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Homework Help: Is this f(x,y) differential at (1,2)?

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f(x,y)=x3 + x2y + y3. Show that f is differentiable at (1,2).

    2. Relevant equations

    A function f(x,y) is differentiable at a point (x0,y0) if the partial derivatives of f(x,y) are defined at (x0,y0) and if the following limiting condition holds:

    lim(x,y)-->(x0,y0) [ f(x,y) - f(x0,y0) - ∂f/∂x|(x0,y0)(x-x0) - ∂f/∂y|(x0,y0)(y-y0) ] / √( (x-x0)2 + (y-y0)2) = 0.

    3. The attempt at a solution

    ∂f/∂x = 3x2 + 2xy ===> ∂f/∂x|(1,2) defined

    ∂f/∂y = 2x2 + 3y2 ===> ∂f/∂y|(1,2) defined

    But I'm having trouble with making lim(x,y)-->(1,2) = 0.

    I got it all the way to

    lim(x,y)-->(1,2) = (x3+x2y+y3+22 -7x-13y)/√((x-1)2+(x-2)2).

    I need to get rid of the denominator. I thought about rationalizing the denominator, but I don't that'll get me anywhere.

    suicide-bear-didnt-even-leave-a-not.jpg

    Suggestions welcome.
     
  2. jcsd
  3. Apr 12, 2010 #2
    Last edited by a moderator: Apr 25, 2017
  4. Apr 13, 2010 #3

    Dick

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    You know in your gut that it's true that it's differentiable, right? So you know that you just somehow MUST be able to factor the numerator as (x-1)*g(x,y)+(y-2)*h(x,y) for some polynomials g(x,y) and h(x,y), right? Divide the numerator by x-1. Take the remainder from that and divide by y-2. Nice decapitation photo. I sympathize.
     
  5. Apr 13, 2010 #4

    Hmmm....

    So I have (x-1)P1(x,y) + (y-2)P2(x,y) = x3 + x2y + y3 + 22 - 7x - 13y.

    But I don't understand your method of solving this. (I'm not looking for a straight-up answer, just a little more detailed explanation on how to simplify this)

    Also, maybe some help with part (b).

    Part (b) reads:

    Check if f(x,y) = (xy2-y3)/(x2+y2) is differentiable at (0,0) using polar coordinates.


    My attempt:

    Obviously, since the numerator is homogeneous of degree 3, and the denominator of degree 2, we will have limr-->0 r * h(ø) = 0, because h(ø) is just kind of there but always -1 ≤ h(ø) ≤ 1 while r sends it to zero.

    Here's the thing, though. I also need to show that "the partial derivatives of f(x,y) are defined at (x0,y0)" (that from part (a)). In this case, of course, I'm using f(rcosø,rsinø) instead of f(x,y). But in dealing with the limit in polar coordinates, I'm only supposed to worry about r. How can their be partial derivatives?
     
  6. Apr 13, 2010 #5

    HallsofIvy

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    Every polynomial in x and y is differentiable for all (x, y).
     
  7. Apr 13, 2010 #6

    Dick

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    Actually my previous suggestion wasn't good enough. Try and write your numerator in the form (x-a)*(x-1)^2+(y-b)*(y-2)^2+(x-c)*(x-1)*(y-2). Expand that and find the constants a, b and c that work. Now it's kind of like your polar coordinate example if you think of |x-1|~|y-2|~r. The denominator is degree r, the numerator is degree r^2.
     
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