Talk:Directional derivative

From Wikipedia, the free encyclopedia

Mathematics Portal

This article is within the scope of WikiProject Mathematics, which collaborates on articles related to mathematics.

Mathematics rating:

Start Class

Low Priority

Field: Analysis

Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments.
Please add to or update the comments to suggest improvements to the article.
Geometry guy 15:44, 10 June 2007 (UTC)

[edit] Directional derivatives along normalized vectors only?

should it be specified that v has to be a normalized vector? --anon

Not necessarily. You can also take the directional derivative in the direction of the zero vector, as no division by zero is involved. Oleg Alexandrov (talk) 03:08, 9 March 2006 (UTC)

Hmmmm...It seems pointless to allow the directional derivative to not be normalized. mathworld seems to specify that the direction ought to be normalized too. What would be the purpose of allowing the directional derivative in the zero vector direction? --anon

One has

$D_{\vec{v}}{f}= \nabla f \cdot \vec{v}$

where $\nabla$ is the gradient.

I see no reason to require that you must do dot products only with unit vectors in the formula above. Oleg Alexandrov (talk) 03:08, 10 March 2006 (UTC)

We aren't talking about ONLY doing dot products with unit vectors, but that the directional derivitive is definied as a gradient of a function dotted with the unit vector of the vector in question evaluated at a point. So long as the unbit point is made, the equation is fine. I can tell that the data in this article came from planet math, which like wikipedia is user editited. Even the the eratta state that the Vector in question is unitary. However, don't take my word for it;

* The directional derivative at Wolfram

* Multivariate Calculus at usd.edu

* The directional derivative at lamar.edu

-- ∞Dbroadwell 19:59, 3 May 2006 (UTC)

Well, if you say "derivative along a vector", that vector does not need to be of length one. If you say "derivative along a direction", then yes, a direction by convention is normalized to length 1. So it makes sense to assume that vectors have length 1, but that is not necessary for the definition to work. Oleg Alexandrov (talk) 21:13, 3 May 2006 (UTC)

I quite agree that it's not necessary for the definition to work, however the standard implementation and usage up till differential equations emphatically states normalized. So, we should at least say so in the definition on the page, as you did. How it stood, it could be mis-read and if someone noted JUST the formula ... they would be wrong on calculus exams. -- ∞Dbroadwell 22:46, 3 May 2006 (UTC)

I've got a question if a function is differentiable for any vector included on X-axis (let,s say v=(1,0)) and differentiable for any vector on Y-axis then the function is differentiable on any direction of the plane (X,Y) i think this is similar to the way in which 'Cauchy-Riemann' equations are obtained.

I think that is important that the norm of the vector must be unitary:

If one takes two parallels non-unitary vectors, $\vec v$ $\vec w$ , such that $||\vec v||\neq ||\vec w||$ , then one has

$D_{\vec{v}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v} = ||\nabla f(\vec{x})|| \; ||\vec{v}|| \cos(\theta)$

and

$D_{\vec{w}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{w} = ||\nabla f(\vec{x})|| \; ||\vec{w}|| \cos(\theta)$

where $θ$ is the angle between the gradient and the vectors (remember they are parallels). The directional derivative depends on direction, then one must verify

$D_{\vec{v}}{f}(\vec{x}) = D_{\vec{w}}{f}(\vec{x})$

that is

$||\nabla f(\vec{x})|| \; ||\vec{v}|| \cos(\theta) = ||\nabla f(\vec{x})|| \; ||\vec{w}|| \cos(\theta)$

$||\vec{v}|| = ||\vec{w}||$

Absurd, it was supposed that $||\vec v||\neq ||\vec w||$ .

In addition, it seems obvious that the directional derivative can`t depend on the vector that one chooses. Because of that I think that is not by convention to choose an unitary vector, if you don´t, your result is a function of the norm of the vector chosen. Sorry about my english, I`m not used to express myself in this language.
--anon

I see where Oleg is coming from, differentiating with respect to 2x is different to differentiating with respect to x, however I don't believe that "derivative along a vector" is an accurate definition of the directional derivative. I believe that it is "derivative in the *direction* of a vector" ... why else would it be called the DIRECTIONal derivative? I would like to follow what every single text book I have ever read says and normalize v.

Oleg, please look at Dbroadwell's links above and if you'd like, find a reference to what you call the directional derivative.129.78.64.101 02:23, 4 October 2007 (UTC)