Directional derivative and normalization of the direction vector.

Question

This issue has been discussed before but I still have some doubt about it. I understand while defining directional derivatives, some authors normalize the direction vector v while some don't. However, intuitively I am thinking directional derivative at a point as the slope of the straight line lying on the tangent plane of the graph at that point in the direction of v. If we don't normalize v, different vectors on the same direction will give different values of directional derivative. However, they all lie on the same straight line on the tangent plane and that line should have the same slope!

Answer 1

Some authors do not normalize the vectors $v$ because they are working with a general version of the directional derivative applied to arbitrary topological vector spaces, and not just to Euclidean vector spaces or normed vector spaces. Without a normed space, there is no notion of normalization to speak of, but having a theory of derivatives for non-normed spaces is still useful/necessary, so we simply get rid of the requirement to normalize. This general version of the derivative is called the Gateux derivative. Simply put, $$dF(x;v)=\lim_{t\to0}\frac{F(x+tv)-F(x)}{t},$$ where, here, $x$ and $v$ are elements of the vector space $X,$ and $t$ is a scalar element (specifically, a real number). The idea now is that if $v=au,$ for some scalar $a$ and vector $u,$ then $$dF(x;au)=a\cdot{dF(x;u)}.$$ Here, $\cdot$ denotes scalar multiplication (because simply using juxtaposition gets confusing at this stage). The geometric interpretation is that the magnitude of the vector multiplies the directional derivative. This makes sense, because a longer directional vector represents a gradient along a direction for a longer distance, so to speak.