How do I tell if a Vector Field is a Gradient Vector Field?
I have the F=-2xi-3yj+5zk
Does it have something to do with it being conservative or not?
not exactly. conservative vector fields are a special kind of vector field. this problem is just asking you whether F is a vector field or not.
if is a vector field, then there is some scalar function (particularly here, one in three variables) such that . you are required to find such a function. can you do it?
well, now you know. it just so happens that such a function f exists if F is a conservative vector field. but there are ways to prove F is conservative without going through the trouble of finding f
just remember that is the gradient vector of a function . it is only appropriate that if the components of are functions, that it be called a gradient vector field, since that's what vector fields are
What are the other ways of determining if it's conservative?
Can't you show that it is conservative by showing that partial P/dx, and partial Q/dy are equal? by Clairaut's Theorem (I don't think I spelled that right)? Are there other way's?
you mean . and there are other things to consider as well. must be defined on an open simply connected region, call it , and and must have continuous first partial derivatives throughout . (of course, here, ).
another way to tell if is conservative is the following:
again, if is on an open simply-connected region, : if is independent of path in (iff for every closed curve in ) then is conservative.
those are the only two ways i remember.
EDIT: Oh, Plato gave another way
Yes I do. I have used the notation for F with classes for years.
The capital I, J, and K are function fields that go naturally with the vector notation i, j, and k.
Conservative means that the curl(F)=0 so the determinate is zero.
Note that in each case the corresponding variable is missing in the partials.
Dose that help?