- #1

- 832

- 30

[tex]x=r \cos\theta[/tex]

[tex]y=r \sin \theta[/tex]

[tex]z=Ar[/tex]

Then I've defined the arc lenght:

[tex]

ds^2=dr^2+r^2d\theta^2+A^2dr^2

[/tex]

So, the arclenght:

[tex]ds=\int_{r_1}^{r_2}\sqrt { 1+A^2+r^2 \left ( \frac{d\theta}{dr}\right )^2 }dr[/tex]

And using Euler-Lagrange equation:

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

[tex]\frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}[/tex]

[tex]\frac{d}{dr}\frac{\partial f}{\partial \dot \theta}=0 \rightarrow \frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}=K

[/tex]

The differential equation which I've arrived is non linear. I don't know if what I've done is fine. I know the problem can also be done using spherical coordinates instead of cylindrical, but I've choossen to do it this way.