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Not sure what square brackets indicate when dealing with partial derivates

by Kushwoho44
Tags: brackets, dealing, derivates, partial, square
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Kushwoho44
#1
Dec12-12, 04:49 AM
P: 10
Hi guys, attached is a picture of my problem and it is also underlined.



I've been reading through this theory and I just don't understand what the square brackets indicate.

I understand that ∇phi is the partial derivative with respect to the scalar function phi.

But what is ∇phi [r(t)] ?

I feel ashamed asking this like I'm going to be laughed at.
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Simon Bridge
#2
Dec12-12, 06:53 AM
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Simon Bridge's Avatar
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##\nabla\phi## is the the gradient of ##\phi##: it is read "grad phi" or "del phi".
So it involved taking the partial derivative of phi with respect to each coordinate axis.
http://en.wikipedia.org/wiki/Gradient

##\phi[\vec{r}(t)]## is just telling you that ##\phi## is a function of ##\vec{r}## which, in turn, is a function of ##t##. What they've done is parameterized the path represented by the C. Having turned ##\phi## into a function of just one variable, the gradient is much simplified.
HallsofIvy
#3
Dec12-12, 07:51 AM
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They are just using "[ ]" in place of "( )" because they are already using "( )" for the "[itex]\vec{r}(t)[/itex]" and don't want to have "))". There is no difference in meaning.


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