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Transform explicit function to polar form

  1. Jun 20, 2011 #1
    Hello,

    I'am new here and happy to find this great forum!
    Here's my first question: there's an explicit function as follows:
    y=2.sin(x)-1

    The transformation to polar form (r=3cos([itex]\theta[/itex]))
    - x=r.cos([itex]\theta[/itex])
    - y=r.sin([itex]\theta[/itex])

    So I get: r.sin([itex]\theta[/itex])=2.sin(r.cos([itex]\theta[/itex]))-1
    Now you see the problem: how can I isolate r in this equation ?
     
  2. jcsd
  3. Jun 20, 2011 #2

    HallsofIvy

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    You can't, or at least not with standard functions.
     
  4. Jun 20, 2011 #3
    I am confused...just because you are used to x and y being Cartesian coordinates, does not mean that they always are...

    what I am trying to say is that x and y in your original equation may not be what you think they are. For example, x in sin(x) is not an x-coordinate from the cartesian plane...for example x is not 42 inches...after all, you cannot take the sin(42 inches)...if x is radians, that's another story, then sin(42radians) is possible...

    so, you need to get your ducks in a row, here, and figure which way you are supposed to do your transformation...as it is, your attempt does not make sense at all...not to me, anyway.

    my 2 cents
     
  5. Jun 21, 2011 #4
    Hello,

    thank you all for quick replies !
    The cartesian equation y = 2.sin(x)-3 has indeed x as argument; ofcourse in radians.
    But what I know form transformation to polar form is that the y x plane transforms to a
    r [itex]\theta[/itex] plane. Being [itex]\theta[/itex] tha angle between r and de x axle. In my case: r=3.cos([itex]\theta[/itex]).

    So I suppose that every variable (also x in radians) in the equation has to be transformed by definition of the polar function. The value of y and x are described by
    x = r.cos([itex]\theta[/itex])
    y = r.sin([itex]\theta[/itex])

    If I can transform f(x) to paramatric function, I also can it transform to polar function (this is essentially a parameteric function). Am I wrong somewhere.

    I do realize that this equation isn't easy to write in polar form!

    greets
     
  6. Jun 21, 2011 #5

    disregardthat

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    You can't get an equation such as r = g(theta) from this, g elementary or not. g won't be well-defined, since there can be many points (x,y) on the original graph on a line with angle theta between it and the x-axis.
     
  7. Jun 21, 2011 #6
    Oké, but there exist an equation in polar form to describe the cartesian equation ? Because there al multiple x,y coordinates, and one of them wil point to the cartesian equation as you explained...

    greetz
     
  8. Jun 22, 2011 #7
    I guess for my example it's not possible, thank for your replies !
     
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