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How can we solve the simultaneous equations:

$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{x\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$

$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{y\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$

I am hoping that the solution is $y=x$, fingers-crossed.

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up vote 3 down vote accepted

Take the difference between the two equations and get

$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x-\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{(x-y)\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$

and the result follows straightforwardly.

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Thanks, I am sure I'm missing something... Does I have to expand the left hand side by the product rule or is it clear by inspection? (Sorry for being stupid, I don't know why I am not seeing it!) – misere Feb 24 '12 at 11:15
Do not worry to ask. Written as a difference, being also symmetric under the exchange of x with y, the conclusion is immediate. – Jon Feb 24 '12 at 11:46
Thank you sooo much, you're great! :) – misere Feb 24 '12 at 12:19
@Jon Although the procedure works, shouldn't we be looking for more solutions than just the trivial one? – Pedro Tamaroff Feb 24 '12 at 23:20 You are right, of course. But I have limited my analysis to the requirements of the OP from a preceding question… . – Jon Feb 25 '12 at 10:20

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