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Calculating polar coordinates

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    B⃗ = -2.0ι^ + 3.0 j^.
    Find the polar coordinates r and theta.
    2. Relevant equations
    n/a

    3. The attempt at a solution
    r=sqrt((-2.0)^2+(3.0^2))
    r = 3.6

    theta = tan^-1(3/-2) = -56 degrees

    The answers seem to be wrong, can I get any guidance on this question please?
     
  2. jcsd
  3. Mar 9, 2016 #2
    Draw a picture. Where is the Cartesian point? Why doesn't your angle make any sense?
     
  4. Mar 9, 2016 #3
    Oh it's in the second quadrant, I see how the angle wouldn't work. So would it then be +56 degrees?
     
  5. Mar 9, 2016 #4

    haruspex

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    Is that in the correct quadrant?
    Do you know what the graph of tan looks like? If you put a horizontal line through it at a random height, what can you say about the intercepts?
     
  6. Mar 9, 2016 #5
    I believe it is on the correct quadrant as -x,+y = quadrant 2. And there should only be one intercept I believe?? I'm confused
     
  7. Mar 9, 2016 #6

    SteamKing

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    Which quadrants have positive angles?
    Where is zero degrees located?
    How many degrees in each quadrant?
     
  8. Mar 9, 2016 #7
    Quadrant I, and III have positive angles I believe There are 90 degrees in each quadrant, and zero is located on the x-axis on quadrant I.
     
  9. Mar 9, 2016 #8

    SteamKing

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    Well 2 out of 3 isn't bad, but it should be three out of three in this case.

    If you start at zero degrees and go counterclockwise to 180 degrees, which quadrants have positive angles?

    If you start at zero degrees and go clockwise to 180 degrees, which quadrants have negative angles?

    The tricky thing about arctan on your calculator is it returns an angle θ such that -π/2 ≤ θ ≤ π/2, and the user is left with deciding in which quadrant the proper angle falls and its measure from zero degrees. That's why you should plot the original cartesian coordinates.
     
  10. Mar 9, 2016 #9

    haruspex

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    Sure, but +56 degrees is not in that quadrant.
    For the intercepts, what range of angles did you consider in saying there is only one intercept?
     
  11. Mar 10, 2016 #10
    Quadrant I would be positive for the first part, and Quadrant IV would be negative on the second question.
    So then what I would do is 180-54=124?
     
  12. Mar 10, 2016 #11
    The answer -56 would be in quadrant IV and +56 would be in quadrant one then right?
     
  13. Mar 10, 2016 #12

    SteamKing

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    You're still guessing here. I don't know why you won't plot the original cartesian coordinates. That would answer your question directly.
     
  14. Mar 10, 2016 #13
    It would be in quadrant II after I plot (-2,3). Would the angle I be measuring start from the x-axis from quadrant I though?
     
  15. Mar 10, 2016 #14
    Never mind, I figured it out, I subtracted 180 with 56 = 124 degrees. Thanks.
     
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