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p(t) is a function of t, and A is a constant.

- Thread starter CSteiner
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- #1

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p(t) is a function of t, and A is a constant.

- #2

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An idea would be to rewrite ##\cos \arctan \frac{\dot{y}}{\dot{x}}## as ##\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}##, and similar for the other equation. Then try to simplify it a bit.

p(t) is a function of t, and A is a constant.

- #3

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Didn't think of that, thanks! I'll keep working on it.

- #4

pasmith

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You may be best served by switching to intrinsic coordinates [itex](s, \psi)[/itex] where [tex]

p(t) is a function of t, and A is a constant.

\dot x = \dot s \cos\psi, \\

\dot y = \dot s \sin \psi.[/tex] Your system is then [tex]

(\dot s - p) \cos \psi = 0 \\

(\dot s - p) \sin \psi = -At.[/tex] Now either [itex]\dot s = p[/itex] or [itex]\cos \psi = 0[/itex]. The first is impossible unless [itex]A = 0[/itex]. The second requires that [itex]\dot x = 0[/itex] and [itex]\dot y = \pm p - At[/itex].

If [itex]A = 0[/itex] then [itex]\dot s = p[/itex] and [itex]\psi[/itex] can be an arbitrary function of [itex]t[/itex].

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Sorry, but it looks like something went wacky with your latex code. Could you retype it? It's hard to understand what your saying.You may be best served by switching to intrinsic coordinates [itex](s, \psi)[/itex] where [tex]

\dot x = \dot s \cos\psi, \\

\dot y = \dot s \sin \psi.[/tex] Your system is then [tex]

(\dot s - p) \cos \psi = 0 \\

(\dot s - p) \sin \psi = -At.[/tex] Now either [itex]\dot s = p[/itex] or [itex]\cos \psi = 0[/itex]. The first is impossible unless [itex]A = 0[/itex]. The second requires that [itex]\dot x = 0[/itex] and [itex]\dot y = \pm p - At[/itex].

If [itex]A = 0[/itex] then [itex]\dot s = p[/itex] and [itex]\psi[/itex] can be an arbitrary function of [itex]t[/itex].

- #6

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It takes time before equations render correctly but eventually they do. What are you seeing? Try a different browser or device.Sorry, but it looks like something went wacky with your latex code. Could you retype it? It's hard to understand what your saying.

Basically he says to substitute ##\dot x = \dot s \cos\psi##, ##\dot y = \dot s \sin \psi##. Then it's very simple to arrive at a solution.

BTW, perhaps the first solution should have been ##\dot x=0##, ##\dot y=p\pm At##.

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