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Directional derivatives

  1. Feb 24, 2014 #1
    Say you are given the equation of a surface f(x,y) and a point (x,y,z) on the surface.

    How would one find the gradient vector in which the directional derivative Duf is equal to zero.
  2. jcsd
  3. Feb 26, 2014 #2
    Given the equation z = f(x,y) and the point (x0,y0,z0) you want to find the direction along which the directional derivative is zero.
    The directional derivative as a function of direction (the latter given by a unity vector n, with components n_x and n_y) can be written as
    $$\frac{\partial f}{\partial n} = \frac{\partial f}{\partial x} n_x + \frac{\partial f}{\partial y} n_y$$
    You can express n as a function of the angle with the coordinate axis, at which point you can equate the expression above to zero, and try to solve it for the angle of n.
  4. Mar 3, 2014 #3
    That's exactly correct. The directional derivative of some function f(x,y,z) at a point x0, y0, z0 in the r direction is just the gradient of f at that point dotted with the unit vector in the r direction. Ie, df/dr = ∇f(x0, y0, z0) [itex]\bullet[/itex] r. You'll want to solve for where df/dr = 0, which might be a little tricky since r has three components. Good luck!
  5. Mar 5, 2014 #4


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    The gradient vector and the direction in which ##D_u(f)=0## are two different things. Which do you want? You have ##\nabla f =\langle f_x, f_y\rangle## and a unit vector ##\hat u =
    \langle a,b\rangle##. Choose ##\vec u## such that ##\nabla f \cdot \hat u## is zero.
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