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Gradient exists but directional derivative does not?

  1. Oct 30, 2009 #1
    First of all, I don't have a concrete example for this, but I hope it's not too hard to understand what I'm trying to get at.

    For a multivariable function of, say, 2 variables x and y, the gradient at a point only depends on the existence of partial x and partial y, right? In other words, if these exist, then as far as I know the gradient exists. One of the definitions of the directional derivative gives it as grad(f) dot product with u, where u is the unit vector in the direction where you want to find the derivative of the function. But what if the function is not continuous in this direction at that point and therefore cannot have a derivative? Using the above definition of the directional derivative would be meaningless, I assume. But in most of the places I've seen that give this formula for the directional derivative, they don't mention that the function has to be continuous/differentiable in this direction for the derivative to exist. Maybe it's so obvious they don't feel like it needs to be mentioned?

    Basically, what I'm getting at is, how can you know if a directional derivative exists in every direction at a certain point in the domain of a function? Is it simply a matter of the function being continuous in every direction from that point, and if so how can you be certain that a given function is continuous in every direction?

    I guess one way to check for the existence of a directional derivative is to use the limit definition, right? I'm refering to the following:

    lim (h->0) of [f(x + h*cos(theta), y + h*sin(theta)) - f(x, y)] / h. And if this limit exists, then the directional derivative exists, correct?

    Is there any other (perhaps easier) method you can use to determine that a directional derivative exists in every direction at a particular point? I'm thinking it may be possible just by examining the equation of the function you're given but I'm not 100% certain.
     
  2. jcsd
  3. Oct 30, 2009 #2
  4. Oct 31, 2009 #3
    Sorry, I'll try to be more clear. What I mean is that for the gradient to exist, the function only has to have partial derivatives for x and y (for a function of two variables). This means that it has to at least be continuous in the x and y directions at any point where the gradient exists. But for the directional derivative to exist in every direction the function has to be continuous in all directions, not just x and y, correct? So the fact that the gradient exists at a particular point (or all points) doesn't tell you that a directional derivative exists in every direction, right?
    If that's true, then you can't just blindly use the formula grad(f) dot product with u to find a directional derivative in some random direction where you don't know if the function is continuous.

    I'm sorry, I really can't think of any example off the top of my head, but imagine a point in the domain of a function where the function is continuous in the x and y directions but not in some other direction (for example corresponding to an angle of 45 degrees from the x axis).
     
  5. Oct 31, 2009 #4

    Office_Shredder

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    So for example the function

    f(x,y)=1 if x or y is zero, and is insanity otherwise (this is a technical term of course).

    At (0,0) the partial derivatives with respect to x and y exist, but the directional derivative does not for any other direction
     
  6. Oct 31, 2009 #5

    LCKurtz

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    A sufficient condition for the directional derivative to exist is differentiability, which is stronger than just fx and fy existing. That is the condition stated in most calculus texts I have seen. But differentiability is not necessary as shown in this link:

    http://www.math.tamu.edu/~tom.vogel/gallery/node17.html
     
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