Results 1 to 8 of 8

Math Help - Gradient vector/optimum path

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    7

    <unsolved> Gradient vector/optimum path

    Hi, i need help in the following question..

    Suppose I am on point (3,-8)

    i am given F(x,y) = 80 - 2x^2 - y^2,

    find the equation of the path i should take (in terms of x and y) if i want to move in the direction of maximum increase of F.

    I know that the level curves are ellipses, and was given a hint to solve it by equating v'(u) = gradient vector of F.. where v(u)=(x(u),y(u)) is the vector representing the path.

    however im stuck at this point, i wonder if anyone can help me? will very much appreciate it!
    Last edited by navelwort; March 24th 2010 at 05:33 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8
    The rate of change of a scalar function f in the direction of a unit vector \mathbf{u} is given by

    \nabla f \cdot \mathbf{u}

    This is the directional derivative

    So, the maximum rate of change is given by

    |\nabla f \cdot \mathbf{u}| = |\nabla f|
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    7
    Quote Originally Posted by Haven View Post
    The rate of change of a scalar function f in the direction of a unit vector \mathbf{u} is given by

    \nabla f \cdot \mathbf{u}

    This is the directional derivative

    So, the maximum rate of change is given by

    |\nabla f \cdot \mathbf{u}| = |\nabla f|
    Hi, thanks for your reply,

    I do know that the maximum change occurs when direction vector u = \nabla f = <-4x, -2y> itself, but i wonder how does it help me to obtain the equation of the path(in terms of x and y) i am suppose to take?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    7
    anyone has any idea?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    Remember, that a line can be described by a direction vector (with components a, b, and c) and a point:

    r(t)=<x_0, y_0, z_0> + t<a, b, c>;

    You have a direction vector (it happens to be the gradient), and you have a point (3, -8).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    7
    Quote Originally Posted by ANDS! View Post
    Remember, that a line can be described by a direction vector (with components a, b, and c) and a point:

    r(t)=<x_0, y_0, z_0> + t<a, b, c>;

    You have a direction vector (it happens to be the gradient), and you have a point (3, -8).
    i agree this makes sense if i have a constant direction vector, and that the path is a straight line, however i believe the path is not a straight line because the path is a curve that is always tangent to the level curves, which are ellipses.

    thus i do not think i am able to use this

    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,405
    Thanks
    1328
    Quote Originally Posted by navelwort View Post
    i agree this makes sense if i have a constant direction vector, and that the path is a straight line, however i believe the path is not a straight line because the path is a curve that is always tangent to the level curves, which are ellipses.

    thus i do not think i am able to use this

    No, the path of fastest increase is perpendicular to the level curves.

    In any case, \nabla f= -4x\vec{i}- 2y\vec{j} so you must have \frac{dx}{dt}= -4x, x(0)= 3 and \frac{dy}{dt}= -2y, y(0)= -8. Solve those equations to get x and y in terms of the parameter t. You can then, if you wish, eliminate t to get y as a function of x.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2010
    Posts
    7
    Quote Originally Posted by HallsofIvy View Post
    No, the path of fastest increase is perpendicular to the level curves.

    In any case, \nabla f= -4x\vec{i}- 2y\vec{j} so you must have \frac{dx}{dt}= -4x, x(0)= 3 and \frac{dy}{dt}= -2y, y(0)= -8. Solve those equations to get x and y in terms of the parameter t. You can then, if you wish, eliminate t to get y as a function of x.
    opps haha i meant perpendicular, yeah thanks! but just wondering, how do i go about solving those equations?

    am i correct to do it like this?

    x(t) = -4xt + C, y(t) = -2yt + D

    x(0) = 3, => C = 3,
    y(0) = -8, => D = -8,

    thus x = -4xt + 3, y = -2yt -8

    solving t, (x-3)/(-4x) = (y+8)/(-2y)

    thus i have -2yx + 6y = -4xy - 32x
    => 3y = xy -16x ?


    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Gradient Vector max/min
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 2nd 2009, 03:54 AM
  2. Gradient vector help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 6th 2009, 04:04 PM
  3. Gradient Vector
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 05:36 PM
  4. need help on path and vector field problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 19th 2008, 08:11 AM
  5. integrating a vector function along a path
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2008, 09:04 PM

Search Tags


/mathhelpforum @mathhelpforum