# Directional derivative Q, stuck.

1. Feb 22, 2015

### ilyas.h

1. The problem statement, all variables and given/known data

In what directions at the point (2, 0) does the function f(x, y) = xy have rate of change -1?

$D_{u}(f)(a,b) = \bigtriangledown f(a,b)\cdot (u_{1}, u_{2})$

$f(x,y) = xy$

$(a,b) = (2,0).$

3. The attempt at a solution
$\frac{\partial f}{\partial x} = y$

$\frac{\partial f}{\partial y} = x$

$\bigtriangledown f(2,0) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})= (y, x) = (0, 2)$

plugging in:

$D_{u}(xy)(2,0) = \bigtriangledown f(0,2)\cdot (u_{1},u_{2}) = -1$

$(0,2)\cdot (u_{1},u_{2}) = -1$

$u_{2} = -0.5$

$u_{1}$ has infinitely many values.

the last line above is the part im confused about. Are there infinitely many values for u1? thanks.

Last edited: Feb 22, 2015
2. Feb 22, 2015

### Ray Vickson

Do you want/need $\vec{u} = (u_1,u_2)$ to be a unit vector? If so, there are only two possible values of $u_1$; if not, there will be infinitely many values of $u_1$. As before, you need to use additional information if you have it.

3. Feb 22, 2015

### ilyas.h

in the equation:

$D_{u}(f)(a,b) = \bigtriangledown f(a,b)\cdot (u_{1}, u_{2})$

the (u1, u2) is a unit vector (according to my lecture notes). So you suggest that u1 has two possible values. How so? and I posted all the information there is in the Q in the thread, so no missing links. im quite confused,

edit: I think I know now, from definition of a unit vector: (u1)^2 + (u2)^2 = 1^2

plug in and you'll get u1 = +/- (root3)/2

Last edited: Feb 22, 2015