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 $\displaystyle \bold{F}$ is a vector field, then there is some scalar function $\displaystyle f$ (particularly here, one in three variables) such that $\displaystyle \nabla f = \bold{F}$. 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 $\displaystyle \nabla f$ is the gradient vector of a function $\displaystyle f$. it is only appropriate that if the components of $\displaystyle \nabla f$ 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 $\displaystyle \frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$. and there are other things to consider as well. $\displaystyle \bold{F}$ must be defined on an open simply connected region, call it $\displaystyle D$, and $\displaystyle P$ and $\displaystyle Q$ must have continuous first partial derivatives throughout $\displaystyle D$. (of course, here, $\displaystyle \bold {F} = P \bold{i} + Q \bold {j}$).
another way to tell if $\displaystyle F$ is conservative is the following:
again, if $\displaystyle \bold{F}$ is on an open simply-connected region, $\displaystyle D$: if $\displaystyle \oint_C \bold{F} \cdot d \bold{r}$ is independent of path in $\displaystyle D$ (iff $\displaystyle \oint_C \bold{F} \cdot d \bold{r} = 0$ for every closed curve $\displaystyle C$ in $\displaystyle D$) then $\displaystyle \bold{F}$ 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?
O.k. so I got $\displaystyle f=-x^2-\frac{3}{2}y^2+\frac{5}{2}z^2$ because del of this would be the F I wanted. Right?
Also, if curlF=0 means that the vector field is conservative, does Del dot F =0 mean anything.
Also, how do I do the integral, Del, and Dot, symbols in this math type?