Directional Derivative Equal to Zero

1. Mar 15, 2013

Parmenides

The problem states:

"In what direction is the directional derivative of $$f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$$ at (1,1) equal to zero?"

I know that $D_uf = \nabla{f}\cdot{{\bf{u}}}$. I believe the problem simply is asking for me to determine what vector ${\bf{u}}$ will yield zero. Thus:
$$\nabla{f} = \left\langle\frac{4xy^2}{{(x^2 + y^2)}^2},\frac{-4{x^2}y}{{(x^2 + y^2)}^2}\right\rangle$$
At the point (1,1), we get $\nabla{f} = \left\langle{1},{-1}\right\rangle$. From here, I think that a vector taken with the dot product of $\left\langle{1},{-1}\right\rangle$ to give zero would be $\left\langle{1},{1}\right\rangle$. I'm not sure if it's that simple though because I could also say that $\left\langle{0},{0}\right\rangle$ gives zero, too. Perhaps my justification is flawed? Much appreciated.

2. Mar 16, 2013

LCKurtz

Assuming your derivative is right it looks good except why would you say <0,0> gives zero? Also besides <1,1>, what about its opposite?

3. Mar 16, 2013

Parmenides

True. I guess I was thinking that the problem was seeking a single answer. As for the former, $\left\langle{0},{0}\right\rangle \cdot \left\langle{1},{-1}\right\rangle = 0$, but I suppose that's not really even a "direction" at all... ><

4. Mar 16, 2013

LCKurtz

While it's true that $\left\langle{0},{0}\right\rangle \cdot \left\langle{1},{-1}\right\rangle = 0$, what point $(x,y)$ have you found where $\nabla f(x,y) = \langle 0,0\rangle$?