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?
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.