Showing that a Vector Field is Conservative

**ILoveTriangles** · March 29th 2012, 03:58 PM

How could I show that the function: F(x,y,z) = <e^(x)*cos(y) +yz, xz-e^(x)*sin(y), xy+z> is conservative?
Much appreciated as I'm lost on where to begin because of 3 space.
An explanation of setting P,Q,R and then what to partial differentiate would be awesome!

(Also what is a potential function?

**HallsofIvy** · March 29th 2012, 07:06 PM

A force field is called "conservative" if it can be derived from a "potential".

Those are, as the reference to "force" might indicate, physics terms. Mathematics terms for the same thing are "exact differential" field which is the derivative of some "anti-derivative".

That is, a vector field, <f(x,y,z), g(x,y,z), h(x,y,z)>, is "conservative" if and only if there exist a function F(x,y,z) such $\frac{\partial F}{\partial x}= f(x,y,z)$ , $\frac{\partial F}{\partial y}= g(x,y,z)$ and $\frac{\partial F}{\partial z}= h(x,y,z)$ .

One way to determine whether or not a given vector field is from such an anti-derivative is to try to find F. Here, $<f, g, h>= <e^xcos(y)+ yz, xz- e^xsin(y), xy+ z>$ so if such an F exists, we must have
$\frac{\partial F}{\partial x}= e^x cos(y)- yz$
Integrating with respect to x (treating y and z as constants) we get
$F= e^x cos(y)+ xyz+ u(y,z)$
Note that, since we are treating y and z as constants, the "constant of integration" may, in fact, be a function of y and z.

Differentiating that with respect to y,
$\frac{\partial F}{\partial y}= -e^x sin(y)+ xz+ u_y= xz- e^x sin(y)$
which tells us that we must have $u_y= 0$ . u is not a function of x and, since its derivative with respect to y is 0 it is not a function of y but it may be a function of z. We now have $F= e^x cos(y)+ xyz+ u(z)$ and differentiating that with respect to z,
$F_z= xy+ u'= xy+ z$
so we must have u'= z and so $u= (1/2)z^2+ C$ where, now, C really is a constant. That means that
$F(x,y,z)= e^x cos(y)+ xyz+ (1/2)z^2+ C$ .

Since the original question was just whether or not the vector field is "conservative" and not what the potential function is, having found that potential function, we can now say, "yes, this is a conservative vector field"!

But because we were only asked whether or not the potential function existed, and not to find it, we could have taken a short cut.

If F is any twice continuously differentiable function, then we have the "mixed derivative" equalities:
$\frac{\partial^2F}{\partial y\partial x}= \frac{\partial^2 F}{\partial x\partial y}$
$\frac{\partial^2F}{\partial x\partial z}= \frac{\partial^2 F}{\partial z\partial x}$
$\frac{\partial^2F}{\partial y\partial z}= \frac{\partial^2 F}{\partial z\partial y}$
where the order of the variables in the 'denominator' indicates the order of the differentiation. In other words, the order of differentiation does not matter.

If there exist F such that $\frac{\partial F}{\partial x}= e^x cos(y)+ yz$ and $\frac{\partial F}{\partial y}= xz- e^x sin(y)$ then we must have
$F_{xy}= -e^x sin(y)+ z$ and $F_{yx}= z- e^x sin(y)$ . Yes, those are the same!

But we have to check the others. If there exist F such that $\frac{\partial F}{\partial x}= e^x cos(y)+ yz$ and $\frac{\partial F}{\partial z}= xy+ z$ then we must have $F_{xz}= y$ and $F_{zx}= y$ . Yes, those are the same!

Finally, if there exist F such that $\frac{\partial F}{\partial y}= xz- e^x sin(y)$ and $\frac{\partial F}{\partial z}= xy+ z$ then we must have
$F_{yz}= x$ and $F_{zy}= x$ . Yes, those are the same also!

Therefore, such an F exists and the vector field is conservative.

**ILoveTriangles** · March 31st 2012, 09:22 AM

Amazingly detailed answer! Thanks so much!
Pretty much helped me on all my potential function questions too

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