Directional Derivative and unit vectors

[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.

 

[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).

 

[Data]

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.

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

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)

 

[HallsofIvy]

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.

So, in other words, if your original vector wasn't a unit vector, just divide your answer by its length:
D_{\vec{v}}f= \frac{\vec{v}\cdot\del f}{|\vec{v}|}[/itex]<br /> <br /> Of course, since \frac{\vec{v}}{|\vec{v}|} <b>is</b> the unit vector in the direction of \vec{v}, that is exactly the same as reducing to a unit vector in the first place.

 

[quasar987]

Strangely, \del doesn't do anything in Latex afaik. \nabla is \nabla and \partial is \partial. I'm sure everyone knew, but here it is anyway,

D_{\vec{v}}f= \frac{\vec{v}\cdot\nabla f}{|\vec{v}|}