# Directional derivative - conventions

"directional derivative" - conventions

So the "directional derivative" in vector analysis - the differential rate of increase of a scalar function φ in a vector direction - is by convention* defined starting from the vector gradient (the 'del' operator): $$\nabla \phi \cdot \hat{u}= \vec{\nabla} \phi(\vec{r}) | _{r_0} \cdot \hat{u}$$
(the directional derivative of φ in the u-direction).

So it's not really fundamental - it's defined in terms of del-φ, which is 'more fundamental' I guess. Big deal.

*"by convention" meaning, in every one of a dozen reference books I looked through...

For example, in cartesian coordinates, a directional derivative (at r) might look like:

$$\vec{\nabla} \phi_r \cdot \hat{u} = \left(\hat{i}\frac{\partial{\phi}}{\partial{x}}| _r + \hat{j} \frac{\partial{\phi}}{\partial{y}} | _r + \hat{k}\frac{\partial{\phi}}{\partial{z}} | _r \right) \cdot \left(u_x \hat{i} + u_y \hat{j} + u_z \hat{k} \right)$$
$$=\frac{\partial{\phi}}{\partial{x}}u_x +\frac{\partial{\phi}}{\partial{y}}u_y + \frac{\partial{\phi}}{\partial{z}}u_z$$

So, the vector u is decomposed into some coordinate axes, and the derivatives of the scalar function are found with respect to those three axes, and things are scalar-multiplied and added together. Your definition of the directional derivatives goes through a coordinate system, of your preference - through the definitions of the del operator.

But why not define this directly, in the usual manner for defining a derivative:

$$\lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}$$

Which is the same thing, without those coordinate decompositions,

$$d \phi = \frac{\partial{\phi}}{\partial{x}}dx + \frac{\partial{\phi}}{\partial{y}}dy + \frac{\partial{\phi}}{\partial{z}}dz$$

$$\mbox{and, } d \vec{u} = \hat{i} d(u_x) + \hat{j}d(u_y) + \hat{k}d(u_z)??$$

Because $d \phi$ looks well-defined to me. And so does $d \vec{u} = \hat{u} d \lambda$. So then, why don't we just say, without ambiguity,

$$\frac{d \phi}{d \vec{u}} \right| _r \equiv \lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}$$

(with the normalized direction $$\hat{u} = \vec{u} / \| \vec{u} \|$$ ) ?!

So as to eliminate any reference to the coordinate system?

I mean, when we're taking the usual partial derivatives, with respect to say the "x" axis, that's just a 'directional' derivative in the $\hat{x}$ direction. So, can we not generalize this to any (normalized) vector direction, and so define the derivative of a scalar function with respect to a vector?

That's my question - is this notation rigorous enough?

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Just to clarify, by "notation", I mean the idea of writing

$$\frac{d \phi}{d \vec{u}}$$

Staff Emeritus
Gold Member
As far as I know, the actual definition of the directional derivative is given as, for a unit vector u:

$$(\nabla_{\vec{u}} f)(\vec{a}) = \lim_{h \rightarrow 0} \frac{f(\vec{a} + h \vec{u}) - f(\vec{a})}{h}$$

So that it may actually defined even if the gradient is not.

amcavoy
I don't know if this helps, but... (2 variables):

Let u=<a,b> so that the line going through u is given by the following:

$$x=x_0+at$$
$$y=y_0+bt$$

$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$

Now from above, it is clear that $\frac{dx}{dt}=a;\quad\frac{dy}{dt}=b$.

Substituting results in the following:

$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}=\frac{\partial f}{\partial x}a+\frac{\partial f}{\partial y}b=\nabla f\cdot \vec{u}$$

Castilla
Apmcavoy, a bit more exactly...

Not $$\frac {\partial f}{\partial t}$$ but $$\frac {d f}{d t},$$ because in your example the only independent variable is $$t.$$

Hurkyl said:
As far as I know, the actual definition of the directional derivative is given as, for a unit vector u:

$$(\nabla_{\vec{u}} f)(\vec{a}) = \lim_{h \rightarrow 0} \frac{f(\vec{a} + h \vec{u}) - f(\vec{a})}{h}$$

So that it may actually defined even if the gradient is not.

Can you direct me to a reference which develops this?

Staff Emeritus
Gold Member
Advanced Calculus (Buck) defines it exactly as I have, except he uses the notation D&beta; for the directional derivative in the direction &beta;.

But frankly, I would expect just about any calculus text book that ever deals with vectors to have this definition in it.

Homework Helper
Gold Member
Dearly Missed
I've never seen a vector calculus book worth its salt lacking Hurkyl's definition of the directional derivitave.
In my opinion, it wouldn't be worth its salt if it lacked a definition like that..

Homework Helper
I agree with Hurkyl, the directional derivative can intrinsicly be defined using the limit-definition, which seems fundamental enough. It's done like that in my Analysis course as well.

If you accept that as definition, it's not so hard to prove that the directional derivative can be calculated using the dot product with the gradient, so it follows as an equivalent way of calculating it.

This definition is given on Wikipedia: Directional Derivative as well.

Homework Helper
I prefer the notation
$$(u\cdot\nabla) f$$
for several reasons
first less confusing, if parenthesis are omited we do not confuse it with the gradient of an inner product, nor become confused should u depend on spatial variables. It is also less confusing should we take a directional derivative of a vector, or a higher order directional derivative.
$$D_u^n f =(u\cdot\nabla)^n f$$
I do not know what you mean by the definition of gradient has reference to the coordinate system. It need not. In any case various operators such as divergence, curl ect. should not be seen as functions of the del operator. Those sugestive notations should only remind us of a way the limit process the operator represents can be realized. i.e.
$$T(\nabla,a)=\lim_{V\rightarrow 0}\frac{1}{V}\oint T(n,a)ds$$
where T(nabla,a) is some combination of nabla and a,
and the integral is over the surface of a simple closed curve having volume V, and n is an outwardly pointed unit normal vector.
thus
$$(u\cdot\nabla)a=\lim_{V\rightarrow 0}\frac{1}{V}\oint (u\cdot n)a \ ds$$
$$(u\cdot\nabla)\times a=\lim_{V\rightarrow 0}\frac{1}{V}\oint (u\cdot n)\times a \ ds$$
and so forth
We can chose a particular curve to get a differential expression such as a cube.
of course we could aways use
$$(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b$$