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Arc Length in Polar Coordinates

  1. May 29, 2010 #1
    In the polar formula for arc length, [tex]ds^{2}=dr^{2}+r^{2}d{\theta}^{2}[/tex], what is the exact meaning of the [tex]r^2[/tex] term multiplying [tex]d{\theta}^2[/tex]? Is it an initial distance from the origin? A final distance from the origin? The change in r from point a to point b? This baffles me to no end and nothing explains it.
     
  2. jcsd
  3. May 29, 2010 #2
    God I hate this stuff. But after you use it, you kind of realize how much effort it saves.

    This is the meaning of that rather "disembodied" statement.

    Let's draw a curve through space. And now let's parametrize that curve with a parameter t. So along this path, [tex]r=\tilda r(t), \ \ \theta = \tilda \theta (t)[/tex].
    Then the arclength satisfies the following 'differential' equation:

    [tex] (\frac{ds}{dt})^2 = (\frac{d\ r (t)}{dt})^2 + r^2 (\frac{d\ \theta (t)}{dt})^2 [/tex]

    I've included the (t) thing to make the method of calculation explicit. You literally differentiate the function r(t) wrt t. r=r(t) is a horrible abuse of notation that actually gets me confused from time to time but...it saves a lot of time too. Haha.
     
  4. May 29, 2010 #3
    Do you mean a geometric reason, or an analytic reason?

    To see how it got there, just consider the following:

    [tex]
    dS = \sqrt{1 + (\frac{dy}{dx})^2}
    [/tex]

    Let [tex] y = r \sin \theta[/tex] and [tex] x = r \cos \theta [/tex] and work it out and you'll get your r^2 factor on the [tex]d\theta[/tex]
     
  5. May 29, 2010 #4
    I mean a geometric reason; the derivation is no problem.
     
  6. May 29, 2010 #5
    It gives the arclength gained by increasing r by dr and theta by dtheta.

    [tex]
    (\frac{ds}{dt})^2 = (\frac{d\ r (t)}{dt})^2 + r^2 (\frac{d\ \theta (t)}{dt})^2
    [/tex]
    So yes, the r^2 corresponds to how far you are from the origin.
    Think about small curves on a sphere. If your r is large, then your ds is going to be larger, for a given dtheta.
     
  7. Jun 7, 2010 #6
    Let's take an example: suppose I am moving from the polar point [tex](2, \frac{\pi}{4})[/tex] to [tex](3, \frac{\pi}{2})[/tex]. Would my distance traveled be [tex]1+\frac{{\pi}^{2}}{4}[/tex]?
     
  8. Jun 8, 2010 #7
    No, the relation you stated in the beginning is a differential relation, meaning it is good only for really close points. (It's like a taylor expansion)

    Otherwise you need to develope a formula using trigonometry or using the definition of distance via cartesian coordinates then substituting with polar coordinates.

    Back to your original question: dtheta denotes a change in the angle. But as you may know, the angle between two rays doesn't depend on their length. So the information of the angle itself doesn't give you a measure of distance. At small distances from the origin, taking a small angle difference will give you a small distance (the arc of a small circle). At large distances, taking a small angle step will give you a larger distance.
    Therefore the r dependence comes in.
     
  9. Jun 10, 2010 #8

    Gib Z

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    http://img716.imageshack.us/img716/3151/arcy.jpg [Broken]

    Differentiable curves exhibit local linearity, so if we zoom up to infinitesimal scales the curve is approximately a straight line. We wish to find the infinitesimal increments of the arc length of the curve between polar coordinates [itex](r,\theta)[/itex] and [itex](r+dr, \theta + d\theta)[/itex].

    By the Pythagorean theorem, [itex]ds^2 = dr^2 + (ab)^2[/itex], but the length of the line segment ab can be approximated by the length of the arc that passes through a and b of the circle centered at the origin of radius r. The length of this arc is given by [itex]r d\theta[/itex].
     
    Last edited by a moderator: May 4, 2017
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