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How to translate from polar to cartesian coordinates:

  1. Jun 5, 2013 #1
    How to translate r = 2 /(2 - cos(theta)) to cartesian coordinates:


    so far:

    r = 2 /(2 - cos(theta))

    r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

    r (2 - cos(theta))= 2

    2*r - rcos(theta) = 2 | know x = rcos(theta)

    2*r - x = 2 | know r^2 = x^2 + y^2

    2*(x^2 + y^2)^1/2 - x = 2 | : 2

    (x^2 + y^2)^1/2 - x/2 = 1 | opens the parentheses

    x^2*1/2 + y^2*1/2 - x/2 = 1

    x + y - x/2 = 1

    y = 1 - x/2 | is this the solution?


    Thank you in advance for help.

    Regards
    sjombol
     
  2. jcsd
  3. Jun 5, 2013 #2

    LCKurtz

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    Is ##\sqrt{a+b} = \sqrt a + \sqrt b##?
     
    Last edited: Jun 5, 2013
  4. Jun 5, 2013 #3

    Mark44

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    It's interesting that you document each step along the way with what you did, and I think that's a good thing. Where you fell down, as LCKurtz notes, is in the step that you describe as "opens the parentheses". The distributive property, which is usually represented something like this -- a(b + c) = ab + ac -- might be what you were thinking. The problem is that that this property applies only to multiplication of a sum. It doesn't apply to expressions raised to powers (including square roots) or other operations.
     
  5. Jun 5, 2013 #4
    Thank you both very much for your reply :-)

    It's been a while since last time I went to school, guess old habits do die sometimes or at least needs refreshing.


    Another go then:

    r = 2 /(2 - cos(theta))

    r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

    r (2 - cos(theta))= 2

    2*r - rcos(theta) = 2 | know x = rcos(theta)

    2*r - x = 2 | know r^2 = x^2 + y^2

    2*(x^2 + y^2)^1/2 - x = 2 | move x to right side and both sides ^2, gives me:

    2^2 * ((x^2 + y^2)^1/2)^2 = (x + 2)^2

    4 * (x^2 + y^2) = (x + 2)^2

    4x^2 + 4y^2 = x^2 + 4x + 4 | move 4x^2 to right side

    4y^2 = x^2 + 4x + 4 - 4x^2

    4y^2 = -3x^2 + 4x + 4 | divide by 4 and takes the square root, giving me:


    y = ((-3x^2 + 4x + 4)/4)^1/2 | the calculator draws this as an half ellipse, above x-axis ?
     
    Last edited: Jun 5, 2013
  6. Jun 5, 2013 #5

    LCKurtz

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    I didn't check all your steps, but if you only took the positive square root, it wouldn't surprise me if half the graph is missing.
     
  7. Jun 6, 2013 #6

    HallsofIvy

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    Why take the square root? You understand that you will be losing half the graph that way, don't you?

     
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