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Homework Help: Calculus III - Multivariate Continuity

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Let
    [itex]\begin{equation*}
    f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases}
    \end{equation*}[/itex]

    Is [itex]f[/itex] continuous at the point [itex](0,0)[/itex]? Are [itex]f_x[/itex] og [itex]f_y[/itex] continuous at the point [itex](0,0)[/itex]?
    2. Relevant equations

    Polar coords

    3. The attempt at a solution

    If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on [itex]\theta[/itex]. I'm just wondering if the derivative of the polar function is dependent on [itex]\theta[/itex] then it isn't continuous and therefore neither are [itex]f_x[/itex] and [itex]f_y[/itex]

    Steps:

    [itex]\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0[/itex]

    Define [itex]g(r) = r (cos^3 \theta - sin^3 \theta)[/itex] then [itex]g'(r) = cos^3 \theta - sin^3 \theta[/itex] and [itex]\displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta[/itex] doesn't exist.

    Cheers.
     
  2. jcsd
  3. Jan 22, 2012 #2

    SammyS

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    How about ∂f/∂x and ∂f/∂y? Are the continuous at (0, 0) ?
     
  4. Jan 22, 2012 #3
    They aren't. I'm asking whether it is a coincident that the derivative of g and the partials of f aren't continuous.
     
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