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Polar Coordinates functional notation.

  1. Jan 14, 2013 #1
    I've always been curious why points in polar coordinates are defined as (r,θ) when all equations (including parametric equations formed from them) are defined as r=f(θ).

    Considering that point in cartesian coordinates are defined as (x,y) where y=f(x).

    Also a,b=(r,θ) ∫1/2[f(θ)]2 further implies that θ is the domain.

    I just find this odd notation wise, and am wondering if anyone can provide me with a reason for this seeming discrepancy.

    :) Thanks!
  2. jcsd
  3. Jan 15, 2013 #2

    Stephen Tashi

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    That's a good question. I conjecture it's just tradition and a matter of convenience.

    Polar coordinates for a point are not unique (even though it's common to hear math people talk about "the" polar coordinates of a point). The point [itex] (r,\theta) [/itex] is the same as the point [itex] (r,\theta + 2 \pi ) [/itex].

    if you want to define points in the cartesian plane that have the polar form [itex] (f(\theta), \theta) [/itex] you have to be careful to make [itex] f(\theta) = f(\theta + 2 \pi ) [/itex]. This happens "naturally" with trigonmetric functions such as [itex] f(\theta) = \sin{\theta} [/itex].

    If you want to define a function by points in the cartesian plane that have the polar form [itex] (r, g(r) ) [/itex] then you have less to worry about. However you can't describe a curve with a set of points having the same radius and several different principal angles.

    I suspect polar coordinates for curves are most often used when we need a shape where two points with the same radii can have different principal angles. These are often written using trig functions so it isn't a problem to insure that [itex] f(\theta) = f(\theta + 2\pi) [/itex].
  4. Jan 15, 2013 #3


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    It is subtle, but a parameterization can map say a value on the real line to that of say [0,2pi) with the simple example being a circle with x = rcos(t), y = rsin(t) for t = [0,infinity).

    Its subtle, but I think its worth noting.
  5. Jan 19, 2013 #4

    But wouldn't a function defined as [itex] (r, g(r) ) [/itex] not be an actual function since [itex] f(\theta) = f(\theta + 2 \pi ) [/itex], so for every r there would be a myriad of possible values of [itex] \theta [/itex] that would be solutions. Just as we can only define inverse trigonometric (sine for example) functions on a limited ([itex] [0,2pi) [/itex] domain.

    Perhaps this is why we can only define polar functions with [itex] \theta [/itex] as the domain?

    Sorry if I seem a little distracted, but I've just been digesting a bunch of Mathematical Grammar and set logic, so my mind is completely scrambled :) haha.
    Last edited: Jan 19, 2013
  6. Jan 19, 2013 #5

    Stephen Tashi

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    My notation (r, g(r)) assumes g(r) is a function from the non-negative real numbers to the real numbers. So each r is mapped to only a single g(r).

    Your question about solving for theta is relevant to the case of ( f(theta), theta). It is correct that f(theta) must be a function with the property that f(theta) = f(theta + 2 pi).

    If we need to write a function whose graph is a spiral, we have to introduce a parameter and make both radius and angle depend on the parameter in the form ( r(t), theta(t)). So it isn't correct to say that there is only one form for a graph in polar coordinates. It's just that (f(theta), theta) is a very commonly encountered form.
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