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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 ?
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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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  • $\begingroup$ Thanks. Can you help me with my first question? $\endgroup$ – paulplusx Aug 2 '18 at 11:44
  • $\begingroup$ @paulplusx Read the first sentence. $\endgroup$ – Ian Aug 2 '18 at 12:49
  • $\begingroup$ I am sorry but I am not able to understand (with the given explanation) what is the exact output of the dot product of gradient and the vector. Could you provide a more intuitive/geometrical answer supported by a bit of maths? (As simple as possible ) $\endgroup$ – paulplusx Aug 2 '18 at 12:59
  • $\begingroup$ @paulplusx It is the directional derivative in the direction of $x$, multiplied by $\| x \|$. The point is that you go $\| x\|$ times further from the start point than you would to estimate the directional derivative for a given $h$, but then you divide by the same $h$, so the ratio is about $\| x\|$ times bigger (which becomes exact for $h \to 0$). $\endgroup$ – Ian Aug 2 '18 at 13:04
  • $\begingroup$ Aaahh. Now I get it. Thanks. $\endgroup$ – paulplusx Aug 2 '18 at 13:23
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"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."

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  • $\begingroup$ Understood. Can you help me with my first question (which is the subject of the question) about non-unit vector dot product with the gradient? $\endgroup$ – paulplusx Aug 2 '18 at 11:45
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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.

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  • $\begingroup$ I meant a scalar function. I have added a note for it now. $\endgroup$ – paulplusx Aug 2 '18 at 17:00

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