1. ## rate of change

What is the rate of change of the function f(x,y)= ln( square root( x^2 + y^2)) at (0, square root(e)) in direction towards the origin?

What is the rate of change of the function f(x,y)= ln( square root( x^2 + y^2)) at (0, square root(e)) in direction towards the origin?
Directional derivative:

$\displaystyle \nabla f \cdot \hat{l}$ where $\displaystyle \hat{l}$ is a unit vector in the direction from $\displaystyle (0, \sqrt{e})$ to (0, 0).

Evaluate this directional derivative at $\displaystyle (0, \sqrt{e})$.

3. The unit vector would be (0,1), but I have not been able to find the $\displaystyle \nabla f$
I thought that I would just plug in $\displaystyle (0, \sqrt{e})$ and (0,0) into the equation, but that doesn't work because you can't when I plug in (0,0) I get ln(0) which isn't possible.
Do you know what I am doing wrong?

The unit vector would be (0,1), Mr F says: Actually it's (0, -1).

but I have not been able to find the $\displaystyle \nabla f$
I thought that I would just plug in $\displaystyle (0, \sqrt{e})$ and (0,0) into the equation, but that doesn't work because you can't when I plug in (0,0) I get ln(0) which isn't possible.
Do you know what I am doing wrong?
$\displaystyle f = \ln \sqrt{x^2 + y^2} = \ln (x^2 + y^2)^{1/2} = \frac{1}{2} \ln (x^2 + y^2)$.

Therefore:

$\displaystyle \frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2}$.

$\displaystyle \frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2}$.

Therefore:

$\displaystyle \nabla f = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j = \frac{1}{x^2 + y^2} \, (x i + y j)$.

Evaluate at $\displaystyle (0, \sqrt{e})$.

5. So it would wind up being (0, yj/(x^2+y^2)) but what does the j represent? Would the j = sq root of e?

6. $\displaystyle j$ is the unit vector in the y direction.