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Polar Coordinates inverse Radius

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    I have to turn this homework in online... I just want someone to check my work
    Convert from Cartesian coordinates to Polar coordinates
    (-1,-sqrt(3))
    if r > 0 and if r < 0.


    2. Relevant equations



    3. The attempt at a solution
    if r > 0 then I believe the answer is
    (r,∅) = (2,4.2)
    if r < 0 then i'm a bit confused, is the only difference I make the 2 negative?
    so
    (r,∅) = (-2,4.2)
     
  2. jcsd
  3. Sep 6, 2013 #2

    eumyang

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    I guess you can't write the angle in terms of pi in the online system?
    No.

    Consider another example. The point (1, 1) (rectangular coordinates) is (sqrt 2, π/4) in polar coordinates. Imagine yourself standing at the origin. (sqrt 2, π/4) would mean that you would turn to face the π/4 "direction" and walk sqrt 2 units forward.

    If r is negative, then you would walk backwards while still facing the direction indicated by θ. So if you look at (-sqrt 2, π/4), you would turn and face the π/4 "direction" and walk sqrt 2 units backwards. You see how this is not the same as (1, 1) in rectangular coordinates? In fact,
    (-sqrt 2, π/4) (polar) = (-1, -1) (rectangular).

    So, if you want r to be negative, what to do? You have to add/subtract π radians, to turn and face the opposite direction. If you look at (-sqrt 2, 5π/4), you face the 5π/4 "direction" and walk sqrt 2 units backwards. You end up at (1, 1) (rectangular).

    Now fix your second point.
     
  4. Sep 7, 2013 #3

    vanhees71

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    What do you mean by "if [itex]r[/itex] is negative"? By definition in polar coordinates [itex]r>0[/itex].

    Further you didn't give the relevant equations, which indeed help to solve the problem. If you have given Cartesian coordinates [itex](x,y)[/itex] of a vector you want to find [itex]r[/itex] and [itex]\varphi[/itex] such that
    [tex](x,y)=r(\cos \varphi,\sin \varphi).[/tex]
    From this you get
    [tex]r=\sqrt{x^2+y^2}[/tex]
    and
    [tex]\varphi=\mathrm{sign} (y) \arccos \left (\frac{x}{r} \right ),[/tex]
    where the range of the angle is [itex]\varphi \in [-\pi,\pi][/itex]. You can always add integer multiples of [itex]2 \pi[/itex] to this [itex]\varphi[/itex].

    NB: Sometimes you find sloppy formulas involving [itex]\tan \varphi[/itex]=y/x. This is dangerous, because you have to think in addition, in which quadrant the original vector is located and then adapt the result accordingly by hand!
     
  5. Sep 9, 2013 #4

    eumyang

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    According to Larson's Precalculus (8th ed., p. 777-778), r is the directed distance from O (the origin) to P (the point), and it proceeds to show examples with a negative r. Paul's Online Math Notes also state that "we also allow r to be negative."
    :confused:
     
  6. Sep 9, 2013 #5
    Well, r can be negative, as 'r' is defined as directed distance ie. wrt angle in question. Just verified it with Thomas' Calculus 11th ed.
     
  7. Sep 9, 2013 #6

    vanhees71

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    These seem to be very strange books then! I've never heard about defining [itex]r<0[/itex] for plane or spherical polar coordinates ever, and it's not necessary!
     
  8. Sep 9, 2013 #7

    CAF123

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    I read it in my book too, but I would agree I have never had to define r < 0. Nor in any polar coordinates application to physics problems.
     
  9. Sep 9, 2013 #8

    Its a pretty redundant concept as far as physics is concerned, though its useful to use it when finding intersection of two curves when only one representation of a point fits the equations, you will often get negative r then.
     
  10. Sep 9, 2013 #9

    LCKurtz

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    If you don't consider ##r<0## when you plot it, you don't get all three leaves of the 3-leaved rose ##r=\sin(3\theta),~0 \le \theta \le\ pi##.
     
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