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Thread: Polar Form of Equation

  1. #1
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    Polar Form of Equation

    Find the polar form of

    a) $\displaystyle 2x-3y=6$
    b) $\displaystyle y^2=4x$

    I can manipulate the equation all I want but I don't know when I've got the polar form. Is there a standard form or something?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by c_323_h
    Find the polar form of

    a) $\displaystyle 2x-3y=6$
    b) $\displaystyle y^2=4x$

    I can manipulate the equation all I want but I don't know when I've got the polar form. Is there a standard form or something?
    Introduce the polar coordinates $\displaystyle \{r,\theta \}$, such that:

    $\displaystyle
    x=r \cos(\theta)$
    $\displaystyle
    y=r \sin(\theta)$,

    then substitute these into your equations.

    RonL
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  3. #3
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    but after i do that i can still manipulate the equation many times. how do i know which equation to write? do i keep going until i can no longer simplify?

    this is my answer for the first one.

    r=6/2cosx-3sinx

    and for the second one

    r^2(sin^2)-4rcos(x)=0
    Last edited by c_323_h; Apr 12th 2006 at 03:52 AM.
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  4. #4
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    Quote Originally Posted by c_323_h
    Find the polar form of

    a) $\displaystyle 2x-3y=6$
    b) $\displaystyle y^2=4x$

    I can manipulate the equation all I want but I don't know when I've got the polar form. Is there a standard form or something?
    a)
    $\displaystyle 2(rcos\theta)-3(rsin\theta)=6$
    $\displaystyle r(2cos\theta-3sin\theta)=6$
    $\displaystyle r=\frac{6}{2cos\theta-3sin\theta}$

    -Dan
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