1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    none

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.

    RGV
     
  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

    User Avatar
    Science Advisor
    Homework Helper

    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).

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook