Participants confirm that the equations represent an ellipse due to the sine and cosine functions involved. One user successfully solves the problem and shares their solution, highlighting the preferred form of the equation as x²/a² + y²/b² = 1 for clarity. The conversation emphasizes understanding parametric equations in the context of particle motion.
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Homework Statement


The x and y coordinates of a particle moving in the x-y plane are x=8sin(t) and y=6cos(t). What is the equation of the path of the particle?


Homework Equations


m=\frac{y_2-y_1}{x_2-x_1}
y-y_1=m(x-x_1)

The Attempt at a Solution


I am stuck on how to approach this problem.
I drew a picture:
7jw9usI.png
. Can I use one point as the origin, (0,0) and the second point as (8sint, 6cost) and use the equation of a line to find the 'path' of the particule? I am confused if the path of the particule means the equation of the line?

Any tips and hints would be great. Thanks!
 
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hi skybox! :smile:
skybox said:
Can I use one point as the origin, (0,0) and the second point as (8sint, 6cost) and use the equation of a line to find the 'path' of the particule? I am confused if the path of the particule means the equation of the line?

no, the path of the particle means the curve joining all the points (8sint, 6cost) :wink:

(if you're still stuck, come back for a hint)
 
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tiny-tim said:
hi skybox! :smile:


no, the path of the particle means the curve joining all the points (8sint, 6cost) :wink:

(if you're still stuck, come back for a hint)

Thanks tiny-tim. After some research, looks like this is a parametric equation. Since it has cosines and sines, it will most likely be a circle or ellipse from 0<=x<=2\pi.

I will try to solve this and post the solution when done. Thanks again!
 
I was able to solve it! Attached is the solution (as an image I did in Word) if anyone is interested.
 

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me! me! i'm interested! :smile:

yes, nicely done :wink:

(btw, for a lot of purposes, the form x2/a2 + y2/b2 = 1 is preferred, so you could have stopped there)
 
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