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Homework Help: Total differential question

  1. Oct 22, 2005 #1
    Hi,

    I have this excercise to do:

    Let's have function f defined as follows:

    [tex]
    f(x,y) = \frac{xy(x^2 - y^2)}{x^2 + y^2},\ \ [x,y] \neq [0,0]
    [/tex]

    [tex]
    f(x,y) = 0,\ \ [x,y] = [0,0]
    [/tex]

    Find out, whether this function has total differential in [0,0].

    Well, first I observed that partial derivatives aren't continuous in [0,0], which is what I expected.

    So I computed them from the definition:

    [tex]
    \frac{\partial f}{\partial x}(0,0) = \lim_{t \rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = 0
    [/tex]

    [tex]
    \frac{\partial f}{\partial y}(0,0) = \lim_{t \rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = 0
    [/tex]

    So, if the total differential exists in [0,0], it must be zero linear transform.

    Now I have to checkout, if such a transform satisfies the limit and thus it is total differential:

    [tex]
    \lim_{||h|| \rightarrow 0} \frac{f((0,0) + h) - f(0,0) - L(h)}{||h||} = 0 ?
    [/tex]

    If I rewrite it in a slightly different way ( I put [itex]h = (h_1,h_2)[/itex] and [itex]h_2 = kh_1[/itex], I get

    [tex]
    \lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1
    [/tex]

    Well, I don't know whether this goes to zero so I tried another way, ie. approaching [0,0] on the x-axis and thus putting [itex]h_2 = 0[/itex]

    The limit I got was zero, so I can't say at this moment that the total differential doesn't exist in [0,0]. Anyway, according to right results our professor gave us, total differential doesn't exist in [0,0].

    How could I prove that?

    Thank you very much.
     
  2. jcsd
  3. Oct 22, 2005 #2

    Tide

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    Homework Helper

    Did your prof give any rationale for his or her assertion?
     
  4. Oct 22, 2005 #3

    HallsofIvy

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    I don't know why you say you don't know whether
    [tex]\lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1[/tex]
    goes to 0- the part involving k is finite for all k while the limit
    [tex]\lim_{h_1 \rightarrow 0} h_1[/tex]
    is clearly 0.

    Another way to handle a problem like this is to put it into polar coordinates. In polar coordinates, this reduces to
    [tex]f(r,\theta)= r^2 sin(\theta)cos(\theta)cos(2\theta)[/tex]
    which clearly is differentiable at the origin.
     
  5. Oct 22, 2005 #4

    Tide

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    Homework Helper

    Halls,

    That's what I did and the trig terms further simplify to [itex]\sin 4 \theta[/itex] (within a factor of 2) which makes the differentiation even easier.
     
  6. Oct 23, 2005 #5
    Well, I don't know polar coordinates. Maybe we sometimes touched them but didn't have them in syllabus.

    So I'm trying to do it in a non-elegant way :)

    So I got to the point where I need to prove the zero limit involving the total differential.

    That is, I need to prove that the following goes to zero as [itex]h = [h_1,h_2] \rightarrow [0,0][/itex]

    [tex]
    \left|\frac{h_1h_2(h_1^2 - h_2^2)}{(h_1^2 + h_2^2)\sqrt{h_1^2 + h_2^2}}\right|
    [/tex]

    Could you please give me some hint how to prove that? It would be sufficient to find there some part that would be bounded for all [itex]h_1,h_2[/itex] going to 0 and multiplying it with some zero part would give me the zero as the result.
     
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