1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    \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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Directional derivatives
  1. Derivative of (Replies: 5)

  2. Directional derivative (Replies: 1)

  3. Direct Current (Replies: 10)

  4. Vector direction? (Replies: 7)