Talk:Directional derivative

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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 $\theta$ 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)[reply]

I'm pretty sure the directional derivative, when defined in the direction of a nonunit vector, is supposed to be the same as the directional derivative in the direction of the unit vector which is parallel to the given vector. This agrees with wolfram mathworld, so I'm editing this page to that effect. I'm also putting the unit vector definition first, as that everyone agrees on. If some authors define the directional derivative to be what the article had before I edited it, could we include that fact with citation, but as an alternate definition rather than as the only one? 75.22.201.232 (talk) 12:55, 24 March 2010 (UTC)[reply]

The article is still a little misleading. I only read the "Generally applicable definition" and was given no indication that v is usually a unit vector. Perhaps a little warning in brackets would prevent misleading? i.e. "along a vector (usually a unit vector)" Muntoo (talk) 00:32, 13 July 2015 (UTC)[reply]

I think you have a point, but the problem may be due to a misleading subheading. The "Generally applicable definition" is one of two nonequivalent definitions. It is really a definition that applies in a wide range of contexts and is used in fundamental physics), whereas the second definition applies is a far more limited range of contexts, and is thus mathematically less useful. Any suggestions as to what to change these two subheadings to? —Quondum 02:28, 13 July 2015 (UTC)[reply]

I've tweaked headings and added explanatory footnotes. By the way, the assumption of "(usually a unit vector)" is not necessary valid; this case is covered by the third subheading, but I suspect that the primary use for this is in grade-school teaching and perhaps in engineering; it should not the the form that is taught in a calculus class even at grade school, for example, because it does not even apply to general multi-variable functions due to the lack of a definition of a unit vector. —Quondum 03:21, 13 July 2015 (UTC)[reply]

Calculus and physics classes in grade school are normally highly restricted to Euclidean vectors in 1-3 dimensions and Cartesian coordinates, for which the Euclidean norm is almost always implied. Even introductory college-level courses on this rarely venture beyond Euclidean space. So I would definitely say that in applications, the vector in question is a unit vector.--Jasper Deng (talk) 03:27, 13 July 2015 (UTC)[reply]

Then they are not covering the important case of a function of many variables. But I suppose that would be dealt with through partial derivatives and the total derivative, and not even given the name directional derivative. However, we should take care not to give undue weight to the school/college level definition. Perhaps the name directional derivative is unfortunate since it does hint at using the direction only, but it definitely is used for the magnitude-dependent version, and the lead refers to this version. What is important here is to make it not confusing to readers including school-goers, so wording suggestions that do not give undue weight to the introductory version would be welcome. Perhaps we should include the third case (to which you refer) in the first definition via an explicit option to restrict the vector to a unit vector? —Quondum 04:00, 13 July 2015 (UTC)[reply]

Maybe the distinction should be mentioned much earlier, and I do agree that the name "directional" derivative is unfortunate because it implies no dependence on the magnitude of a vector pointing in that direction. At the elementary level, the gradient is often taught as the vector whose magnitude is the steepest possible directional derivative and points in the direction of that directional derivative, but such an interpretation can only be done for the unit vector restriction. Also, on non-normed vector spaces, I'm not even sure if we have a notion of "magnitude" dependence, although we can still talk about a scalar multiple of the vector.--Jasper Deng (talk) 04:29, 13 July 2015 (UTC)[reply]

Okay, I'll try to bring the distinction in near the start of the definition.

I'm not sure how to interpret your last statement. Even in the absence of a concept of magnitude, the directional derivative is well-defined, and scales with the vector along which it is taken. —Quondum 05:58, 13 July 2015 (UTC)[reply]

Of course it's always well-defined. I was just being pedantic about the word "magnitude" (which usually implies norm), which of course is besides the point: any scalar multiple of the same vector will yield a directional derivative "along the same diretion". Thanks for doing this.--Jasper Deng (talk) 07:26, 13 July 2015 (UTC)[reply]

Please clarify

D_{\vec {v}}{f}({\vec {x}})=\nabla f({\vec {x}})\cdot {\vec {v}}=\nabla _{v}f({\vec {x}})

Is the second equal sign true? I assume, but I didn't see it explicitly mentioned in the article. User:Nillerdk (talk) 17:01, 29 January 2008 (UTC)[reply]

Thanks for pointing out the inconsistency in notation. The issue has been addressed. Silly rabbit (talk) 17:07, 29 January 2008 (UTC)[reply]

--> Just a question, how would i go about determining the second order directional derivative, if the answer were to be fuu=fxx.a^2 +fyy.b^2, isn't there going to be cases that dont hold. for instance if i take a sheet and fix a point in it's middle, pull the corners upwards and pull the midpoints of each edge down wards. Would i have not created a case where fxx,fyy= -ve and fuu = +ve?

