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Directional derivative and gradient concepts

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data

    A series of true/false questions. I guess I don't understand the concepts of this very well:

    1. If you know the directional derivative of f(x,y) in two different directions at a point P, we can find the derivative with respect to the x and y axes and thus we can determine the derivatives at this point P.

    2. If [tex]f(x,y) = x^{2} + y^{2}[/tex] then
    [tex]\nabla f[/tex][tex]\bot [/tex] [tex]graph(f)[/tex].

    For the level curves in the figure and point P:
    a) If u is a unit vector and the level curves of f(x,y) are given as shown, then at the point P, we have

    [tex]f_{u} = D_{u}f = \nabla f[/tex].

    b) For the same f and the unit vector v shown,
    [tex]f_{v} = D_{v}f =[/tex](I can't get this to come out right, but it's supposed to say that the magnitude of the gradient of f times cos theta)

    c) There is a function z = f(x,y) and a point P so that the maximum rate of change in f as you move away from P is 7 and the minimum rate of chang ein f as you move from P is -5.

    2. Relevant equations

    3. The attempt at a solution

    1. Not sure about this, but I want to guess that it has something to do with the chain rule for many variables?

    2. I want to say true, because I know that the gradient is always perpendicular to the level curve, but I'm not sure if that is what "graph(f)" refers to?

    a) Not a clue really here. I want to say false because I don't think that the directional derivative is equal to the gradient, unless they're parallel? In any case I don't think that they would be equal to the partial derivative...

    b) I kind of want to say false again because, because where did the unit vector go? The dot product of the gradient and the unit vector u should be equal to the directional derivative, but the u has disappeared, unless u is equal to 1? Which I don't think it is.

    c) So they're trying to say that the magnitude of the gradient is 7 and the negative of that is -5? I think that's wrong, because isn't the maximum and minimum the same except that the minimum is the negative of the magnitude? So it can't be 7 and -5?

    Most of my answers here are complete guesswork. Help would be appreciated.

    Attached Files:

    Last edited: May 3, 2010
  2. jcsd
  3. May 3, 2010 #2


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    Homework Helper

    do most of these assume a conitinuous differentiable function?

    start with the directional derivative using the dot product, assuming f is differentiable at the point, then think vectors & linear independence
    not sure what graph(f) is either....
    once again use the dot product form of the directional derivative? note that u looks perpindicular to the level curves at that point... (i'd make that assumption)
    doesn't make sense.... aren't we onto v & cos(theta)? anyway write out the directional derivative...
    your reasoning is good here,

    It helps me to think of the plane that is tangent to the function at a given point. The rate of maximum change on the plane is the direction of the gradient, and as you point out moving in the negative direction will have the same magnitude but a negtive rate of change (think of the dot product form of the directional derivative again)

    when you move perpindicular to the gradient you're moving horizontally along the plane & the rate of change is zero (think dot product again)
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