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I have seen several different starting points for definition the directional derivative of a function $f$ at a point $p$. Ultimately though, they can all be reduced to the equivalent definition via the gradient:

$$ D_v f(p) = \langle \nabla f(p), v \rangle $$

What is not clear though is why some texts only allow $v$ to be a unit vector and why other texts have no such restriction. If $v$ is not a unit vector one can always be produced by dividing $v$ by its norm; however, strictly speaking, the two definitions (one which requires a unit vector and one which doesn't) will differ by a scaling factor.

So, my question is, is there any reason to restrict the definition to a unit vector? What is the motivation for some texts to allow only unit vectors?

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Unit vectors are directions. –  Qiaochu Yuan Mar 27 '12 at 18:05
@QiaochuYuan Ok, can't argue with that. But non-unit vectors have direction too, so I don't really understand the point of your comment. –  AFX Mar 27 '12 at 18:09
That definition involving gradient is only correct if $f$ is (Fréchet) differentiable. It's possible for $f$ to have directional derivatives in all directions and for that formula to still not be valid. –  Chris Eagle Mar 27 '12 at 18:13
@AFX Unit vectors are in one-to-one correspondence with directions. Calculating a directional derivative with respect to a non-unit vector is akin to finding $\frac{dy}{d(2x)}$ in one-variable calculus. You could do it, but it doesn't tell you anything you didn't already know from the unit vector case. –  Brett Frankel Mar 27 '12 at 18:13
@AFX: think about one dimension. We only care about the derivative $f'(x)$ instead of all of the possible directional derivatives $a f'(x)$ because $f'(x)$ already encapsulates the information we care about, namely how fast $f$ is changing in the positive $x$-direction. I'm not saying I agree with this reasoning, by the way (removing the restriction gives you a linear map and more structure is generally better in my book). –  Qiaochu Yuan Mar 27 '12 at 18:13

3 Answers 3

up vote 2 down vote accepted

The motivation is that some authors want to focus entirely on the direction for the construction of the directional derivative, so there is no extra "speed" taken into account. The limit definition,

$$D_v f(x)=\lim_{\epsilon\to0} \frac{f(x+\epsilon\, v)-f(x)}{\epsilon},$$

divides the difference of $f$ at $x$ and a point precisely $\epsilon$ units away by the length of the displacement, which is $\epsilon$. If you allow $v$ to be non-unit length, then the quotient will divide by a quantity (which is, again, $\epsilon$) that is proportionate but not equal to the length of the underlying displacement (which is $\epsilon\|v\|$ in size). This discrepancy means that the more generally defined directional derivative is more than just direction-dependent, it is also determined by the "size" associated to the direction.

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The importance of the first definition is when you take the "derivative" of a function with respect to a vector field instead of just a simple vector. When your space is not Euclidean, (for example a sphere) this distinction is very important, as there doesn't always exist a smooth vector field, defined on the tangent space of the surface, (more generally a differentiable manifold) that is always non-zero (see Hairy Ball theorem). This means that you can't just divide the vector field by it's norm at each point. The distinction seems trivial because the underlying geometry of Euclidean space.

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When you say "not Euclidean" do you just mean its a space where the distance can't be measured with the "standard" distance formula $d(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$ ? –  AFX Mar 27 '12 at 19:12
There's A LOT of different formulations, but at the simplest level its a "space" in which the axioms of Euclidean Geometry do not hold. For example if you are on the north pole of the planet, and just fly/swim/walk/whatever in one (any) constant direction you will eventually reach the north. That is to say on a sphere a "line," or a geodesic, there exists points such that there exists infinitely many lines connecting them. –  J Mann Mar 27 '12 at 21:38
If you want to know more I would pick up any book on Riemannian Geometry. I would suggest Boothby's "An Introduction to Differentiable Manifolds and Riemannian Geometry." Also Peterson's "Riemannian Geometry" is great if you're looking for more of a challenge. –  J Mann Mar 27 '12 at 21:41

If you define $D_v|_p(f)=\frac{1}{|v|}\left<\nabla f(p),v\right>$, you have that the directional derivative is independent to the magnitude of $v$ as everyone hope, but in this way the function $v\mapsto D_v|_p(\cdot)$ isn't linear, infact: $$\frac{1}{|u+v|}\left<\nabla f(p),u+v\right>\neq \frac{1}{|u|}\left<\nabla f(p),u\right>+\frac{1}{|v|}\left<\nabla f(p),v\right>$$

We decide to treat only unit vectors and to define $D_v|_p(f)=\left<\nabla f(p),v\right>$ so as to preserve both the independence of the derivative to the magnitude of $v$ and the additivity of the above function.

In differential geometry, if $v$ is a general vector, the directional derivative is $$D_v|_p(f)=\left<\nabla f(p),v\right>=\frac{d}{dt}\big|_{t=0}f(x+tv)$$ so we allow the value of directional derivative to change with the magnitude of $v$.

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