How To Tell If Vector Field Is Conservative

In calculus, conservative vector fields take a number of important properties that greatly simplify calculations, including path-independence, irrotationality, and the ability to model phenomena in existent life, such as Newtonian gravity and electrostatic fields. Checking if a vector field is conservative or not is therefore a useful technique to aid with calculations.

1
Use Clairaut's theorem. This theorem states that mixed fractional derivatives commute, given that they are continuous.
- In other words, ${\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}={\frac {\partial }{\fractional 10}}{\frac {\fractional }{\partial y}}.$ Notation that these are 2d derivatives.
two

Consider the part. For our convenience, permit's label $P=2xy^{two}-y^{2}+2{x}$ and $Q=2x^{2}y-5-2xy.$
iii

Compute the partial derivatives.
iv

Check to come across that the mixed partials commute. Our instance patently does. Our vector part is continuous (well-behaved), so this field is conservative. Near fields that yous will deal with, especially in physics, volition merely need to satisfy Clariaut'southward theorem to exist conservative. However, in pure math, this is not always quite the example.

1
Relate conservative fields to irrotationality. Conservative vector fields are irrotational, which means that the field has zilch coil everywhere: $\nabla \times \mathbf {F} =0.$ Considering the curl of a slope is 0, we can therefore limited a conservative field every bit such provided that the domain of said function is simply-connected.
- $\nabla \times \nabla f=0$
- The last condition highlights an important limitation for functions that are not well-behaved. Although all conservative fields are irrotational, the antipodal is not true. Fifty-fifty if the function satisfies Clairaut'southward theorem, it may still not be bourgeois if there exists discontinuities or other singular points.
2

Consider the "vortex" role $\mathbf {v}$ . Above is a visualization of the vortex.
3

Check if this part satisfies Clairaut'southward theorem. It is worth noting that the calculations in this footstep are equivalent to checking if the part is irrotational. Both methods involve the evaluation of the quantity ${\frac {\partial P}{\fractional y}}-{\frac {\partial Q}{\partial ten}},$ or the $\mathbf {thou}$ component of the curl.
4

Verify path-independence using a loop integral. If this field is indeed conservative, then we can say that a loop integral enclosing whatsoever part of the domain is 0. Consider the path of the unit circle in this field.
5
Check if the domain is simply-continued.
- In lodge for a domain to just be connected, whatever 2 points must be able to be connected past a continuous line. The vortex satisfies this, and so its domain is continued.
- To be only-connected, every closed loop in the domain must also have its interior in the domain likewise. The vortex fails this. Since the part is non defined at the origin, the unit circumvolve we made as the closed loop does not take all of its interior within the role's domain.
- Some other way of maxim this is that any closed loop of an arbitrary shape in the domain can be topologically plain-featured to a point in the domain. In other words, we tin can squeeze the loop down to a point. Because the origin is non in the domain of the vortex role, the domain is non just-connected.
- Nosotros accept given an example of a role that satisfies Clairaut's theorem, only ended upwardly declining path-independence anyway. And so for a function to be conservative, its domain must too be but-continued every bit well.