# Directional Derivative and unit vectors

1. Jul 23, 2006

### Gott_ist_tot

What happens if a unit vector is not used in calculating the directional derivative. From when I worked it out the directional derivative is multiplied by a scalar if a unit vector is not used. So I gather that the directional derivative must be calculated by a unit vector. Because it is still calculating a rate of change and a rate of change multiplied by a number is not the same answer. I was just curious if my answer was on the right track or way off. Thanks.

2. Jul 23, 2006

### quasar987

Here's a more useful interpretation IMO.

In the same way as $f(x+a) \approx f(x) + af'(x)$ (sometimes known as Euler's approximation) in single variable calculus,

$$f(\vec{r}+\vec{A}) \approx f(\vec{r}) + \nabla f (\vec{r}) \cdot \vec{A}$$

gives a first order approximation of the value of f(r+A).

Last edited: Jul 23, 2006
3. Jul 23, 2006

### Data

A directional derivative is just the projection of a function's gradient along some direction. To get the projection of a vector in a particular direction you take the dot product of the vector with a unit vector in the direction you're looking at.

If you don't use a unit vector, then your directional derivative will be multiplied by the length of the vector that you do use.

quasar, I think you need to fix a few errors in your post :yuck:

$$f(x+a) \approx f(x) + af^\prime (x)$$

$$f(\vec{r}+\vec{A}) \approx f(\vec{r})+\vec{A} \cdot \nabla f(\vec{r})$$

(note, this is basically just taking a Taylor expansion up to first-order)

Last edited: Jul 23, 2006
4. Jul 24, 2006

### HallsofIvy

Staff Emeritus
So, in other words, if your original vector wasn't a unit vector, just divide your answer by its length: