Finding the Differential Equation for Straight Lines in Polar Coordinates

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littleHilbert
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Homework Statement



Problem from Arnold's "Mathematical Methods of Classical Mechanics" on page 59.

Find the differential equation for the family of all straight lines in the plane in polar coordinates.

Homework Equations



[tex]\Phi=\displaystyle\int^{t_2}_{t_1} \sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}\,dt[/tex]

[tex]\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}}-\frac{\partial{L}}{\partial{q_i}}=0[/tex]

The Attempt at a Solution



L is the integrand.

We have two equations:

[tex]\frac{d}{dt}\frac{\dot{r}}{\sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}}=\frac{r\dot{\phi}^2}{\sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}}[/tex]

[tex]\frac{d}{dt}\frac{r^2\dot{\phi}}{\sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}}=0[/tex]

The second gives: [tex]\frac{r^2\dot{\phi}}{\sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}}=c[/tex]

If c=0 we have the derivative of phi zero, that is phi would be constant and we are (essentially) done. If phi is constant, we have a bundle of lines paasing through the origin. But what about r? Do we get problems in the origin? Polar coordinates are not defined there, are they?
If c is not zero, then the first equation can be rewritten as:
[tex]\frac{d}{dt}\frac{\dot{r}}{r^2\dot{\phi}}=\frac{\dot{\phi}}{r}[/tex]

I cannot get an idea of how to solve it for r-dot AND phi-dot. Tried to differentiate the left side and see what happens, but somehow nothing attractive comes out.
So how to proceed? Is it the right way? Or is there anything I can't see at a glance that helps?
 
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on Phys.org
Just take this,
[tex]\frac{r^2\dot{\phi}}{\sqrt{{\dot{r}}^2+<br /> r^2{}\dot{\phi}^2}}=c[/tex] and solve by separation of variables. I.e. write it as f(phi)d(phi)=g(r)d(r). Then integrate. To tell if you are correct, look up general representations of a line in polar coordinates, for example, r=p*sec(phi-phi0) defines a line for various choices of p and phi0.
 
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He is supposed to find the differential eqn for the family of all st lines. There should be no constants there. E.g., in cartesian co-ordinates, it should be y''=0.

It would be simpler to treat one of the co-ordinates, say phi, as the independent co-ord and form the integral in terms of r'=dr/dphi.

[tex]\Phi=\displaystyle\int^{\phi_2}_{\phi_1} \sqrt{{\dot{r}}^2+r^2{}\dot{\phi}^2}\,d\phi[/tex]

Now he can apply:

[tex]\frac{d}{d\phi}\frac{\partial{L}}{\partial{\dot{r}}}-\frac{\partial{L}}{\partial{r}}=0[/tex]

to get one 2nd order differential eqn.