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Homework Help: Gradient of a Simple Surface at a point (without knowledge of derivative)

  1. Aug 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The statement that the simple surface F has gradient at the ordered pair ((x,y),F(x,y)) of F means that there exists only one ordered number pair (p,q) such that if c is a positive number then there exists a rectangular segment , S, containing (x,y) such that, if (u,v) is in S and in the domain of F then
    F(u,v) - F(x,y) = (u-x)p + (v-y)q + ε|(u,v)-(x,y)|
    where ε is a number between -c and c

    2. Relevant equations


    3. The attempt at a solution

    I realize that this is not a 'problem' per se, however I am simply trying to make sense of the definition of gradient without resorting to derivatives. I would like to prove that if F has gradient at a point then it is continuous at the point, again without using any notion of derivative, just the definition provided. But I would like to attempt to prove the implication of continuity myself.

    I am simply looking for anyone to assist me to break down the definition of gradient given and understand it a little better. I have been staring at it for a few days and I am just not getting anywhere. I don't understand where the positive number c comes into play and therefore the ε as well.

    Any assistance/insight/guidance would be appreciated, thanks!
    Last edited: Aug 7, 2012
  2. jcsd
  3. Aug 7, 2012 #2

    Ray Vickson

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    Essentially, it claims that the pair (p,q) is the gradient. Note that if you take y=v and let u → x you get F(u,y)-F(x,y) = p(u-x) + ε|u-x|. Presumably, the 'c' referred to can be any chosen positive number (and after choosing c, we get to specify the 'rectangle'). If that is what the author meant, then, of course, p will just be the partial derivative ∂F(x,y)/∂x, because [tex] \frac{F(u,y)-F(x,y)}{u-x} = p + \epsilon \frac{|u-x|}{u-x} \rightarrow p [/tex] as [itex] u \rightarrow x.[/itex] This holds because for any positive number c we have
    [tex] \left|\frac{F(u,y)-F(x,y)}{u-x} -p\right| \leq |\epsilon| \leq c [/tex] for any u sufficiently near x.

  4. Aug 7, 2012 #3
    It basically means that for any desired degree of "accuracy", which is denoted by c, the difference from any point close enough to (x, y) can be approximated by a linear function of the displacement. Which is very similar to the differential concept, except that the differential deal with infinitesimal displacements.
  5. Aug 7, 2012 #4
    RGV - thanks for the quick reply. Your response is very helpful to me.

    Can you elaborate a bit on the rationale behind letting y=v as well as how we know that the following is true:

  6. Aug 7, 2012 #5

    Ray Vickson

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    I want to know if p is ∂F(x,y)/∂x, which is the x-derivative with y held fixed. So I don't want to change y to v ≠ y; I just want to change x by itself. Certainly the horizontal line segment from (x,y) to (u,y) is in the rectangle S, so I am allowed to write the first equation.

    The second equation is obtained from the first one by dividing through by u-x (assuming u ≠ x, which I neglected to say).

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