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Help describing a region in polar coordinates

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data
    If (r, θ) are the polar coordinates of a point then describe the region defined by the restrictions
    -1 < r < 0, π/2 < θ < 3π/2

    2. Relevant equations
    No clue

    3. The attempt at a solution
    I tried drawing the curve in a polar grid by starting at π/2 and finishing at 3π/2. I was careful to draw it according to the negative values of r. I ended up getting a shape that looked like half of a heart. So my answer was that it was a cardioid.
     
  2. jcsd
  3. Apr 22, 2017 #2

    BvU

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    Hi,

    Very unusual definition of polar coordinates (negative distance ? ?:) )
    Strange. Can you post it ?
     
  4. Apr 22, 2017 #3

    Mark44

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    The r coordinate can be negative. For example, ##(2, 5\pi/4)## and ##(-2, \pi/4)## identify the same location.

    No, that isn't right. The shape you get should be half a circular disk.

    @Nidum, in answer to the question you posted, but deleted, yes, that is what is meant.
     
  5. Apr 22, 2017 #4

    Nidum

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    Here it is again :

    pi chart old.jpg

    I think though that using ( r , theta + pi ) would be preferable to using ( - r , theta ) for the coordinates . It's difficult to visualise what - r means in polar coordinates .
     
  6. Apr 22, 2017 #5

    Nidum

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    pi chart.jpg
     
  7. Apr 22, 2017 #6

    Mark44

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    It's really not that difficult to visualize. The ray in the first quadrant has r > 0. For a negative r with theta being the same, just go out the opposite direction as r is in. This works the same way as it does for vectors, where u and -u point in opposite directions.
     
  8. Apr 22, 2017 #7
    With that negative ##r##, I don't how you will plot the curve easily since most graphing calculator assume r to be positive.

    Also it is impossible to get -ve ##r## from cartesian coordinates since ##\sqrt{x^2 + y^2}= r##.
     
  9. Apr 22, 2017 #8

    Nidum

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    No calculator needed . In fact no calculation needed .

    Once you work out what is actually meant by the problem statement you can draw the required shape using simple straight lines and arcs .
     
    Last edited: Apr 23, 2017
  10. Apr 22, 2017 #9

    Mark44

    Staff: Mentor

    But if the relationship is instead ##r^2 = x^2 + y^2##, then you can get negative values of r.

    From wikipedia, https://en.wikipedia.org/wiki/Polar_coordinate_system:
    Emphasis added
     
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