# Not sure what square brackets indicate when dealing with partial derivates

by Kushwoho44
Tags: brackets, dealing, derivates, partial, square
 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.
 Homework Sci Advisor HW Helper Thanks ∞ PF Gold P: 10,913 ##\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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,877 They are just using "[ ]" in place of "( )" because they are already using "( )" for the "$\vec{r}(t)$" and don't want to have "))". There is no difference in meaning.

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