Talk:Directional derivative

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[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.
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