Fastest approach path to a moving object Say you have a point $A$ which's coordinates at the time $t$ are given by $(0, tv_0)$ for some constant $v_0$.
You have another point $B$ with coordinates given by the function $x$ with $x(0)=(x_0,y_0),|x'(t)|=v_1$ for some constant $v_1$.
How would you find the function $x$ if you know that the time it takes for $B$ to reach $A$ is minimal?
There are some cases where, depending on $v_0$ and $v_1$, $B$ will never reach $A$. You can assume those are not the case.
I always assumed that the best tactic was figuring out the intersection if the point $B$ moves linearly and then just take that path, but my friend that studies physics said that $\cosh$ is the best function to solve some problem similar to this one.
(I would appreciate if someone would edit the tags to make them more appropriate for this question)
 A: I believe you are right; the optimal path is a straight line.  We can prove this by contradiction.  Suppose $x(t)$ is an optimal (curved) path with $x(0) = B$, $|x'(t)| = v_1$, and that $x(T) = P = (0, Tv_0)$.  Then we will construct a better path $y(t)$ such that $y(T') = Q = (0,T'v_0)$ and $T'<T$. 
 
Indeed, let $y(t)$ start with a straight-line path from $B$ to $P$.  This path arrives at $P$ before $x$ does, because straight lines are the shortest paths between points.  So we can let $T_P < T$ be the time when $y(T_P) = P$.  After $T_P$, let $y(t)$ turn around and go straight down the vertical axis, so $y(t) = (0, Tv_0 - (t-T_P)v_1)$ for $t>T_P$.  It's clear to see that $y(t)$ will intersect the point at $Q$ before time $T$.  Therefore, $x(t)$ was not optimal.
A: Here's my proof that a straight line is the best approach.
At any time $t$, the farthest point $B$ can move is to the edge of a circle of radius $tv_1$. We know that the precise position of $A$ at time $t$ is $(0, tv_0)$. If this point isn't inside the circle, clearly there is no way to for $B$ to reach $A$ by that point in time. 
As $t$ increases, so does the radius so there will be a time $t_1$ when $A$ will be on the edge of the circle. This is the earliest possible moment when $B$ can reach $A$ and the only way to do that is to go straight to that point of intersection.
