Depending on the source, the definition of the directional derivative does not include the restriction that the direction vector be of unit length.
In this case, it seems to me that we can then in particular take the derivative in the vanishing direction, and that this should, just straight from the definition, be $0$: $$D_{0}f=(f\circ \alpha)'=f'(\alpha)\alpha'=f'(\alpha)0=0$$ So is this indeed usually taken as being defined and allowed? I'm asking this because one of my homework questions becomes quite trivial under these conditions, so that makes me doubt that this is usually allowed/defined.
Please do not answer this question, it is just to demonstrate what I mean by it being trivial
Let $M$ be a compact surface in $\mathbb{R}^3$, and $f:M\to\mathbb{R}$ be a smooth function.
- Shwo that there exists a unique vector field $grad(f)$ s.t. $$I_p(grad(f),v)=df(v)$$ for all $p\in M$, $v\in T_pM$, and $I$ is the first fundamental form
- Show that there exists atleast one point $p\in M$ s.t. $(\nabla_{grad(f)}grad(f))(p)=0$. Here $\nabla$ is covariant differentiation.
Now for $2$ I think that by compactness of $M$, $f$ attains it's maximum, there $grad(f)=0$, and hence trivially $(\nabla_{grad(f)}grad(f))(p)=0$.
Again, please don't tell me if this is correct/incorrect, just indicate is this is an acceptable answer.
Indeed it is common to speak about directional derivative without requiring the "direction" to be a vector of unit length. This is the only way to speak about directional derivatives of a function on a vector space that doesn't even have a norm, for example.
In that case, the map $v\mapsto D_vf$ for fixed $f$ ought to be a linear transformation, which implies that $D_0 f$ must be $0$ if it exists at all. And since there is no particular reason not to let it exist, that's what it is.