Interpretation of directional derivative without unit vector I understand that using a unit vector (of a vector say $\vec{a}$ ) and computing the directional derivative gives the slope  (or rate of change of the function) in the direction of the vector.
I have three questions :


*

*If I use the vector itself rather than it's unit vector what will
I get when I compute it's dot product with the gradient of the
function? It wouldn't give the slope of the curve(formed by the
slicing of the function with the plane containing the vector $\vec{a}$ ), would it?


Note: The function is scalar.
Also going by it's formal definition:
$\displaystyle \nabla _{\mathbf {v}}{f}({\mathbf {x}})=\lim _{h\rightarrow 0}{\frac {f({\mathbf {x}}+h{\mathbf {v}})-f({\mathbf {x}})}{h}}$
$\mathbf {v}$ is a vector
Quoting from Wikipedia 

This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.



*What does that mean?


Also quoting from Wikipedia:

If the function f is differentiable at x, then the directional derivative exists along any vector v, and one has

$\displaystyle \nabla _{\mathbf {v} }{f}({\mathbf {x} })=\nabla f({\mathbf {x} })\cdot {\mathbf {v} }$

Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v with respect to time, when moving past x.



*Why is it mentioned with respect to time isn't it with respect to
the change in x (or/and y )  in the direction of the vector ?

 A: If you define $\nabla_x f(x_0)=\lim_{h \to 0^+} \frac{f(x_0+hx)-f(x_0)}{h}$, then you have the identity $\nabla_x f(x_0)=\| x \| \nabla_{x/\| x \|} f(x_0)$. (I will remark that this notation clashes with notation elsewhere in math, but I will stick with it here.) That is, the derivative "along $x$" is the directional derivative multiplied by the norm of $x$. In effect instead of just moving in a direction and measuring the change in $f$ relative to the distance you traveled in that direction, you are moving in a direction at a particular rate in time and measuring the change in $f$ relative to that change in time. The speed is the conversion factor between these measurements.
This definition of $\nabla_x$ doesn't depend on there being such a thing as the norm of $x$, whereas the directional derivative does. But for your purposes you can ignore this remark for now.
I said this in the first paragraph, but just to directly address your third question, let me add one more thing. The directional derivative does not really have a notion of time, it is really a change in $f$ with respect to distance traveled in the specified direction. Your generalized notion $\nabla_x$ effectively involves time after you identify $\| x \|$ as a speed and $h$ as a time, so that $hx$ is a displacement and $h \| x \|$ is a length.
A: The following definition should explain the directional derivative of a non-unit vector; we can call it the general directional derivative. It is similar to the 'unit vector' definition of the directional derivative given in most intro-calculus textbooks, but scaled by the magnitude of the vector.
$$\displaystyle \nabla _{\mathbf {v} }{f}({\mathbf {x} })=
\nabla f({\mathbf {x} })\cdot {\mathbf {v} } = \nabla f({\mathbf {x} })\cdot \left( {\frac{\mathbf {v}}{\| \mathbf{v} \| } \|\mathbf{ v}\| }\right)  = \|\mathbf{ v}\| ~\nabla f({\mathbf {x} })\cdot  \left( {\frac{\mathbf {v}}{\| \mathbf{v} \| }  } \right)$$
In the case that $\mathbf {v}$ is already a unit vector, this simplifies to $\nabla f({\mathbf {x} })\cdot {\mathbf {v} }$.
A: "This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.
What does that mean?"
Not all vector spaces have a defined inner product space or a norm for that space.  But, returning to first principles we can define a directional derivative.
"Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v with respect to time, when moving past x.
Why is it mentioned with respect to time isn't it with respect to the change in x (or/and y ) in the direction of the vector ?"
If we are looking at the changes in $f(\bf{x})$ as $\bf x$ traverses some path, we may find that we parameterize that path, and might like to think of that parameter as "time."
A: Its always better to understand vectors and their derivatives in some context taken from physics. 


*

*directional derivative of distance w.r.t time gives you velocity in the respective direction (like x or y axis/direction). Its a differentiation w.r.t to time. Also, the vector remains a vector after this operation (both distance and velocity have components on the axes in space).

*gradient of voltage (where we differentiate w.r.t distance) gives you electric field in a particular direction. Here, the operation converts a scalar to a vector. Though voltage is dependent on the position in space, it has a value but no direction. (Just like its hotter when closer to a furnace, there's higher voltage when closer to a positive charge).
So, the two derivatives 


*

*are not the same

*are used for different reasons

*best understood in a given context (because most often math is a means to an end, and to what end?... right?).


Answers to your questions : 
1) dot product of (vector, gradient of the function) ... please note that you can't compute gradient of a vector.
2) the 'h' mentioned must be infinitesimal time being multiplied with 'v' (velocity in the x direction). Hence, it is indeed differentiation w.r.t to time.
3) This is answered in 2.
