always points in the direction of greatest increase of the function U. Therefore, points in the direction of greatest decrease.
Basically it isn't asking you to prove that s(t) is a flux line, it's asking that IF s(t) is a flux line with the gradient field of F, then is U(s(t)) a decreasing function.
Well, let me see! You're saying that s(t) is a decreasing function imagine it is... Why Does ir imply that U(s(t)) is a decreasing function aswell? Can't get there. Bur I would lime you to tell me aswell Why the hell is s(t) a decreasing function. Isn't ir just a curve?
No one has said that s(t) is a decreasing function! In fact, if U is a function of two or three variables, s(t) maps R into or which are not ordered so "increasing" and "decreasing" make no sense for s(t). I am beginning to think you are completely confused as to what this question is asking.
(By the way, you will have to take "decreasing" in the more general sense of "non-increasing". For example, if U is a constant function the U(s(t)) will be constant for any s.)
I gess I've not commited a mistake saying someone told it was decreasing...someone said that s(t) is decreasing I'm not mad, not yet...So you're stating simply that s(t) is decreasing, so any function, U(s(t)) is decreasing.
And why would that statement be right? imagine U(s(t))=-s(t) then if s(t) is increasing U(s(t)) is decreasing. Maybe you're not explaning things clearly...or maybe (and most probably, I'm missing something) doesn't it have a formal proof?
The crucial point is, as I said before that [tex]\nabla U[tex] points in the direction of fastest increase and so points in the direction of fastest decrease.
F = -grad(U), s(t) is a curve such that its velocity vector( derivative) s'(t) = F. Let's compute the derivative of U(s(t)) as a function of t.
d( U(s(t)) )/dt (using the chain rule, use <,> to denote inner product) = <grad(U), s'(t)> = <-F, F> = -||F||^2 < 0. DONE.