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Math Help - Polar Coordinate Problem

  1. #1
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    Polar Coordinate Problem

    Hello, I'm having a problem involving polar coordinates and writing an equation using rectangular coordinates. Here are the directions and the problem:

    The letters r and θ represent polar coordinates. Write each equation using rectangular coordinates (x,y).

    r=(3)/[3-cos(θ)]

    I attempted to work on the problem, but I eventually got stuck. Here is my attempt:

    r = (3)/[3-cos(θ)]
    r[3-cos(θ)] = 3
    3r - rcos(θ) = 3
    [3r - rcos(θ)]2 = 32
    9r2 - 6r2cos(θ) + r2cos(θ) = 9
    9r2 - 5r2cos(θ) = 9

    After that, I'm not exactly sure on what I need to do. I would turn the r2 into x2 + y2, but I'm not sure what to do with the 5r2cos(θ).
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  2. #2
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    Re: Polar Coordinate Problem

    Quote Originally Posted by Smugleaf View Post
    Hello, I'm having a problem involving polar coordinates and writing an equation using rectangular coordinates. Here are the directions and the problem:

    The letters r and θ represent polar coordinates. Write each equation using rectangular coordinates (x,y).

    r=(3)/[3-cos(θ)]

    I attempted to work on the problem, but I eventually got stuck. Here is my attempt:

    r = (3)/[3-cos(θ)]
    r[3-cos(θ)] = 3
    3r - rcos(θ) = 3
    [3r - rcos(θ)]2 = 32
    9r2 - 6r2cos(θ) + r2cos(θ) = 9
    9r2 - 5r2cos(θ) = 9

    After that, I'm not exactly sure on what I need to do. I would turn the r2 into x2 + y2, but I'm not sure what to do with the 5r2cos(θ).

    the coordinate transform equations are

    x = r \cos(\theta)

    y = r \sin(\theta)

    so

    \cos(\theta)= \frac{x}{r} =x / sqrt(x^2+y^2)

    LaTex interpretation is being dysfunctional tonight sorry.
    Last edited by romsek; February 3rd 2014 at 08:05 PM.
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  3. #3
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    Re: Polar Coordinate Problem

    Hello, Smugleaf!

    You squared prematurely . . .


    \text{Write in rectangular coordinates: }\:r\:=\:\frac{3}{3-\cos\theta}

    r(3-\cos\theta) \:=\:3

    3r - r\cos\theta \:=\:3

    3\sqrt{x^2+y^2} - x \:=\:3

    3\sqrt{x^2+y^2} \:=\:x+3

    \text{Square: }\:9(x^2+y^2) \:=\:x^2 + 6x + 9

    . . . . . . 9x^2 + 9y^2 \:=\:x^2+6x+9

    . . 8x^2 - 6x + 9y^2 \:=\:9
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