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Tangential derivative

  1. Jul 9, 2009 #1
    Dear all,

    How could one find a tangential derivative of a function z=f(x,y) at a give point (x0,y0)? Is it always zero?

    I tried to go through in the following way. Well, if I need a tangential derivative f_t of a function f(x,y) I have to use a directional derivative in tangent direction, i.e. f_t=grad(f) . t, where t is the tangential vector. How one could find tangential vector t at a given point? I think, one has to find a tangential plane at a given point. The tangential plane at a given point (x0,y0) is fx(x-x0)+fy(y-y0)+f(x0,y0)=0. All the tangential vectors live on the tangential plane, i.e. they are perpendicular to the gradient of f(x,y) what means that f_t=grad(f) . t=0.

    Do I have a gap in my understanding.

    Thanks a lot in advance!
  2. jcsd
  3. Jul 9, 2009 #2
    no, not all of the lines in the tangent plane will be perpendicular to the gradient vector. (otherwise the tangent lines wouldn't form a tangent plane.) If all of the directional derivatives are zero, that would mean that the function is constant, but that is obviously not true for all functions. The vectors that are perpendicular to the gradient will be tangent to a level curve of the function.
  4. Jul 9, 2009 #3
    Could you please give me an example of a tangent vector which is not perpendicular to the gradient at a given point?
  5. Jul 9, 2009 #4


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    You may be talking about different things here. The gradient vector of what?

    The graph of the function z= f(x,y) can be thought of as a "level surface" for the function F(x,y,z)= z- f(x,y) since z= f(x,y) is the same as z- f(x,y)= 0. The gradient of F(x,y,z), [itex]<f_x, f_y, 1>[/itex], is perpendicular to that surface and so perpendicular to the tangent plane at that point. Therefore, that gradient is perpendicular to the tangent plane and every line in the plane is perpendicular to it. However, the two dimension gradient of f itself, [itex]<f_x, f_y>[/itex] is NOT perpendicular to any three diemensional surface.

    Since F(x,y,z)= z- f(x,y) is identically equal to 0, yes, the derivative of that function in any direction is 0.

    But the derivative of z= f(x,y), in a given direction, the rate at which we would move up or down moving along f(x,y) in that particular direction, is NOT always 0. It is [itex]\partial f/\partial{x}cos(\theta)+ \partial{f}/\partial{y}sin(\theta)[/itex] where [itex]\theta[/itex] is the angle the direction makes with the positive x-axis.
    Last edited: Jul 9, 2009
  6. Jul 9, 2009 #5
    Thanks a lot for the detailed answer! It seems I found the gap in my understanding.

    Well, a tangent vector to f(x,y) lives on the tangent plane z=fx(x-x0)+fy(y-y0)+f(x0,y0) and it is perpendicular to the gradient of F(x,y,z) but, of course, the gradient of F(x,y,z) is not the gradient of f(x,y). Well, I know a tangent vector to f(x,y) which is a 3D vector, but to compute the tangential derivative of f(x,y) I need a 2D tangent vector, what should I do? Do I have to project a 3D vector to a 2D plane and take that projection?
  7. Jul 9, 2009 #6
    Thanks a lot to everyone who helped! I think I fathomed it out!
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