- #1
Parmenides
- 37
- 0
The problem states:
"In what direction is the directional derivative of [tex]f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}[/tex] 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:
[tex]\nabla{f} = \left\langle\frac{4xy^2}{{(x^2 + y^2)}^2},\frac{-4{x^2}y}{{(x^2 + y^2)}^2}\right\rangle[/tex]
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.
"In what direction is the directional derivative of [tex]f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}[/tex] 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:
[tex]\nabla{f} = \left\langle\frac{4xy^2}{{(x^2 + y^2)}^2},\frac{-4{x^2}y}{{(x^2 + y^2)}^2}\right\rangle[/tex]
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.