- #1

mp6250

- 3

- 0

He defines a simple example in which

[tex] ƒ(x^j) = x^1 [/tex]

and then goes on to write the directional derivative along a vector [itex] v[/itex] as:

[tex] v ⋅ ∇ ƒ(x^j) = v ⋅ ∇ x^1 = v^i ⋅ \delta^1_i = v^1 [/tex]

Next the author says that all we need to do to get the vector out is to feed the corresponding component into the directional derivative

[tex] v^i = v ⋅ ∇ x^i [/tex]

I don't understand what is being said here at all. This formula only works for the first component and only when the function is is [itex] f(x^j) = x^1 [/itex]. If we wanted to get the second component out of the vector we would end up with 0 no matter what the vector actually was. If our function was [itex] f(x^j) = (x^1)^2 [/itex] then we would have

[tex] v ⋅ ∇f(x^j) = v ⋅ (2x^1, 0, 0) [/tex]

and the first component of our vector would depend on the point of evaluation.

Directional derivatives are scalars, I don't understand how you could equate them with vectors. Directional derivatives need context (a function, a point of evaluation, and a direction from that point) but a vector alone needs none of that.

I'm very confused by this section and I think i completely mis-interpret what is trying to be said. What is meant by all this?