Conservative Fields and Path Independence
A conservative field has a potential whose gradient equals the field, so vector line integrals depend only on endpoints. Testing mixed partials requires smoothness and domain assumptions.
What a learner can do afterwards
- Apply the mixed-partial test with its smoothness and domain assumptions
- Recover and verify a potential function, including the freedom to add a constant
- Explain why the integral around any closed loop of a conservative field is zero
1 · Read
You have integrated F dot dr along a path. A potential f satisfies grad f = F throughout the domain. Here grad f = (f_x, f_y), the two partial derivatives in x, y order. A field with a potential is conservative. In this lesson fields are continuous on open, connected regions. Paths have finitely many smooth pieces and stay in the domain.
Write F = (P, Q), P_y = dP/dy and Q_x = dQ/dx. Suppose P and Q have continuous first partial derivatives on an open region. If every loop there can shrink to a point inside the region, then P_y = Q_x everywhere proves conservativeness. The whole xy-plane meets this domain condition. Unequal mixed partials anywhere rule out a potential on the whole region.
For F = (6xy, 3x² + 4y), integrate P in x: f = 3x²y + g(y). Then f_y = 3x² + g'(y). Matching Q gives g'(y) = 4y, so g(y) = 2y² + C. Thus f = 3x²y + 2y² + C. Check both partials: f_x = 6xy and f_y = 3x² + 4y. They reproduce F everywhere on the plane.
For these polynomial fields, integrate P in x while holding y fixed. Include g(y), whose x derivative is zero. Differentiate the candidate in y and match Q. Integrate the resulting g'(y), then verify both partial derivatives against F throughout the domain. This verification already proves conservativeness; a separate preliminary test is optional. Adding a constant leaves the gradient unchanged.
For F = grad f, the fundamental theorem of line integrals gives integral from A to B = f(B) - f(A). Evaluate f at both endpoints and subtract start from end. Reversing the path changes the sign. On a closed loop, A = B, so the result is zero. Add integrals for consecutive path pieces. Any intermediate potential values cancel, as do constants added to f.
Take f = xy + 2y, whose gradient is F = (y, x + 2). From A = (1, 0) to B = (2, 3), f(A) = 0 and f(B) = 2 × 3 + 2 × 3 = 12. The integral is 12 on every allowed path from A to B. Returning along any path contributes -12, giving zero for the full loop.
Path independence compares every pair of paths with the same endpoints. Equivalently, every closed-loop integral is zero. One zero loop does not prove this. On the plane with the origin removed, a loop surrounding the missing point cannot shrink inside the domain. Equal mixed partials alone then do not prove a global potential. A field there may still be conservative if a potential can be verified throughout its domain.
In a fixed Newtonian gravity field, work done by gravity on an object of fixed mass depends only on its starting and ending positions. Friction and other forces can make total work depend on the route. For a conservative force, physical potential energy U satisfies F = -grad U. Its work from A to B is U(A) - U(B). The mathematical potential f uses F = grad f, so f = -U is a valid choice.
Check the derivative test's assumptions, or verify a potential directly. For a conservative field, the integral is the ending potential value minus the starting value.
Lesson, questions, worked solutions and flashcards written for this dataset. OpenStax and Khan Academy records informed the mathematical checks; WeBWorK records informed question design.
Readings, extra questions, worked solutions and flashcards written for this dataset. Linked videos: Khan Academy, "Path independence for line integrals" and "Closed curve line integrals of conservative vector fields".
More to read
Check the domain before using the test
Let F = (P, Q), with continuous first partial derivatives on an open region in the plane. Compute P_y and Q_x throughout that region. If they differ anywhere, F cannot have a potential valid on the whole region.
For equality everywhere to prove conservativeness, the usual sufficient test also requires that every loop can shrink to a point without leaving the region. The whole plane meets this condition. The plane with the origin removed does not. A loop surrounding the missing point cannot shrink to a point while avoiding it.
A missing point does not make every field nonconservative. It means that matching mixed partials alone no longer settles the global question. For instance, f = x + y has gradient (1, 1) on the whole plane and remains a valid potential after the origin is removed. Directly checking a potential throughout the stated domain can prove conservativeness even when the sufficient test's domain condition is unavailable.
Closed loops give another characterization: a continuous field on an open, connected region is conservative exactly when every closed-loop vector line integral is zero. Finding zero around one chosen loop does not establish this universal claim. In particular, a path followed immediately in reverse has cancelling integrals for any continuous field, whether or not the field is conservative.
Recover the missing part of a potential
A potential must reproduce both field components. For F = (P, Q), the requirements are f_x = P and f_y = Q throughout the domain.
Consider F = (4xy, 2x² + 3) on the whole plane. Integrating the equation f_x = 4xy in x gives f = 2x²y + g(y). Differentiating g(y) in x gives zero, so the first component cannot determine this function of y.
Now differentiate the candidate in y. The result is f_y = 2x² + g'(y). To match Q, set 2x² + g'(y) = 2x² + 3. Subtracting 2x² gives g'(y) = 3, and integration gives g(y) = 3y + C.
Thus f = 2x²y + 3y + C. Check the result: its x derivative is 4xy, and its y derivative is 2x² + 3. This verifies a potential on the whole plane and proves conservativeness without a separate preliminary test.
Different values of C give the same gradient. They also give the same endpoint differences: adding C at both endpoints contributes C - C = 0. A candidate that matches only one component is not yet a verified potential.
Subtract endpoint values in the right order
Suppose f has continuous first partial derivatives and F = grad f. For a path with finitely many smooth pieces, from A to B inside the domain, the fundamental theorem of line integrals gives integral of F dot dr = f(B) - f(A).
Take f = x² + 2y, A = (1, 1) and B = (3, 2). The endpoint values are f(A) = 3 and f(B) = 13. The integral is 13 - 3 = 10. Any allowed return path gives 3 - 13 = -10. Joining the outward and return paths produces a closed loop with total integral zero.
For a path split at an intermediate point M, add the two contributions: [f(M) - f(A)] + [f(B) - f(M)]. The two f(M) terms cancel, leaving f(B) - f(A). This explains why splitting a path does not change its integral.
The sign convention differs for physical potential energy. A conservative force satisfies F = -grad U. Setting f = -U gives work = U(A) - U(B), so positive work by that force corresponds to a decrease in its potential energy.
In a fixed Newtonian gravity field, this gravitational work is independent of the route for an object of fixed mass. Friction and other forces may still make total work depend on the route. Equal gravitational work does not mean equal effort.
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