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Arc length in polar form θ = f (r)

  1. Sep 7, 2008 #1
    hi! i need some help here, do you have any available example on how to find the arc length in polar form θ = f (r)? using integral calculus, i mean. i searched the internet but i only got the r= f(θ) example. i hope you can help me. thanks!:)
  2. jcsd
  3. Sep 7, 2008 #2
    Let's take this as a starting point: When ever we have a path defined with mapping [itex]t\mapsto (x(t),\; y(t))[/itex], the length of the path is given by

    \int dt\;\sqrt{(\dot{x}(t))^2 + (\dot{y}(t))^2}.

    Now you have the path defined with a formula [itex]\theta=f(r)[/itex]. A systematic way to find the length of the path is to somehow return it to the parametrization in Cartesian coordinates. This is how it can be done. We think of the [itex]r[/itex] as the parameter that replaces the [itex]t[/itex] in the previous formula, and then define the Cartesian components with formulas

    x(r) = r\cos(f(r))

    y(r) = r\sin(f(r))

    Then the length of the path is given by

    \int dr\; \sqrt{(\dot{x}(r))^2 + (\dot{y}(r))^2} = \cdots

    and the task left to be done is to compute [itex]\dot{x}(r)[/itex] and [itex]\dot{y}(r)[/itex] with the usual derivation rules.
  4. Sep 7, 2008 #3
    jostpuur- thanks for the immediate reply. :) by any chance, do you have an example regarding that? i mean, i'm looking for an example on how to solve the arc length using integral calculus if the given is in the form of θ = f (r). my professor in calculus gave us the formula to that.. (i don't know how to post an equation here. sorry i made it very inconvenient.) and it is different from what you post.
  5. Sep 7, 2008 #4
    Actually I proceeded in the calculation further on the paper, but I intentionally just wrote the dots [itex]\cdots[/itex] and didn't write it all here, because I wasn't sure if it would be okey to write it now. But are you saying that you already know the formula for the length of the path defined with [tex]\theta = f(r)[/tex]? A formula, which does not contain the functions x(r) and y(r), but only f(r) (actually its derivative)? If so, I can write more details about how it can be derived.

    To write equations, write for example [ tex ] 1+1=2 [ /tex ], but don't put the spaces after and before characters "[" and "]". If you click an equation with the mouse button, you can see the code that has been used for it.
  6. Sep 7, 2008 #5
    i still can't write equations. i tried to use that [ tex ] but it turned out that it can't read the symbols i used such as the integral sign with the upper and lower limit as theta 2 and theta 1.

    in words, the equation is like this.

    S= integral (upper limit: theta 2, lower limit: theta 1) of square root (1+ (r squared)*(dr/dtheta)^2) dr.
    Last edited: Sep 7, 2008
  7. Sep 7, 2008 #6
    I think there is something wrong with you equation. You mean this?

    \int\limits_{\theta_1}^{\theta_2} \sqrt{1 + r^2\big(\frac{dr}{d\theta}\big)^2} dr

    (Click the equation to see how it can be typed)

    You are integrating with respect to r, but the integration limits are thetas. The derivative [tex]dr/d\theta[/tex] seems to be of a function [tex]r(\theta)[/tex], but we don't have this kind of function now.

    In any case, this is very close to the expression I derived couple of minutes ago, so I'm guessing that you just made some mistakes when concentrating in the latex problems. It could be we are talking about the same equation still.
  8. Sep 7, 2008 #7
    I would still encourage you to see where this start

    will lead. I'm sure that this is one way to derive the formula you are talking about.

    According to the usual derivation rules we get

    \dot{x}(r) = \cos(f(r)) - r f'(r) \sin(f(r))

    Then derive same kind of formula for [tex]\dot{y}(r)[/tex], and compute what

    (\dot{x}(r))^2 + (\dot{y}(r))^2 = \cdots

  9. Sep 7, 2008 #8
    oh, yes. i got your point there. thanks for the help, and thanks for the info on how to write an equation.
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