Results 1 to 9 of 9

Math Help - gradient field

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    58

    gradient field

    how can I prove that if s(t) is a flux line of a gradient field F=-1*nabla(U)
    tehn U(s(t)) is a decreasing function of t
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,372
    Thanks
    1314
    \nabla U always points in the direction of greatest increase of the function U. Therefore, -\nabla U points in the direction of greatest decrease.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2010
    Posts
    58
    that I was gessing that was going to be the starting point. I just can't get any further...how does that prove what I want to prove?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2010
    Posts
    6
    Quote Originally Posted by Mppl View Post
    how can I prove that if s(t) is a flux line of a gradient field F=-1*nabla(U)
    tehn U(s(t)) is a decreasing function of t
    Since F=-1(\nabla U) is the direction of greatest decrease (as said above), then any U(s(t)) would be a decreasing function of t.

    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.
    Last edited by moderata; December 13th 2010 at 09:09 PM. Reason: Fixed some latex code.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    Posts
    58
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,372
    Thanks
    1314
    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 R^2 or R^3 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.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2010
    Posts
    58
    So you're stating simply that s(t) is decreasing, so any function, U(s(t)) is decreasing.
    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...

    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,372
    Thanks
    1314
    Quote Originally Posted by moderata View Post
    Since F=-1(\nabla U) is the direction of greatest decrease (as said above), then any U(s(t)) would be a decreasing function of t.

    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.

    So you're stating simply that s(t) is decreasing, so any function, U(s(t)) is decreasing.
    Okay, I didn't see this last sentence. Of course, it isn't correct. As I said, it, in general, makes no sense to say that s(t) is decreasing and, even if it were, it does not follow from that that U(s(t)) is decreasing.

    The crucial point is, as I said before that [tex]\nabla U[tex] points in the direction of fastest increase and so -\nabla U points in the direction of fastest decrease.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    Quote Originally Posted by Mppl View Post
    ............... doesn't it have a formal proof?
    Here goes one:
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Condition for F to be a gradient field
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 1st 2009, 06:25 PM
  2. Using gradient vector field to find value of f
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2009, 08:53 AM
  3. Field Gradient
    Posted in the Calculus Forum
    Replies: 12
    Last Post: April 26th 2009, 12:16 PM
  4. Gradient Vector Field?
    Posted in the Calculus Forum
    Replies: 15
    Last Post: February 23rd 2008, 05:43 PM
  5. gradient field
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 22nd 2007, 08:28 PM

Search Tags


/mathhelpforum @mathhelpforum