The second order derivative along u = (a,b) is given by

D^{2}f\{\mathbf {u} ,\mathbf {u} \}=a^{2}f_{xx}+2abf_{xy}+b^{2}f_{yy}

provided f is twice continuously differentiable. The formula you gave is missing the crossterm. siℓℓy rabbit (talk) 14:27, 15 June 2008 (UTC)[reply]

in my text book they jumped to the conclusion that [let d = delta] dz= (fx+e1)dx + (fy +e2)dy, (not that its hard to believe). instead, i proved the directional derivative equation by using continuity and following a pathway along f(x,y)dx and f(x+dx,y)dy. is there a similar method for deriving the second derivative. let me know if you have a link etc. i think it would be nice to see a wikipedia page on the second derivative test one day, i have not been able to find one. —Preceding unsigned comment added by 122.110.28.250 (talk) 09:29, 22 June 2008 (UTC)[reply]

See Second partial derivative test and Second derivative test. siℓℓy rabbit (talk) 13:36, 22 June 2008 (UTC)[reply]

Image

Hello,

Here is a link to an image which may be useful.

By the way, the french definition does not utilize normalized vector, and, more important, I do not understand why the limit is calculated with h taking only positive values

Sorry for my bad english —Preceding unsigned comment added by 90.43.213.76 (talk) 21:03, 25 March 2010 (UTC)[reply]

I think the French version does use normalized vectors. The French version of the article says, "On parlera de dérivée directionnelle de f au point u dans la direction de h lorsque le vecteur h est unitaire." So it's only willing to use the name "directional derivative" when the vector is unit. Otherwise it uses the term "La dérivée de f au point u selon le vecteur h," i.e. the derivative of f at u along the vector h.

As for the restriction to positive values - I've seen both definitions, but it's arguable that the directional derivative of f in the direction h should take into account how much f changes when you move in the direction h, but not in the reverse direction. That's all the restriction to positive values does.

In any case, I think what's going on here is that there are multiple different conventions. In the case of a differentiable function and a unit vector, they all agree, but otherwise there are slight differences between conventions.

Also, your English was flawless. 75.22.201.232 (talk) 19:26, 27 March 2010 (UTC)[reply]

Thank you for this convincing answer; I had corrected the French version, after various readings. (Asram) —Preceding unsigned comment added by 90.43.214.46 (talk) 23:48, 3 April 2010 (UTC)[reply]

normalisation

The definition

\nabla _{\vec {v}}{f}({\vec {x}})=\nabla f({\vec {x}})\cdot {\frac {\vec {v}}{|{\vec {v}}|}}

appearing at the beginning of the article is in direct contradiction with the definition given in the "differential geometry" section. Tkuvho (talk) 10:01, 3 May 2010 (UTC)[reply]

Indeed, directional derivatives are not covariant derivatives which are denoted as

\nabla _{V}

for V a vector field 2018-11-18 16:42 (UTC) — Preceding unsigned comment added by 92.189.150.15 (talk)

Normal derivative

I can't make sense of

{\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =Df(\mathbf {v} )

I suppose there has something gone wrong --Trigamma (talk) 21:57, 28 May 2010 (UTC)[reply]

The partial derivative-like notation on the left means the gradient of the scalar-valued function f, which happens to be the total derivative of a scalar-valued function with respect to all its variables. If f were vector-valued, it would be instead the Jacobian, which is the total derivative of one vector with respect to the other. The other notations in that equation make it clear that this is what is meant.--Jasper Deng (talk) 08:33, 23 November 2013 (UTC)[reply]

slight clean up

Minor changes:

add a couple more sources,
introduce better numbering scheme (indented and non-bracketed),
made the main properties stand out more,
made vector notation consistent all the way through (it was a mix of bold + overarrow, typographically bold is slightly easier)

F = q(E+v×B) ⇄ ∑_ic_i 13:23, 13 April 2012 (UTC)[reply]

Verifiability of definition

As may be seen from the threads above, there seems to be a lot of confusion about the definition of the directional derivative. It may be noted that the citations given are very weak.

