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I was wondering if is there a way to attack with Euler-Lagrange equation the following problem.

Suppose $B$ is moving in straight line with costant velocity $\mathbf{u}=u\,\hat{x}$. What is the fastest path for an $A$, which can move with costant speed $v>u$ (and variable velocity $\mathbf{v}=v\hat{\mathbf{e}}$), to catch $B$? Suppose $A(0)=(0,0)$ and $B(0)=(b_1,b_2)$.

I didn't do much progress. What I think is that, since the speed of $A$ is costant, the best path must be the one with minimal lenght. However, the fact that $B$ is moving is quite disorienting and I don't know how to set up the problem.

Do you have any idea?

pppqqq
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  • What are the relative positions of A and B and what is the direction of the velocity vectors? – response May 29 '13 at 17:31
  • Please see edited question, I've added some details. $B$ is moving in $x$ direction, $A$ is moving in some variable direction with costant speed. – pppqqq May 29 '13 at 17:47

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You can do it without calculus. The coordinates of $B$ are $(b_1+ut,b_2)$. For $A$ to catch $B$, it should move in a straight line to the intercept point. The distance from the origin to $B$'s location is $\sqrt{(b_1+ut)^2+b_2^2}$. We can equate this to $A$'s range to get $$vt=\sqrt{(b_1+ut)^2+b_2^2}\\v^2t^2=b_1u^2t^2+2b_1ut+b_1^2+b_2^2\\(v^2-b_1^2u^2)t^2-2b_1ut-(b_1^2+b_2^2)=0$$ which can be solved for $t$ using the quadratic equation. Then plug that $t$ into $B$'s trajectory and you have the intercept point. $A$ should move straight toward that point.

Ross Millikan
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  • I may be missing some obvious point, but I don't understand why do you say: "For A to catch B, it should move in a straight line to the intercept point". Why couldn't $A$ follow some different trajectory, for example, always pointing towards $B$? – pppqqq May 29 '13 at 18:03
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    @pppqqq: That will lengthen the trajectory, which will lengthen the time. If A can catch B on some other trajectory, it could move to the intercept point and arrive before B. Then it could find another intercept point closer (hence shorter time) to the origin. – Ross Millikan May 29 '13 at 18:25
  • Hell, yeah. Maybe you could add this comment to the original answer. – pppqqq May 29 '13 at 18:28