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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.

Thankyou

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# Homework Help: Calculus of Variations

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