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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.

1 Answers 1

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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!)2012-02-24
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    Do not worry to ask. Written as a difference, being also symmetric under the exchange of x with y, the conclusion is immediate.2012-02-24
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    Thank you sooo much, you're great! :)2012-02-24
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    @Jon Although the procedure works, shouldn't we be looking for more solutions than just the trivial one?2012-02-24
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    @PeterT.off: You are right, of course. But I have limited my analysis to the requirements of the OP from a preceding question http://math.stackexchange.com/questions/112542/infimum-length-of-curves .2012-02-25
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    @Jon Oh, I see.2012-02-25