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Directional derivatives

  1. Jun 20, 2005 #1
    Let [itex]f(x,y) = \begin{array}{cc}
    \frac{xy}{\sqrt{x^2 + y^2}} &, (x,y) \neq(0,0) \\
    0 & ,(x,y) = (0,0) \\

    Show that the directional derivatives at (0,0) in directions [itex]a\mathbf{i} + b\mathbf{j}[/itex] with [itex]a\neq 0[/itex] and [itex]b\neq 0[/itex], do not exist.

    Let [itex]\mathbf{u} = a\mathbf{i} + b\mathbf{j}[/itex]

    D_{\mathbf{u}}f(0,0) & = \lim_{h\rightarrow 0}\frac{f(0+ ha,0 + hb) - f(0,0)}{h}
    & = \lim_{h \rightarrow 0} \frac{\frac{h^2ab}{\sqrt{h^2a^2 + h^2b^2}}}{h}
    = \frac{ab}{\sqrt{a^2 + b^2}}

    which if I'm not mistaken, exists. How do I show that this doesn't exist? Also, in order to show that f(x,y) is everywhere continuous, will it suffice to say that xy/sqrt(x^2 + y^2) is continuous when (x,y) != 0 and that the limit of xy/sqrt(x^2 + y^2) as (x,y) tends to (0,0) along the x-axis is 0?
    Last edited: Jun 20, 2005
  2. jcsd
  3. Jun 21, 2005 #2
    Anyone have any ideas? I'm really confused about this.
  4. Jun 21, 2005 #3


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    I'm not sure about this, but maybe they want you to take the directional derivative the usual way (dot the gradient into the direction vector), and then take the limit as (x,y)->0. I tried this, and it seemed like I got two different answers depending on how I approached 0. I think this would mean the derivative doesn't exist, even though both answers were finite. If you want to try this, I used polar coordinates and I think it made it easier.
  5. Jun 21, 2005 #4


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    \lim_{h \rightarrow 0} \frac{\frac{h^2ab}{\sqrt{h^2a^2 + h^2b^2}}}{h}
    = \frac{ab}{\sqrt{a^2 + b^2}}

    You sure about that?
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