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Vector Calculus - gradient geometry

  1. Apr 21, 2013 #1
    Hello. I can't seem to wrap my head around the geometry of the gradient vector in ℝ3

    So for F=f(x(t),y(t)), [tex]\frac{dF}{dt}=\frac{dF}{dx}\frac{dx}{dt}+\frac{dF}{dy}\frac{dy}{dt}[/tex]
    This just boils down to
    [tex]\frac{dF}{dt}=∇F \cdot v[/tex]

    Along a level set, the dot product of the gradient vector and velocity vector equals zero. But going uphill/downhill, the dot product no longer equals zero.

    To my understanding, the velocity vector lies on the tangent plane, which is perpendicular to the gradient vector. By this logic, the dot product should always equal zero.

    What am I missing here? What is the geometry of the gradient vector and velocity vector in relation to one another?

    Edit: Is it possible that I'm just confusing the gradient vector with the normal vector? In that case, the dot product of the normal vector and velocity vector will always equal zero, right?
     
    Last edited: Apr 21, 2013
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  3. Apr 21, 2013 #2

    dx

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    The gradient vector and the velocity vector have nothing to do with each other. The velocity can be in any direction, and so can the gradient.

    The velocity is tangent to the curve t → (x(t), y(t)), not to the level surface of F
     
  4. Apr 22, 2013 #3
    Thank you. So just to clarify, the normal and velocity vectors are always perpendicular, but the gradient and normal vectors are not necessarily the same?
     
  5. Apr 22, 2013 #4

    dx

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    Normal to what? Normal to the level sets? The velocity can be in any direction, so it does not have to be perpendicular to anything.

    The normal to the level sets is in the same direction as the gradient, but does not have the same magnitude. The normal is defined to have magnitude 1.
     
  6. Apr 22, 2013 #5
    Oops, I meant the normal to the tangent plane. Sorry about that.
     
  7. Apr 22, 2013 #6

    dx

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    Say you have a scalar field F, and a particle moving through space.

    Now how the particle moves does not depend at all on what F is. It can move any way it wants to. So the velocity of the particle is not related to what F is, or what the gradient of F is. Depending on what F is, the gradient of F can be perpendicular to the velocity, or it can be in the same direction as the velocity, or any other direction.

    The normal vector is the unit vector perpendicular to the tangent plane (I assume you're talking about the plane tangent to the level sets), which again depends on what F is. It too can be in any direction, unrelated to the velocity. Without changing how the particle moves, you can change F to make the normal vector point in any direction you want.
     
    Last edited: Apr 22, 2013
  8. Apr 22, 2013 #7
    Would all of this be the same if the particle is moving along a parametric surface in ℝ3 defined by [tex]r(x,y)= <x,y,f(x,y)>[/tex]

    I thought that [tex]n=\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}[/tex]

    And the partial derivatives represent the x- and y-components of the velocity vector. Would this mean that the velocity vector lies on the tangent plane and is therefore perpendicular to the normal vector to the plane?

    Regarding the gradient vector for a parametric surface, is it's geometry still arbitrary for a parametric surface?

    (I apologize if I was using the incorrect terminology before, and if I did not give proper context to my question)
     
  9. Apr 22, 2013 #8

    dx

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    Ah I see, so you were talking about a particle constrained to move on a particular surface, and by 'tangent plane' you meant a plane tangent to the constraint surface.

    In that case, yes the normal to the tangent plane is always perpendicular to the velocity.

    The gradient however is still arbitrary. It does not have any special relationship with the velocity.
     
  10. Apr 22, 2013 #9
    Thank you very much! Those were the answers I needed, sorry it took so long for me to correctly phrase my initial question.
     
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