Directional Derivatives - Geometric intuition I just have a question about directional derivatives.
I know that for a unit vector $v$, it is $\nabla f \cdot v$ (where $v$ is the direction we want to find the derivative of). 
The proof of this using limits makes sense, but how would you look at this geometrically? 
The gradient vector is the direction in which the function increases the most, but then having a dot product of this with a unit vector apparently gives the derivative going in that direction. 
I'm not quite sure how to interpret the dot product geometrically.
Thank you!

Say, we have $f:\Bbb R^n\to\Bbb R$, a $v\in\Bbb R^n$ unit vector and a point $x_0\in\Bbb R^n$.
Then look at the graph of $f$ in a one dimension higher space.
$$\Gamma_f:=\ \{({\bf x}, f({\bf x}))\,\mid\,{\bf x}\in\Bbb R^n \}\ \subseteq\ \Bbb R^{n+1}$$
Let $e_{n+1}$ be the last element of the standard basis of $\Bbb R^{n+1}$ (that is, $e_{n+1}=(0,0,\dots,0,1)^T$).
The directional derivative $\partial_vf(x_0)\ =\nabla f(x_0)\cdot v$ is the slope of the tangent line of ($\Gamma_f$ intersected with the 2d plane planned by $v$ and $e_{n+1}$).
In other words, take values of $f$ only over the line of direction $v$, which contains $x_0$.
