Directional Derivative of F: Same for V & 2V?

[seeker101]

A very basic question:
Is the directional derivative of a scalar function F at some point 'a' in the direction V the same as the directional derivative of F at 'a' in the direction 2V?

Going by http://en.wikipedia.org/wiki/Directional_derivative#Definition" definition of a directional derivative (using a non-normalized direction vector), they appear to be different!

 

[HallsofIvy]

A very basic question:
Is the directional derivative of a scalar function F at some point 'a' in the direction V the same as the directional derivative of F at 'a' in the direction 2V?

Going by http://en.wikipedia.org/wiki/Directional_derivative#Definition" definition of a directional derivative (using a non-normalized direction vector), they appear to be different!

?? The vectors V and 2V have the same direction. I cannot find anywhere on the Wikipedia page that implies thee directional derivative will be different.
For any vector V, the derivative of f in the direction of vector V is
[tex]\frac{\nabla f \cdot V}{||V||}[/itex]<br /> Of course, for vector 2V that becomes <br /> $$\frac{\nabla f \cdot 2V}{||2V||}= \frac{2\nabla f \cdot V}{2||V||}= \frac{\nabla f \cdot V}{||V||}$$<br /> exactly the same as before.[/tex]

 

[seeker101]

So, are you saying http://en.wikipedia.org/wiki/Directional_derivative#Definition" is flawed? (I'm familiar with the (standard) definition you have mentioned above. But I just wanted to clarify the definition used on that wiki page - which btw references Apostol's Analysis text)

The definition of a directional derivative given there is:
$$\nabla_{\vec{v}}f(\vec{x})=\nabla f(\vec{x}) \cdot \vec{v}$$

The wiki page goes on to say: "Usually directions are taken to be normalized, so $$\vec{v}$$ is a unit vector, although the definition above works for arbitrary vectors."

Wouldn't this mean
$$\nabla_{2\vec{v}}f(\vec{x})=\nabla f(\vec{x}) \cdot 2\vec{v}$$?

 

[AUMathTutor]

It certainly seems to be flawed. The way they have it written, it looks to me like, by an appropriate choice of |v|, you can make just about any directional derivative equal to any number you want. Except, well, for perpendicular directions, in which case that formula will actually give the correct answer for v, 2v, 3v, etc.

I mean, you can obviously see the Wiki page is wrong. I'm sure it happens more than you'd think.

 

[Ben Niehoff]

It depends what you want it to mean. If I just want "the derivative in the direction of v", then I would divide by the norm of v.

But sometimes you actually want "the derivative along v", in which case you want distinct answers for v, 2v, 3v, etc.