# Directional derivative

1. Dec 10, 2008

### Master J

I have a function of 2 variables. I know it increase most rapidly in the direction of the gradient, but how about in wht direction is it not increasing?

I am thinking that the gradient (dot product)(direction in which it is not increasing) = 0

Any hints?

2. Dec 10, 2008

### CompuChip

Well, since you mentioned the word "directional derivative" anyway: you could check for which $\vec v$
$$(\vec\nabla f(x, y)) \cdot \vec v < 0$$
?

3. Dec 10, 2008

### HallsofIvy

Staff Emeritus
Yes, it is true that $\vector{\nabla f}\cdot \vector v$ is the directional derivative in the directional derivative in the direction of $\vec{v}$ (for $\vec{v}$ of length 1). And that tells you the derivative is 0 perpendicular to the gradient.

(CompuChip, surely you didn't mean "<"?)

4. Dec 11, 2008

### CompuChip

Err, no comment? :)

5. Dec 11, 2008

### Master J

That is what is was thinking, since of course cos(pi/2) = 0. So the vector that is at a right angle to the gradient is in the direction of zero increase. But how do I go about finding this vector?