Solving Dynamical Systems with Polar Coordinates

[Juggler123]

I've been solving some dynamical systems and have had to convert some of the ode's to polar co-ordinates, I've been using the formula

r\dot{r} =x\dot{x} y\dot{y}

to easily solve for \dot{r}

I'm just wondering if there's an easy to use formula to find \dot{\theta}

I've been trying to formulate one but with no real luck!

Any suggestions?

 

[tiny-tim]

Hi Juggler123!

(you missed out a + )

tanθ = y/x, so … sec²θ dθ = … ?

 

[Juggler123]

Ooops yeah missed that +!

I've worked on from tanθ = y/x but can't end up a nice formula for theta dot. Just wondering if one exists!?

 

[tiny-tim]

Show us how far you've got.

 

[Juggler123]

tex function dosen't seem to be working for me which is annoying...

theta-dot*r^2=x^2*d/dt(y/x)

which I think gives

theta-dot*r^2=x^2*[y-dot*x^(-1) - y*x-dot^(-2)]

not particularly nice though..

 

[tiny-tim]

Looks ok to me …

r²θ' = xy' - yx' ​

 

[inknit]

\frac{48}{2}(9+3)

 

[DarthPickley]

There are about 4 pages of this stuff in my english notebook... (my physics notebook was full)

This is what I got:

θ' = (xy' - yx')/r
r' = (xx' + yy')/r

of course its possible that I made a mistake. most of my work from then on dealt with trying to derive formulas for θ'' and r''.

I got r'' = (x''x + y''y + (θ')² )/r

oh, and I did all this because I was trying to figure out like Newton's Laws and trying to get a differential equation that would end up with like r = r₀/( 1 + e cos θ ) to validate kepler's 1st law, and also get a parametric equation for r and θ in terms of time, but I wasn't able to since I haven't taken math that high in school yet, and an upperclassman told me that there was a better way, so geometric differentiation I guess, because I know that Newton did a lot of that.

 

[tiny-tim]

Welcome to PF!

HI DarthPickley! Welcome to PF!

θ' = (xy' - yx')/r

No, that can't be right, the dimensional analysis is wrong …

the LHS is T^-1, but the top line of the RHS is L²T^-1, and the bottom line is only L

(similarly your r'' must be wrong …

the best way of getting r'' would be to differentiate r² = x² + y² twice )

 

[DarthPickley]

I thought I had probably made a mistake.

but anyways I just learned about Feynman's lost lecture so I have another way of understanding the Kepler stuff, but anyways,
I think that here, for my derivations, I will redo things:

having a vector v such that |v| = r, and then you have v + dv, with |v+dv| = r+dr. to get what dr is, you have to project dv onto v, so you have dv ∙ v / |v| = (x*dx + y*dy)/r. to find dr / dt, you get (x*dx/dt + y*dy/dt)/r = ( x*x' + y*y' ) / r.
to get r * sin dθ [$], which is the angular component of dv, you have to find the part of dv that is perpendicular to v. since the vector perpendicular to v would be < -y , x >, you can project dv onto this vector to get r * sin dθ. so r * sin dθ = dv ∙ <-y,x> / r = ( -y*dx + x*dy ) / r. sin dθ = (x*dy - y*dx) / r² also, for very small dθ, sin dθ --> dθ. (they are about equal) so, dθ/dt = (x*dy - y*dx )/r² / dt = (x*dy/dt - y*dx/dt)/r² = ( x*y' - y*x' ) / r² Thus, the REAL final answer is θ' = (xy'-yx')/r² [$] -- this is where I messed up before (probably). I think I said something silly like that θ is the same as the tangential component. Thanks for pointing this out, I puzzled over it awhile before realizing.

also, this is just the same thing that tiny-tim gave.