Re: Generalized Pursuit Curve Problem

From: Narasimham G.L. (mathma18_at_hotmail.com)
Date: 09/06/04 

Date: 6 Sep 2004 13:43:35 -0700

pikalaw <tsang@cims.nyu.edu> wrote in message news:<I5R_c.7708$Wv5.1576@newsread3.news.atl.earthlink.net>...
> pikalaw wrote:
> >
> > Clearly, grad q != Q-P for -1 <= t <= 0.
> >
> Ooops, I made a mistake. Q *is* in the direction of P in your program:
> grad q = (2A, -2Bt) = 2(A, -Bt)
> and
> P-Q = (-At, Bt^2) = -t(A, -Bt).
> Since t is negative, these two are in the same direction.

No, we can find poining better by: (-2Bt/2A) = (Bt^2/-At)= tan(theta),
for all t, irrespective of sign. Did you run the QBASIC program ?

> So, you are right if Q does not begin at A. Of course, a simpler
> counter-example is:
> Q starts at the position x=0, and
> P starts at A located at x=1 and goes the straight path toward B
> located at x=2.
> Clearly, the path of Q has length 2 while the path of P has length 1.
>
> However, if Q does start at A, intuitively the Q's path cannot be longer
> than P's. Of course, initially when both P and Q are at A, Q's
> direction toward P is the zero vector and Q will not move until a later
> time when P has left A.
>
> -pikalaw

A general proof may still be worked out... But you have never
explained how in the first place you had this hunch.