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Mathematics
Calculus
Directional derivative: identity
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[QUOTE="zinq, post: 5628051, member: 462505"] "I think they're just saying that these are different [I]notations[/I] for the same concept. There is no proof involved in a notation, it's just a convention." No, it is a theorem (though not at all a difficult one), which is in fact proved in the same post that the above quote appears in. At first, we can speak of the derivative of a function, say [INDENT]f(x, y, z), [/INDENT] along a parametrized path [INDENT]A(s) = (x(s), y(s), z(s)),[/INDENT] as defined by d/ds f(A(s))[SIZE=6]|[/SIZE][SUB]s=0[/SUB] = d/ds (f(x(s), y(s), z(s))[SIZE=6]|[/SIZE][SUB]s=0[/SUB] . At this point, for all we know this depends on more than just the derivative A'(0) at s=0 of the path A(s). But then by doing the calculation in #3 ("using the chain rule"), we find out that this derivative depends only on A'(0) (and of course on f(x, y, z) near A(0)). So it is the dot product of the gradient ∇f and the tangent vector A'(0) to the path A(s) at s=0. [/QUOTE]
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Mathematics
Calculus
Directional derivative: identity
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