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Polar coordinates

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Find two other pairs of polar coordinates of the given polar coordinate, one with r > 0 and one with r < 0. Then plot the point.

    (2, 5π/3)


    2. Relevant equations

    I don't there are any.

    3. The attempt at a solution

    I'm not completely sure of how to do this actually.

    I know that we can add 2 pi to the answer to get the coordinates for r > 0 this gives an answer of [itex] (2,\frac{11\pi}{3} [/itex]

    But I'm not sure about the r<0 option. Would I simply have to subtract 2 pi from 5 pi/3? In addition, would the r value change to the negative sign?

    Thanks for your help.
     
    Last edited by a moderator: Mar 21, 2013
  2. jcsd
  3. Mar 18, 2013 #2

    tiny-tim

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    hi stunner5000pt! :smile:
    are you sure that isn't θ > 0 and θ < 0 ? :confused:
     
  4. Mar 20, 2013 #3
    I chekced again and yes it is R and not theta ...
     
  5. Mar 20, 2013 #4

    eumyang

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    r can be negative. Think of yourself being in the origin of the coordinate plane, facing the positive y direction. If you walk in the positive y direction 5 units, then where you end up will correspond to
    [itex]\left( 5, \frac{\pi}{2} \right)[/itex]
    in polar coordinates.

    Think of yourself back in the origin, facing the positive y direction. Now walk backwards 5 units. By "walking backwards" your r is negative. Since you are still facing the positive y direction, this point will correspond to
    [itex]\left( -5, \frac{\pi}{2} \right)[/itex]
    in polar coordinates. But this particular point is the same as facing the negative y direction and walking forwards 5 units, or
    [itex]\left( 5, \frac{3\pi}{2} \right)[/itex]

    Now see if you can answer your question.
     
  6. Mar 20, 2013 #5
    Thanks for your advice

    based on what you have said, in the case of (2, 5pi/3), the moment arm for the 'backward' motion is in the second quadrant

    If we 'walked' backward that would be [itex] (-2, \frac{5\pi}{3}) [/itex]
    for the 'positive' motion it would be [itex] (2, \frac{2\pi}{3}) [/itex]

    How does this sound?
     
  7. Mar 21, 2013 #6

    eumyang

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    Not quite. The three points you state are not the same. The three points I used in my example were not the same either - the last two formed a separate example from the first.

    The question is this: which direction would you have to "face", if, when walking "backwards" 2 units, you end up at the same point as (2, 5π/3)? In other words,
    [itex]\left( 2, \frac{5\pi}{3} \right) = \left( -2, ? \right)[/itex]
    Then you also need to consider coterminal angles.
     
  8. Mar 21, 2013 #7
    if we walked backward, then the angle would be pi/3

    As for the answer with r > 0, the answer wuld be (2, -pi/3)

    Is that correct?
     
  9. Mar 21, 2013 #8

    SammyS

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    See the red comments.
     
  10. Mar 21, 2013 #9

    eumyang

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    Sorry, that's still not correct. It may help if you looked at a polar coordinate grid like the one on this site.
     
    Last edited: Mar 21, 2013
  11. Mar 21, 2013 #10
    Hmm... ok (2, -pi/3) is correct. This is because if we have a positive moment arm, the angle would be measured negative as we rotate clockwise

    for the negative angle, wouldn't the angle be the same? But since the r is negative, would that imply that if we rotate clockwise, we get a positive angle?
    would that mean that for r = -2, the angle should be + pi/3?
     
  12. Mar 21, 2013 #11

    eumyang

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    No. See the attached image. Point A is the point you want, [itex]\left( 2, \frac{5\pi}{3} \right)[/itex]. Point B is [itex]\left( -2, \frac{\pi}{3} \right)[/itex], the point that you are saying that is the same as point A. They are not the same.
     

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  13. Mar 21, 2013 #12
    In polar coordinates, the radial coordinate r is always positive.
     
  14. Mar 21, 2013 #13
    I am so confused

    SO lets start from the top

    if we consider a point [itex] \left( 2,\frac{5\pi}{3} \right) [/itex]
    then for r> 0 [itex] \left( 2,\frac{11 \pi}{3} \right) [/itex]
    and for r < 0 [itex] \left( -2,\frac{2\pi}{3} \right) [/itex]

    obtained r > 0 by simply rotating once (+2 pi)
    obtained r < 0 by rotating half circle (- pi)
     
  15. Mar 21, 2013 #14
    Who says?
    Bingo! :smile:
     
  16. Mar 21, 2013 #15
    Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).
     
    Last edited: Mar 21, 2013
  17. Mar 21, 2013 #16
    Well apart from at least one other person on this thread saying so, and at least one link on this thread to a website saying so...

    Even the Wiki page for Polar Coordinates says so. (Although, I will agree that Wiki is not always Gospel.)
     
  18. Mar 21, 2013 #17

    LCKurtz

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    Plot ##r = \sin (3\theta)## for ##\theta## from ##0## to ##\pi## and you will see an nice 3 leaved rose with one leaf plotted with ##r<0##.
     
  19. Mar 21, 2013 #18

    SammyS

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    I don't blame you.

    It's clear that this argument concerning polar coordinates is not helping you understand how to solve this problem you have been given.



    Try looking at the case of r = -2 as follows.

    In Cartesian coordinates, the point you're working with is [itex]\displaystyle \ (x,\, y)=\left(1,\,-\sqrt{3}\right)\ .[/itex]

    In general, x = r cos(θ) and y = r sin(θ).

    For this point, you then have [itex]\displaystyle \ 1=(-2)\cos(\theta)\ \text{ and }\ -\sqrt{3}=(-2)\sin(\theta) \ .[/itex]

    This gives you [itex]\displaystyle \ \cos(\theta)=-\,\frac{1}{2}\ \text{ and }\ \sin(\theta) =\frac{\sqrt{3}}{2} \ .[/itex]

    Thus the angle, θ, is in the second quadrant.
     
  20. Mar 22, 2013 #19

    eumyang

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    Precalculus by Larson (8th ed.) refers to r as a "directed distance from O to P" (p. 777) and it states that "another way to obtain multiple representations of a point is to use negative values for r" (p. 778).

    Precalculus: Graphical, Numerical, Algebraic by Demana/Waits/Foley/Kennedy (7th ed.) states that "r is the directed distance from O to P" and that "if r < 0 then P is on the terminal side of θ + π" (both on p. 534)

    That's two references, and I think it's safe to say that you are mistaken.
     
  21. Mar 22, 2013 #20
    Actually, Wiki is a little schizophrenic on this. First the say "The radial distance ρ is the Euclidean distance from the z axis to the point P." This can only be positive. But then they say something about allowing negative values of ρ to locate a point.

    In a 40 year career as an engineer/mathematician, I have never seen negative values of ρ being used in practice. So maybe negative values of ρ are only found in Ivory Tower land.
     
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