Well, here we are. It's my turn.
I know this is a heavily disussed topic on this site.
Nevertheless, I can't stop myself asking this question. Indeed, my reasearches have left me mostly unsatisfied and I'm still scratching my head around this problem.

My question basically wants to focus on the reasons why we often conveniently choose to keep our directional derivatives vectors as they are. We don't need them to be scaled down.

I might go in the wrong direction :), but I want to offer what is my current thought on this topic.
I'm going to present a scenario (hopefully relevant) where I'll compare the interpretation of the normalized and not-normalized version of a direcitonal derivative.

Let's consider a position-altitude graph:

  • Here the directional derivative vector is normalized. The directional derivative along the vector $\displaystyle \vec{u}$ represents the slope of the graph in that direction.

    enter image description here

  • Here the directional derivative vector is not-normalized. We can see how for each point in the plane there is an associated vector $\displaystyle \vec{v}$ denoting the velocity of an object passing by that point and going in a specific direction. The directional derivative along the specific vector $\displaystyle \vec{v}$ represents the rate of change of the altitude considering how fast the object is moving.

enter image description here

Notice (read "I imagined") that we can consider the first case just a "special case" of the second. Indeed, that would be the case in which the velocity vector module $\displaystyle \vec{v}$ is always the same and equal to 1 (the object is moving with a constant velocity of 1 for all the time and in all directions).
As the object moves with constant velocity 1, it travels the distance of 1 meter in 1 second. So, the rate of change of the altitude with respect to time is the same of the rate of change of the altitude with respect to distance (the slope).

In the second case, the velocity is not constant neither equal to 1, but changes in different points and directions. Now, the rate of change of the altitude with respect to time is different from the rate of change of the altitude with respect to distance. The directional derivative (which is, then, usually considered as a rate of change with respect to time) takes account of this fact by scaling up/down its value according to the velocity of the object travelling in its direction.

I hope that these intuitions are relevant and not particularly misguided. Any other example is appreciated.

  • 3
    $\begingroup$ Possible duplicate of why normalize and the definition of directional derivative $\endgroup$ – Brian Mar 18 '19 at 18:27
  • $\begingroup$ @Brian Stated at the beginning that is an heavily discussed topic, I think my question is better exposed and useful for the community $\endgroup$ – Gabriele Scarlatti Mar 18 '19 at 18:31
  • 3
    $\begingroup$ In short, there are two conventions, one that normalizes and one that doesn't. The advantage of not-normalizing is the resulting derivative is linear in the direction vector $u$. Also, the non-normalized derivative answers the question of how approximately much the function changes if you move exactly along $u$, incorporating it's length. $\endgroup$ – Alex R. Mar 18 '19 at 18:33
  • $\begingroup$ Basically, the normalized version see the vector as only specifying a direction, and is measuring how fast the function changes in that direction. The unnormalized version interprets the vector as a change, and is interested in the total change it induces in the function. Which is preferable depends on the applications one is making. $\endgroup$ – Paul Sinclair Mar 19 '19 at 3:27

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