Help me understand this:


by aaaa202
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#1
Feb26-12, 06:09 AM
P: 991
My teacher often writes for a scalar function F:

dF = [itex]\nabla[/itex]F [itex]\bullet[/itex] dr

Why is it you are allowed to do this. Shouldn't you use the pythagorean theorem?
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#2
Feb26-12, 08:13 AM
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Quote Quote by aaaa202 View Post
My teacher often writes for a scalar function F:

dF = [itex]\nabla[/itex]F [itex]\bullet[/itex] dr

Why is it you are allowed to do this. Shouldn't you use the pythagorean theorem?
Hi aaaa202!

It's an identity that follows from the definition of the derivative of a function of multiple coordinates.
Consider that the total derivative is:
$$dF = {\partial F \over \partial x}dx + {\partial F \over \partial y}dy + {\partial F \over \partial z}dz$$
And the right hand side is:
$$\nabla F \cdot d\mathbf{r} = \begin{bmatrix}{\partial F \over \partial x}\\{\partial F \over \partial y}\\{\partial F \over \partial x}\end{bmatrix} \cdot \begin{bmatrix}dx\\dy\\dz\end{bmatrix} = {\partial F \over \partial x}dx + {\partial F \over \partial y}dy + {\partial F \over \partial z}dz$$

I guess you could say that he pythagorean theorem is not involved because any application of it cancels left and right in the identity.
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#3
Feb27-12, 04:20 AM
P: 991
hmm yes I kinda got this already, but I'm just unsure how to interpret the term:

dF = dF/dx * dx + dF/dy * dy + dF/dz *dz

What does this infinitesimal bit of represent? Surely its not a vector, since the result is a scalar. Surely it can't be an infinitesimal part of its length either, since dF =√(dFx^2 + dFy^2 + dFz^2)
So what does it mean this dF?

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Feb27-12, 09:44 AM
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Help me understand this:


Quote Quote by aaaa202 View Post
hmm yes I kinda got this already, but I'm just unsure how to interpret the term:

dF = dF/dx * dx + dF/dy * dy + dF/dz *dz

What does this infinitesimal bit of represent? Surely its not a vector, since the result is a scalar. Surely it can't be an infinitesimal part of its length either, since dF =√(dFx^2 + dFy^2 + dFz^2)
So what does it mean this dF?
You can interpret infinitesimals as small delta's.

dF is the change in F(x,y,z) if you change the coordinates by (dx,dy,dz) to F(x+dx,y+dy,z+dz).
You can write this as: dF=F(x+dx,y+dy,z+dz)-F(x,y,z).
This is the (scalar) difference in F between 2 points in space.
As such dF =√(dFx^2 + dFy^2 + dFz^2) does not apply, since it is not a vector.
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#5
Feb27-12, 11:41 AM
P: 991
Okay yes, but I just don't find the idea of adding up the changes of the function in respectfully the x-, y- and x- direction very interesting.
So far I've seen it used, but rather length has been used (or maybe I'm wrong) - for instance in continuity considerations you require that f(x,y) is well defined inside the cirfumference of an infinitesimal circle.


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