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Homework Statement
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]
Homework 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]
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...