# Directional derivative of a surface

1. Sep 4, 2012

### Poley

1. The problem statement, all variables and given/known data
What is the directional derivative of the function z = x3 - y at the point (1, 2, -1) and in the direction of a vector (1,1,1)?

2. Relevant equations

3. The attempt at a solution
If f(x,y) = x3 - y, then ∇f = (3x2, -1) which equals (3, -1) at the given point. Now I understand I have to take the dot product of the gradient with the unit vector (1/√3, 1/√3, 1/√3) but I'm not quite sure how to...

Can a function like z = f(x,y) have a directional derivative in the direction of a three dimensional vector?

Thanks for any help!

2. Sep 5, 2012

### christoff

You are right, you cannot take the directional derivative of f(x,y) in the direction of a vector in three dimensions (unless the z component is zero, in which case you sort of can).

3. Sep 5, 2012

### HallsofIvy

The surface $z= x^3- y$ is given by $x\vec{i}+ y\vec{j}+ (x^3- y)\vec{k}$ the gradient of that vector gives the normal vector and you can take the dot product of that with a unit vector in the direction of $\vec{i}+ \vec{j}+ \vec{k}$- but I would NOT call that a "directional derivative".