> Also, how is A(t) and B(t) are parallel not the same as "the curves A > and B are parallel at t"?
Well, suppose A and B are lines with endpoints (0,0), (2,0) for A and (0,1),(2,1) for B. Obviously, for all t, A and B are parallel at t. However let t = 0.5. Then A(t) = (1,0) and B(t) = (1, 1). The vectors (1,0) and (1,1) are not parallel, so saying A(t) || B(t) is the same as saying that there exists c such that (1,0) = c*(1,1), which isn't true. However, A'(t)=(2,0) and B'(t)=(2,0), and the vectors (2,0) and (2,0) are parallel. Does this make more sense? Regards, Denis. ----- "Jim Graham" <james.gra...@oracle.com> wrote: > On 10/20/10 7:54 AM, Denis Lila wrote: > >> In #2, you have a bunch of "I'() || B'()" which I read as "the > slope > >> of the derivative (i.e. acceleration) is equal", don't you really > mean > >> "I() || B()" which would mean the original curves should be > parallel? > >> Otherwise you could say "I'() == B'()", but I think you want to > show > >> parallels because that shows how you can use the dxy1,dxy4 values > as > >> the parallel equivalents. > > > > Not really. I've updated the comment explaining what || does, and > > it should be clearer now. Basically, A(t) || B(t) means that > vectors > > A(t) and B(t) are parallel (i.e. A(t) = c*B(t), for some nonzero > t), > > not that curves A and B are parallel at t. > > I'm not sure we are on the same page here. > > I'() is usually the symbol indicating the "derivative" of I(). My > issue > is not with the || operator, but with the fact that you are applying > it > to the I'() instead of I(). > > Also, A(t) = c*B(t) is always true for all A and B and all t if you > take > a sample in isolation. Parallel means something like "A(t) = c*B(t) > with the same value of c for some interval around t", not that the > values at t can be expressed as a multiple. > > Again, I'() || B'() says to me that the derivative curves are > parallel, > not that the original curves are parallel... > > ...jim
