[Math] Interpretation of directional derivative without unit vector

multivariable-calculus

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 :

  1. 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.

  1. 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.

  1. 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 ?

Best Answer

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.

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