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Arc-length integral with curve giving extremal value

  1. Mar 11, 2013 #1

    Leb

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    1. The problem statement, all variables and given/known data
    View attachment B3.bmp
    For whoever does not want to read the attached problem:

    Firstly, I need to express the arc-length from given [itex]x=r\cos\theta, y=r\sin\theta z = f(r), \text{ where } f(r) \text{ is an infinitely differentiable function and } r=r(\theta) \text{ i.e. parameter is } \theta[/itex] I begin with expressing it as [itex]ds = \sqrt{ r^{2} + r'^{2} + f'^{2} }[/itex] is that correct ?

    I then need to show that the solutions of those curves are given by:
    [itex]r ' ^{2} = r^{2}\frac{\left( \frac{r^{2}}{C} - 1\right)}{1 + f ' ^{2)}}[/itex]



    2. Relevant equations

    Lagrangian is independent of theta in this case, so I think I can use
    [itex]\frac{d}{d\theta}\left( L - r' \frac{\partial{L}}{\partial{r'}} \right) = 0 [/itex]

    3. The attempt at a solution


    Using the above, the term in parenthesis is equal to a constant. Differentiating L w.r.t. r' and rearranging I get [itex]r'^{2} = \frac{r^{4}+2f'^{2}r^{2} + f'^{4}-Br^{2} - Bf'{2}}{B}[/itex] where B is a constant. I thought that this uses the good old multiply by "1" trick, but when I tried it it got no cancellations.

    I can see the [itex]r^{2}\left( \frac{r^{2}}{C} - 1\right)
    [/itex] term in there, but do not know how to get the rest...
     
  2. jcsd
  3. Mar 11, 2013 #2
    For the line element ds the following is true ##ds^2 = dx^2 + dy^2 + dz^2##. Since your curve is on a surface parametrized by ##r## and ##\theta##, and since ##r## itself is a function of ##\theta##, you should end up with ##ds = g(\theta)d\theta##.
     
  4. Mar 11, 2013 #3

    Leb

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    Thanks for the reply !
    Do you mean I should express [itex]\frac{df}{d\theta} \text{ as } \frac{df}{dr}\frac{dr}{d\theta}[/itex] ?
     
  5. Mar 11, 2013 #4
    What I mean is that the expression for ds you got is incorrect. You can derive the correct equation from the expression for ds in the Cartesian coordinates by using the chain rule.
     
  6. Mar 11, 2013 #5

    Leb

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  7. Mar 11, 2013 #6
    [tex]
    ds^2 = dx^2 + dy^2 + dz^2
    = (d(r \cos \theta))^2 + (d(r \sin \theta))^2 + (d(f(r)))^2
    = (dr \cos \theta - r \sin \theta d\theta)^2
    + (dr \sin \theta + r \cos \theta d\theta)^2
    + (f'dr)^2
    \\
    = dr^2 + r^2 d\theta^2 + f'^2 dr^2
    = (1 + f'^2)dr^2 + r^2d\theta^2
    \\
    = (1 + f'^2)(dr(\theta))^2 + r^2d\theta^2
    = (1 + f'^2)(r'd\theta)^2 + r^2d\theta^2
    = ((1 + f'^2)r'^2 + r^2)d\theta^2
    \\
    ds = \sqrt {(1 + f'^2)r'^2 + r^2}d\theta
    [/tex]
     
  8. Mar 11, 2013 #7

    Leb

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    Thanks for taking the time to type that, but if we parameterize w.r.t. theta, why do we have d \theta ?
     
  9. Mar 11, 2013 #8
    Parametrized by ## \theta ## means that the length is a function of ## \theta ##: ## s = s(\theta) ##, thus ## ds = s'd\theta ##.
     
  10. Mar 11, 2013 #9

    Leb

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    So I take it that the previous link is rubbish ? Thanks again !
     
  11. Mar 11, 2013 #10
    No, it is not rubbish. If you look carefully, the formula I derived equals their formula if f = const. Your problem has three dimensions, not two as they have.
     
  12. Mar 11, 2013 #11

    Leb

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    Last edited: Mar 11, 2013
  13. Mar 11, 2013 #12
    Why should it disappear? Take a very simple example: ## y = x^2 ##. You can say that "y is parametrized by x". Then ## dy = 2x dx ##; x does not disappear. Even if ## y = x ##, you still have ## dy = dx ##.
     
  14. Mar 11, 2013 #13

    Leb

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    Well, I am a numpty and always had particular trouble with notation of derivatives and using the "implicit" chain rule.

    Anyway, thanks very much for bothering to reply !
     
  15. Mar 11, 2013 #14

    SteamKing

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  16. Mar 11, 2013 #15

    Leb

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    Hehe, reparametrization... I always viewed the parameters as an "arbitrary" choice when we are talking about simple curvy curves (not circles or ellipses).

    Thanks for the link !
     
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