Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transform explicit function to polar form

  1. Jun 20, 2011 #1

    I'am new here and happy to find this great forum!
    Here's my first question: there's an explicit function as follows:

    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


    User Avatar
    Science Advisor

    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

    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!

  6. Jun 21, 2011 #5


    User Avatar
    Science Advisor

    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...

  8. Jun 22, 2011 #7
    I guess for my example it's not possible, thank for your replies !
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook