Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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
    
@PeterT.off: You are right, of course. But I have limited my analysis to the requirements of the OP from a preceding question math.stackexchange.com/questions/112542/… . –  Jon Feb 25 '12 at 10:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.