Register to reply

Not sure what square brackets indicate when dealing with partial derivates

by Kushwoho44
Tags: brackets, dealing, derivates, partial, square
Share this thread:
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.
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
Simon Bridge
#2
Dec12-12, 06:53 AM
Homework
Sci Advisor
HW Helper
Thanks ∞
Simon Bridge's Avatar
P: 12,436
##\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
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,327
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.


Register to reply

Related Discussions
Partial Derivates - Chain Rule Calculus & Beyond Homework 2
What do these wierd square brackets mean? [| |] Calculus & Beyond Homework 3
Partial Derivates using Chain Rule Calculus & Beyond Homework 1
Two square brackets? Quantum Physics 2
Partial Derivates Calculus & Beyond Homework 1