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Another polar / rectangular simplification

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Convert into rectangular coordinates:

    [tex]
    r = \frac {1}{1-cos(theta)}
    [/tex]



    2. Relevant equations



    3. The attempt at a solution

    [tex]
    r = \frac {1}{1-cos(theta)}
    [/tex]


    [tex]
    r –r(cos) = 1
    [/tex]
    (why can't I get a minus sign to display correctly? - I'm trying to show r - r*cos = 1)

    I used
    [tex]
    r = \sqrt {x^2+y^2}
    [/tex]
    and

    [tex]
    cos = \frac {x}{r}

    x = (r)(cos)
    [/tex]

    resulting combinations gives:
    [tex]
    \sqrt {x^2+y^2} – x = 1
    [/tex]
    (I think) - and again - I'm having a problem (maybe just on my end) displaying a minus sign. I'm seeing "8211" on the screen for the minus sign.

    I'm trying to display sqrt(x^2+y^2) - x = 1.

    The book gets

    [tex]
    y^2} = 1 + 2x
    [/tex]

    I‘ve tried some various algebra stuff but am not getting close to the book’s answer.

    Thanks
    -Sparky
     
    Last edited: Jan 6, 2009
  2. jcsd
  3. Jan 6, 2009 #2
    Never mind, I got it - I don't know how to delete the post (if I'm allowed)

    I moved x over and squared both sides.

    [tex]
    \sqrt {x^2+y^2} = 1 + x
    [/tex]

    [tex]
    {\sqrt {x^2+y^2}}^2 = (1 + x)^2
    [/tex]

    [tex]
    {x^2+y^2} = 1 + 2x + x^2
    [/tex]

    [tex]
    y^2} = 1 + 2x
    [/tex]
     
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