TITLE: Why isn't the directional derivative generally scaled down to the unit vector?
QUESTION [37 upvotes]: I'm starting to learn how to intuitively interpret the directional derivative, and I can't understand why you wouldn't scale down your direction vector $\vec{v}$ to be a unit vector.
Currently, my intuition is the idea of slicing the 3D graph of the function along its direction vector and then computing the slope of the curve created by the intersection of the plane. 
But I can't really understand how the directional derivative would be a directional derivative if it were not scaled down to be a change in unit length in the direction of $\vec{v}$. Is there an intuitive understanding I can grasp onto? I'm just starting out so maybe I haven't gotten there yet.
Note, I think there may be a nice analogy to linearization, like if you take "twice as big of a step" in the direction of $\vec{v}$ , then the change to the function due to the change in this step is twice as big. Is this an okay way to think about it?

REPLY [2 votes]: Here's a decent reason to have the thing we call the directional derivative.
And not require the reference vector to be of unit magnitude. 
Start with the case of $f:\mathbb{R}\to\mathbb{R}$. Write the directional
derivative with respect to some arbitrary constant $h\ne0$, and apply
first principles
$D_{h}f\left[a\right]=\lim_{t\to0}\frac{f[a+th]-f[a]}{t}$
$=h\lim_{t\to0}\frac{f[a+th]-f[a]}{th}$
$=hf^{\prime}\left[a\right]$.
Neither surprising nor particularly interesting. It's just another
way of saying what we already knew.
$x\left[t\right]=a+th$
$g\left[t\right]=f\left[x\left[t\right]\right]$
$g^{\prime}\left[0\right]=f^{\prime}\left[a\right]x^{\prime}=hf^{\prime}\left[a\right].$
But it gets a bit more interesting if we use it to build up a Taylor
series for a real-valued function of a real argument. I'll leave the
notation to speak for itself.
$D_{h}D_{h}f\left[a\right]=D_{h}^{2}f\left[a\right]=h^{2}f^{\prime\prime}\left[a\right]$.
$D_{h}^{r}f\left[a\right]=h^{r}f^{(r)}\left[a\right]$.
$D_{h}^{0}f\left[a\right]=f\left[a\right]$.
$f(a+h)=\sum_{r=0}^{k}\frac{D_{h}^{r}f\left[a\right]}{r!}+\frac{D_{h}^{k+1}f\left[\zeta\right]}{(k+1)!}$.
I think that's kind of pretty.
Now lets call $f:\mathbb{R}^{n}\to\mathbb{R}$. I use this non-standard
notation as a shorthand:
$\{\mathbb{J}\}=\left\{ \left\{ j_{i}\geq0\right\} _{n}|\sum_{i=1}^{n}j_{i}=k\right\} $.
Think of the $\mathbb{J}^{'s}$ as multi-indices. Each $\mathbb{J}$
is a set of non-negative integers $\{j_{1},\dots,j_{n}\}$ such that
$j_{1}+\dots+j_{n}=k$.
Now use those indecies to write the multinomial expansion of a sum
of n terms:
$\left(x_{1}+\dots+x_{n}\right)^{k}=\sum_{\mathbb{J}}\begin{pmatrix}k\\
j_{1}\dots j_{n}
\end{pmatrix}x_{1}^{j_{1}}\dots x_{n}^{j_{n}}$.
Here the coefficients are the multinomial generalization of the binomial
coefficients:
$\begin{pmatrix}k\\
j_{1}\dots j_{n}
\end{pmatrix}=\frac{k!}{j_{1}!\dots j_{n}!}=\begin{pmatrix}k\\
\mathbb{J}
\end{pmatrix}$.
Now write the Taylor series as before, but using $\mathfrak{h}\in\mathbb{R}^{n}$.
$f(\mathfrak{a}+\mathfrak{h})=\sum_{r=0}^{k}\frac{D_{\mathfrak{h}}^{r}f\left[\mathfrak{a}\right]}{r!}+\frac{D_{\mathfrak{h}}^{k+1}f\left[\vec{\zeta}\right]}{(k+1)!}$
$=P_{k}\left[\mathfrak{h}\right]+R_{k}\left[\mathfrak{h}\right].$
The directional derivative is:
$D_{\mathfrak{h}}f\left[\mathfrak{a}\right]=\mathfrak{h}\cdot\nabla\left[f\left[\mathfrak{a}\right]\right]=\left(h_{1}D_{1}+\dots+h_{n}D_{n}\right)\left[f\left[\mathfrak{a}\right]\right].$
So:
$D_{\mathfrak{h}}^{r}f\left[\mathfrak{a}\right]=\left(h_{1}D_{1}+\dots+h_{n}D_{n}\right)^{r}\left[f\left[\mathfrak{a}\right]\right].$
So the Taylor polynomial of degree $k$ is:
$P_{k}\left[\mathfrak{h}\right]=\sum_{r=0}^{k}\sum_{\mathbb{J}}\begin{pmatrix}r\\
j_{1}\dots j_{n}
\end{pmatrix}\left(h_{1}D_{1}\right)^{j_{1}}\dots\left(h_{n}D_{n}\right)^{j_{n}}\left[f\right]$.
The $\left[f\right]$ indicates that the beast to the left is an opperator
being applied to $f$.