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I Show that a particle moves over a circumference

  1. Mar 7, 2016 #1
    The problem asks me to show that a particle moves over a circumference with its center at the origin.

    The position vector of a moving particle is:
    ljiZbVp.png

    I've tried using the x2+y2=r2 formula of the circumference, squaring both components of the vector function but I couldn't figure out what to do with the result.

    How should I go in order to show this?
     
  2. jcsd
  3. Mar 8, 2016 #2

    Svein

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    Simple algebra: [itex]x^{2}+y^{2}=(\frac{1-t^{2}}{1+t^{2}})^{2}+(\frac{2t}{1+t^{2}})^{2}=\frac{1-2t^{2}+t^{4}+4t^{4}}{(1+t^{2})^{2}}=\frac{1+2t^{2}+t^{4}}{1+2t^{2}+t^{4}}=1 [/itex].
     
  4. Mar 8, 2016 #3

    andrewkirk

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    Those formulas are from the 't formula' that is sometimes useful in trigonometry. If you read the first part of the link, you'll see why the particle travels around a circle.
     
  5. Mar 8, 2016 #4
    Oh my god, that's it. I did the math but for some reason I had 4t2 instead of 2t2. I should double check my math more often. Thanks for the reply.

    The problem also asks about the direction the particle moves as t increases, if it's counter or clockwise, and if there are points in the circumference that are not occupied by the particle when t goes from negative infinity to infinity.

    Do I just plug in t values and see how the function behaves to find out the direction?
     
  6. Mar 8, 2016 #5

    Mark44

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    There's a typo in one of the expressions - ##\frac{1-2t^{2}+t^{4}+4t^{4}}{(1+t^{2})^{2}}## should be ##\frac{1-2t^{2}+t^{4}+4t^{2}}{(1+t^{2})^{2}}##
     
  7. Mar 8, 2016 #6
    Looks to me that the original formula has a typo: The numerator of the y-coordinate should be 2t2, not 2t.

    Then you actually get

    x(t)2 + y(t)2 = 1​

    for all t.

    Hint: You might notice that

    y(t) / x(t) = tan(θ),​

    so

    θ(t) = arctan(y(t) / x(t))​

    will tell you the angle θ(t) (or θ(t) ± π) around the circumference. In either case dθ/dt is the rate the angle is increasing.
     
  8. Mar 8, 2016 #7

    Samy_A

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    How so?
    Looking only at the numerators:
    ##(1-t²)²+(2t)²=1-2t²+t^4+4t²=1+2t²+t^4=(1+t²)²##, and that is equal to the denominator squared.
     
    Last edited: Mar 8, 2016
  9. Mar 8, 2016 #8

    andrewkirk

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    If you use the 't formula' link from post 3, you'll see that the formulas for x and y can be expressed as sin and cos of an angle (which in the link they call ##x## but for this purpose would be better thought of as ##\theta##), which is the angle travelled around the circle in the conventional direction for polar coordinates.
     
  10. Mar 8, 2016 #9
    Samy, thanks again for the correction. It seems I ought to join up so I have more time to delete my stupid posts.
     
  11. Mar 8, 2016 #10

    Math_QED

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    These are indeed the t-formulas: Let tan(x/2) = t
    =>
    (1-t^2)/(1+t^2) = cos x
    2t/(1+t^2) = sin x
     
  12. Mar 8, 2016 #11
    I see that now. Thank you for your help!!!

    The problem also asks if there are points that are not occupied by the particle as t goes from negative infinity to positive infinity.
    Since the vector function (cosθ, sinθ) is defined at every t from -infinity to +infinity, is it safe to say that all the points are occupied by the particle as t changes?

    The answer says that the point (-1,0) is not occupied, by I couldn't figure out why. I think it has something to do with the r2 x2+y2=r2 being ±1
     
  13. Mar 8, 2016 #12

    andrewkirk

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    The function is defined for every ##t\in(-\infty,\infty)##. But ##t## must be real, so cannot take on a value of ##\pm\infty## as that is not a real number. Hence points on the circle that would require an infinite ##t## are never attained - only asymptotically approached.
     
  14. Mar 8, 2016 #13
    And why is the point (-1,0) not occupied by the particle if the circle is always defined for all real numbers?

    Is it because making x = (1-t^2)/(1+t^2) = -1 would give me 2 = 0, therefore not possible.
    And plugging x= -1 in x2+y2 = -1 would give me y = 0?

    Sorry if it's confusing, my english is a bit bad
     
  15. Mar 8, 2016 #14

    andrewkirk

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    Yes
     
  16. Mar 8, 2016 #15
    Got it. Thank you so much for your help!
     
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