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Why when we compute the directional derivative we have to use the unit vector $\vec{\bf{u}}$? I know that using $2\vec{\bf{u}}$ would change the directional derivative as the second point is further apart. But why not use $\frac{\vec{\bf{u}}}{2}$ then? Wouldn't that gradient be "more precise"?

NAND
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Sorfosh
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    For the directional derivative in a coordinate direction to agree with the partial derivative you must use a unit vector. If you don't use a unit vector the derivative is scaled by the magnitude of the vector. – Seth Oct 19 '15 at 01:00
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    That is a way to calculate directional derivatives when the gradient exists, but directional derivatives can be defined without this. In those definitions, there’s no need to use a unit vector. – amd Oct 19 '15 at 01:30
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    Yup got it, thank you guys so much ^^. A bit of stimuli helped me out a lot – Sorfosh Oct 19 '15 at 01:56
  • @JohnMa Agreed. The OP found that there's a definition that is restricted to unit vectors, which seems to be what his teacher is using, as well as the more general one that you were talking about, so it appears that he's resolved his own question. – amd Oct 19 '15 at 03:51
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    @Sorfosh Now that you've figured it out, you should post an answer to your own question. – amd Oct 19 '15 at 03:53
  • As it seemed to me that you reached a conclusion I purged the long comment exchange a bit as a system flag summoned me here. If you absolutely need something undeleted @-ping me. – Jyrki Lahtonen Oct 19 '15 at 05:31
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    Does this answer your question? why normalize and the definition of directional derivative – NAND Apr 20 '21 at 23:27

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At the class we were given the definition of a directional derivative when the direction vector is a unit vector. Because of that I got confused. The real definition can be found here: https://en.wikipedia.org/wiki/Directional_derivative (Variation using only direction of vector for euclidean space). Here we can see that using the unit vector just simplifies things and is not necessary. It is very similar to 2D derivative as the distance between the two point $f(x+hv)-f(x)$ is the length of vector $\vec{\bf{v}}$.

NAND
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Sorfosh
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