The only sensible definition (i.e. one that is mathematically useful and elegant) does not involve any reference to normalization. My impression is that the most notable definition (i.e. generally used definition in authoritative texts) is this one, though I do not have many of these texts to hand. There are many reasons for this claim, to the extent that I would argue for the relegation of the use of the normalized vector to a footnote. A major argument for this is that in the context of functions over a set of coordinates and related contexts, the norm of a vector is usually not defined but the directional derivative remains an extremely useful concept. Even where an (indefinite) metric tensor is defined as in the physics of general relativity, the definition runs into serious difficulties with no apparent benefits.

My guess is that the weaker definition arose as the tail wagging the dog: a term was needed for the very useful construct that generalizes the partial derivative. An obvious term would be the "directional derivative", even if this may be a partial misnomer, which then was interpreted literally.

I would appreciate authoritative texts being cited by those who have access to them. — Quondum 07:15, 30 October 2012 (UTC)[reply]

I have an (elementary) textbook defining it using unit vectors, but it only considers steady vector fields and does not go beyond ordinary 3-dimensional space.--Jasper Deng (talk) 08:34, 23 November 2013 (UTC)[reply]

Error in the first plot

It is a small error but I think it should be corrected anyway. In the first plot (the one with the contour plot), the gradient has an index 'u' in its name, but a 'u' in index means that we talk about directional derivative (that is not a vector by the way). I don't know how to replot it but I would do it myself if I knew. Thanks --Jotwo (talk) 18:16, 4 April 2014 (UTC)[reply]

Quantities in the "continuum mechanics" section

I think some of the quantity types are wrong. For example, the directional derivative of a vector field in the direction of any vector should again be a vector field, not a tensor. Since the directional derivative is the limit of a sequence of quotients of the original function's values by scalars, I think the directional derivative must match the original function's type.--Jasper Deng (talk) 17:48, 29 November 2015 (UTC)[reply]

Add example for beginner users?

I think it would be a good idea to add a simple example how to typically calculate the directional derivative for beginners. But in what section should I add this example?

Example

Thus, if the function $f$ and vector $\mathbf {v}$ is defined as:

f(\mathbf {x} ,\mathbf {y} )=x^{2}+y^{2}

and:

\mathbf {v} =(1,2),|\mathbf {v} |={\sqrt {1^{2}+2^{2}}}={\sqrt {5}}

The directional derivative in point $(x,y)$ will be:

$\nabla _{\mathbf {v} }{f}={\frac {\partial {f}}{\partial {\mathbf {x} }}}\cdot {\frac {\mathbf {v_{x}} }{|\mathbf {v} |}}+{\frac {\partial {f}}{\partial {\mathbf {y} }}}\cdot {\frac {\mathbf {v_{y}} }{|\mathbf {v} |}}=2x\cdot {\frac {1}{\sqrt {5}}}+2y\cdot {\frac {2}{\sqrt {5}}}$

An example. Wow!

At last there is someone with some empathy with the readers of Wikipedia. Thanks, mate/mam! His question "where to put it?" tells the whole story about this article.

Inconsistent typography of vectors (x,v), in the "Definition"

The "Definition" section uses both mathrm and mathbf fonts for the vectors x and v. This needs to be harmonized. As most of the rest of the article uses the mathbf font for vectors, I would suggest going with that.

Also, I don't see any sense in this definition for $h(t)$ : "This follows from defining a path $h(t)=x+tv$ ." Probably supposed to be something like: "This follows from defining a path $\mathbf {x} +t\mathbf {v}$ ." One could use the standard notation from differential geometry of defining $\mathbf {v}$ as the vector tangent to the path ${\boldsymbol {\gamma }}(t)$ at $t=0$ . But that doesn't fit the spirit of the rest of the definition, so not advisable, unless more differential geometry infrastructure is built into the "Definition".