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Conservative vector fields also have a special property called the cross-partial property . This property helps test whether a given vector field is conservative.
Let F be a vector field in two or three dimensions such that the component functions of F have continuous second-order mixed-partial derivatives on the domain of F .
If is a conservative vector field in then If is a conservative vector field in then
Since F is conservative, there is a function such that Therefore, by the definition of the gradient, and By Clairaut’s theorem, But, and and thus
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Clairaut’s theorem gives a fast proof of the cross-partial property of conservative vector fields in just as it did for vector fields in
[link] shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won’t have equal cross-partials.
Show that rotational vector field is not conservative.
Let If F is conservative, then the cross-partials would be equal—that is, would equal Therefore, to show that F is not conservative, check that Since and the vector field is not conservative.
Is vector field conservative?
Let and If F is conservative, then all three cross-partial equations will be satisfied—that is, if F is conservative, then would equal would equal and would equal Note that so the first two necessary equalities hold. However, and so Therefore, is not conservative.
We conclude this section with a word of warning: [link] says that if F is conservative, then F has the cross-partial property. The theorem does not say that, if F has the cross-partial property, then F is conservative (the converse of an implication is not logically equivalent to the original implication). In other words, [link] can only help determine that a field is not conservative; it does not let you conclude that a vector field is conservative. For example, consider vector field This field has the cross-partial property, so it is natural to try to use [link] to conclude this vector field is conservative. However, this is a misapplication of the theorem. We learn later how to conclude that F is conservative.
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