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Homework Help: Calculus of Variations

  1. Nov 21, 2005 #1
    I have this question,

    Express the length of a given curve [tex]r = r(\theta)[/tex] in radial co-ordinates. Using the Variational principle derive the shortest path between two points is a line.

    Ive drawn a picture with two angles (measured from the x-axis) [tex]\theta_1[/tex] and [tex]\theta_2[/tex] so that [tex]r(\theta_1) = r_1[/tex] and [tex]r(\theta_2) = r_2[/tex].

    I found the legth of the infinitesimal [tex]dl[/tex] to be

    [tex]dl = \sqrt{r'^2 + r^2} \; d\theta[/tex] so that

    [tex]\displaystyle{L = \int^{\theta_2}_{\theta_1} \sqrt{r'^2 + r^2} \; d\theta}[/tex]

    So my functional is [tex]f = \sqrt{r'^2 + r^2}[/tex]

    I have calculated [tex]\displaystyle{\frac{\partial f}{\partial r} = \frac{r}{\sqrt{r^2 + r'^2}}}[/tex] and [tex]\displaystyle{\frac{\partiall f}{\partial r'} = \frac{r'}{\sqrt{r^2 + r'^2}}}[/tex]

    Also [tex]\displaystyle{\frac{\partial f}{\partial \theta} = 0}[/tex]

    I have tried setting up

    [tex]\displaystyle{\frac{d}{d\theta} \frac{\partial f}{\partial r'} - \frac{\partial f}{\partial r} = 0}[/tex] but this is leading to a nasty nonlinear second order differential equation which i presume is not correct.

    I know that [tex]\theta[/tex] is cyclic so a conservation law exists but im not sure how to go about using this.

    If someone could give me a plan of how to do the next couple of steps it would be great.

  2. jcsd
  3. Nov 21, 2005 #2

    Physics Monkey

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    I think the equation you should obtain from varying the path length is
    \frac{d}{d\theta}\left(\frac{r'}{f}\right) = \frac{r}{f},
    where [tex] f(r,r') = \sqrt{r^2 + r'^2} [/tex]. First, let me note that a circle ([tex] r [/tex] constant) does not satisfy this equation. You should expect this since the arc of a circle is not the shortest distance between two points. Second, you should probably write out the equation for a line in polar coordinates. It is kind of messy in general, but you could pick a representative type of line and show that it solves the equation. Once you know this, you can always rotate your coordinates to show that all other lines satisfy the differential equation too.

    I worked it out myself so it can be definitely be done. Let me know if you have any more trouble.
    Last edited: Nov 22, 2005
  4. Nov 22, 2005 #3
    In cartesian coordinates you have:
    [tex]\int{\sqrt{1 + \dot{y}^2}dx}[/tex]
    [tex]\frac{\partial f}{\partial y} - \frac{d}{dx}\displaystyle{(}\frac{\partial f}{\partial \dot{y}}\displaystyle{)} = 0[/tex]
    ... thus:
    [tex]f = \sqrt{1 + \dot{y}^2}[/tex]
    [tex]\frac{\partial f}{\partial y} = 0; \frac{\partial f}{\partial \dot{y}} = \frac{\dot{y}}{\sqrt{1 + \dot{y}^2}}[/tex]
    [tex]\frac{d}{dx}\displaystyle{[}\frac{\dot{y}}{\sqrt{1 + \dot{y}^2}}\displaystyle{]} = 0[/tex]
    [tex]\frac{\ddot{y}}{\sqrt{1 + \dot{y}^2}^3} = 0[/tex]
    ... thus:
    [tex]\ddot{y} = 0[/tex]
    ... and:
    [tex]y = ax + b[/tex]
    Last edited: Nov 22, 2005
  5. Nov 22, 2005 #4
    Thankyou very much for your time and thoughts. I have solved the problem in polar co ordinates and if anyone is particularly interested i will write it up for them. I will check back tomorrow.
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