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Problem on differentiability

  1. Oct 14, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##f:\mathbb ℝ:→ℝ^2## be a function defined as:
    ##f(x,y)=\frac{x^2y-2xy+y} {(x-1)^2+y^2} \forall (x,y)≠(1,0)## and ##f(1,0)=0##.
    Prove that for any curve ##α:(-ε,ε)→ℝ^2## of class ##C^1## (where ##ε>0##) such that ##α (0)=(1,0)## and ##α(t)≠(1,0)## for every ##t≠0##, the derivative ##(foα)'(0)## exists; but that ##f## is not differentiable at the point ##(1,0)##.

    The attempt at a solution:

    I didn't have any problems to prove that f isn't differenitable at ##(1,0)##. First, I proved that the partial derivatives exist. Both of them gave 0, then supposed ##f## was differentiable at that point. If this was the case, the limit ##\lim_{(x,y)\rightarrow (1,0)} \frac{ |f(x,y)-(f(1,0)+<∇f(1,0),(x-1,y)>|} {||(x-1,y)||}## would have to equal 0. If I consider the curve ##ψ(t)=(t+1,t)##, then the limit is ##\frac{\sqrt(2)} {4}≠0##.

    I got stuck in proving the first part of the statement "for any curve ##α:(-ε,ε)→ℝ^2##..., the derivative ##(foα)'(0)## exists". I mean, I can't show this using the limit definition: ##\lim_{h\rightarrow 0} \frac {foα(h)-foα(0)} {h}##, because I don't have enough information to calculate it. Am I doing something wrong? What should I do to prove that part of the statement?
     
    Last edited: Oct 14, 2013
  2. jcsd
  3. Oct 14, 2013 #2
    You know what f(x,y) is and you have a(t) = (x(t), y(t)). So you are looking at f(x(t),y(t)). Differentiate with respect to t, being careful about the chain rule. You will get some expression involving derivatives coming from f which you can see are okay, and a'(t) which by definition was ##C^1##.
     
  4. Oct 14, 2013 #3
    ##{\frac{\partial f}{\partial y}}(1,0)## and ##{\frac{\partial f}{\partial x}}(1,0)## exist and are both ##0##. So, what you've suggested is to apply the chain rule. By the chain rule ##(foα)'(0)=<∇f(α(0)),α'(0)>=<(0,0),(x'(t),y'(t)>=0## Is this ok? I thought that in order to apply the chain rule I needed both ##f(x,y)## and ##α(t)## to be differentiable, that's why I didn't try to solve it this way earlier. Instead, I've tried to proved differentiability by using the limit definition.
     
  5. Oct 14, 2013 #4

    Dick

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    Science Advisor
    Homework Helper

    f isn't differentiable in the two dimensional sense of the word at (1,0). But it does have well defined directional derivatives. Another way to think about this is that if you take a C^1 curve then the derivative along the parameter t is the same as the derivative along a tangent line to the curve at (1,0). Just check all tangent lines through (1,0) have a well defined derivative at (1,0).
     
  6. Oct 15, 2013 #5
    Re the chain rule, it applies fine to partial derivatives and is used that way all the time.

    Your computation looks plausible but I think it is not quite correct. You have shown that the expression ##∇f(a(0)) \cdot a'(0)## exists and is 0. This is not the same as showing the ## \lim \frac {∇f(a(t)) \cdot a'(t) - ∇f(a(0)) \cdot a'(0)}{t}## exists. There might be some function f where ##∇f(a(0)) \cdot a'(0)## = 0, and yet the limit itself might not exist.

    Unless you are prepared to show that scenario cannot happen -- and I suspect that it can -- you need to look specifically at the f you were given, and work out the f'(a(t)) based on that function. Then take the limit as t ##\rightarrow ## 0.

    Your expression ##\psi(t) ## is a perfect example of the kind of a(t) you are looking at. In this case you can see that f(##\psi(t) ##) has a derivative as a function of t at 0. The purpose of the a(t) is to analyze the situation in general.

    I think what this problem is getting at is that having partial derivatives and being differentiable are not the same thing in any dimension > 1, and you seem to have grasped that at least in terms of formally writing down your definition of differentiable.

    What differentiable at a point W = (##x_0, y_0## ) really means is that in some small neighborhood of W f is nearly linear. "Nearly" is what your definition of differentiable nails down.
     
    Last edited: Oct 15, 2013
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