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Directional derivatives and the gradient vector

  1. Oct 15, 2014 #1
    If the unit vector u makes an angle theta with the positive x axis then we can write u = <cos theta, sin theta>

    Duf(x, y) = fx(x,y) cos theta + fy(x,y) sin theta

    What if I am dealing with a function with three variables (x, y, z)?

    How can I find the directional derivative if I have been given an angle?
     
  2. jcsd
  3. Oct 15, 2014 #2

    Matterwave

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    In three dimensions, a unit vector will be distinguished by 2 angles. It could be a polar angle ##\theta## and an azimuthal angle ##\phi## or some other two numbers to define a direction (sometimes the designation is flipped). In spherical polar coordinates a unit vector will be ##\hat{u}=(\sin\phi\cos\theta,\sin\phi\cos\phi,\sin\phi)##
     
  4. Oct 15, 2014 #3
    How should I approach this question?

    Find the directional derivative of f(x,y,z)=xz+y2 at the point (3,1,2) in the direction of a vector making an angle of −π/4 with ∇f(3,1,2).

    I found the gradient to be <2,2,3>
     
  5. Oct 15, 2014 #4

    Matterwave

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    I don't understand what it means: "a vector making an angle of −π/4 with ∇f(3,1,2)."? In 3-D space, there are an infinity of vectors making a specific angle with respect to another vector...they form a cone.
     
  6. Oct 15, 2014 #5
    Is there a way to obtain a vector (out of many)?
     
  7. Oct 16, 2014 #6
    You do not have to think about angles to understand the idea of a directional derivative. Let x and v be in Rn, v be a unit vector and t in R . The directional derivative of f at a in the direction v is just the derivative of the single variable function h(t) = f(a + tv) at t=0 (that is h'(0) ).

    This gives the definition of the directional derivative without discussing the gradient. And in fact some directional derivatives can exist without f having a defined gradient. Say f is differentible with respect to x but not with respect to y. However if f is differentible then the direction derivative of f at a in the direction v is grad(f)(a)*v.
     
  8. Oct 16, 2014 #7

    HallsofIvy

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    Given function f(x, y, z), with gradient [itex]\nabla f[/itex], we can talk about the "directional derivative" in the direction of unit vector [itex]\vec{v}[/itex] as the dot product: [itex]\nabla f\cdot \vec{v}[/itex].

    In three dimensions, a direction cannot be specified by a single angle. We need two angles such as the "[itex]\theta[/itex]" and "[itex]\phi[/itex]" used in spherical coordinates. Or we can use the "direction cosines", the cosines of the angles the direction makes with the three coordinate axes: [itex]\theta_x[/itex] is the angle a line in the given direction makes with the x-axis, [itex]\theta_y[/itex] the angle it makes with the y-axis, and [itex]\theta_z[/itex], the angle it makes with the z-axis. Of course, those three angles are not idependent. We can show that we must have [itex]cos^2(\theta_x)+ cos^2(\theta_y)+ cos^2(\theta_z)= 1[/itex] which means that the vector [itex] cos(\theta_x)\vec{i}+ cos(\theta_y)\vec{j}+ cos(\theta_z)\vec{k}[/itex] is the unit vector in that direction.

    That is, the "directional derivative" of f(x, y, z) in the direction that makes angles [itex]\theta_x[/itex], [itex]\theta_y[/itex], and [itex]\theta_z[/itex] with the x, y, and z axes, respectively, is given by [itex]\nabla f\cdot (cos(\theta_x)\vec{i}+ cos(\theta_y)\vec{j}+ cos(\theta_z)\vec{k})= f_x cos(\theta_x)+ f_y cos(\theta_y)+ f_z cos(\theta_z)[/itex].
     
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