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Homework Help: Parameterization of a Circle Question

  1. Sep 15, 2011 #1
    This isn't really a HW question, it's just something that's been confusing me in my Calc class.

    We recently went over how to find curvatures of curves in 3D space. In lecture, the professor went over a simple example: a circle of radius 3 at any given point.

    Maybe it's because I don't remember my Calc II class that well, but he parameterized the equation of a circle,

    x2 + y2 = 9

    to become r(t) = 3cos(t)i + 3sin(t)j .

    So from:

    x2 + y2 = r2

    If x = cos(t) and y = sin(t), the identity cos2(t) + sin2(t) = 1 still holds true.

    If the radius is 3,

    x2 + y2 = 32

    3(x2 + y2) = 9(12)

    9cos2(t) + 9sin2(t) = 9(12)

    9(cos2(t) + sin2(t)) = 9(12)

    is how I understand it. If there's an easier way to picture it, please let me know!


    So then r(t) = 3cos(t)i + 3sin(t)j describes the circle when "traced."


    My question: why does the "x" part and the "y" part of the parameterization have to be the respective cos, sin trig functions? In class, one student said that x was sin; but the prof said that no, it's cos. Why does x have to be cos, rather than sin? Does it have to do something with the direction the curve "goes in?" How can I determine the "direction" the curve goes in? Or did I just hear my professor wrong...?
     
  2. jcsd
  3. Sep 15, 2011 #2
    Try drawing a picture. Do you remember your unit circle? For a right-handed coordinate system, x = rcos[itex]\theta[/itex] and y the sine function. Positive theta corresponds to counter-clockwise motion from the positive x-axis.
    Now when you parametrize the equation [itex]\theta[/itex] becomes [itex]\dot{\theta}[/itex]t, where [itex]\dot{\theta}[/itex] is how quickly you are going around the circle. In your case this is simply one.
    Strictly speaking, x doesn't HAVE to be the cosine. It's just convention to use a right handed coordinate system. If you measured [itex]\theta[/itex] from a different axis or you switched positive x being in the vertical, your sines and cosines may change.
     
  4. Sep 15, 2011 #3

    LCKurtz

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    Yes, it has to do with the orientation of the curve and where you want it to be when t = 0. Parameterizations are in general not unique. If the curve is specified to start at (1,0) and go counterclockwise then R(t) = < cos(t), sin(t) > is a natural choice. If it went clockwise you might use <cos(t), -sin(t)>. If nothing is given about the orientation of the curve then you could use <sin(t), cos(t)> or other choices; they are all equally correct.
     
  5. Sep 15, 2011 #4

    HallsofIvy

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    They don't have to be. There exist an infinite number of parametric equations describing any curve. [itex]x= 3sin(t)[/itex], [itex]y= 3cos(t)[/itex] are perfectly good parametric equations and, in fact, have the same "direction" as [itex]x= 3cos(t)[/itex] [itex]y= 3sin(t)[/itex]: with the first, as t goes from 0 to [itex]\pi/2[/itex], (x, y) goes from (0, 3) to (3, 0), counter-clockwise around the circle and the second goes from (3, 0) to (0, 3), also counterclockwise.

    But before you go telling your teacher that we say he/she is wrong, note that they do have different points for specific values of t. The parametric equations [itex]x= cos(t)[/itex], [itex]y= sin(t)[/itex] "start" (t= 0) at (1, 0) and then go around the circle. Perhaps for some reason, that is important.
     
  6. Sep 15, 2011 #5
    Wow, thank you all for the absolutely fantastic replies. It REALLY cleared it up for me, big time.

    Thanks again!
     
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