Questions: Conservative Vector Fields and Potential Functions

Question 1

A vector field F = ⟨P, Q⟩ is defined on ℝ² minus the origin. A student checks that ∂P/∂y = ∂Q/∂x everywhere on the domain and concludes that F must be conservative. Is this reasoning valid?

• A. Yes — the mixed partial test is both necessary and sufficient for conservativeness
• B. No — the domain has a hole; the mixed partial test is necessary but not sufficient on non-simply-connected domains
• C. No — the mixed partial test only works in three dimensions, not two
• D. Yes — as long as F is smooth, equal mixed partials guarantee path independence on any domain

Answer: (B)

The mixed partial test (∂P/∂y = ∂Q/∂x) is sufficient only on simply connected domains — regions with no holes where every closed loop can be shrunk to a point. ℝ² minus the origin has a hole. On such domains, a field can satisfy the mixed partial condition everywhere yet still fail to be conservative, because a loop encircling the hole cannot be contracted. The classic example is F = ⟨−y/(x²+y²), x/(x²+y²)⟩, which passes the mixed partial test but gives ∮ F · dr = 2π around the origin.

Question 2

You have confirmed that F is conservative with potential function f. You want to compute the work done by F along a curve from A = (1, 0) to B = (3, 4). Which statement is correct?

• A. You must parameterize the specific curve and evaluate ∫ F · dr directly
• B. The work equals f(B) − f(A), regardless of which path connects A to B
• C. The work is zero, because conservative fields do no work on any path
• D. You need to choose the shortest path to minimize the work integral

Answer: (B)

Path independence is the defining property of a conservative field: the work integral ∫_C F · dr depends only on the endpoints, not the path. With a potential function f in hand, the Fundamental Theorem for Line Integrals gives the work as f(B) − f(A) — no parameterization needed. Option C confuses 'conservative' with 'zero work'; a conservative field does zero work on closed loops, not on all paths. Option A describes the brute-force approach that works for any field but misses the elegance of the conservative case.

Question 3

If F = ∇f is a conservative vector field, then ∮_C F · dr = 0 for every closed curve C in the domain.

• T. True
• F. False

Answer: True

This is one of the four equivalent characterizations of a conservative field. Since going around a closed loop returns to the starting point, and the work equals f(end) − f(start), we get f(A) − f(A) = 0 for any closed curve. This is why conservative fields are physically associated with energy conservation — lifting a mass and returning it to the same height costs zero net work against gravity, which is a conservative field.

Question 4

A conservative vector field has a unique potential function f — there is exactly one f such that F = ∇f.

• T. True
• F. False

Answer: False

The potential function is determined only up to an additive constant. If F = ∇f, then F = ∇(f + C) for any constant C, since ∇C = 0. This means there is an infinite family of valid potential functions differing by constants. To specify a unique potential, you must impose a normalization condition — for example, requiring f = 0 at a reference point. This is directly analogous to indefinite integration, where ∫f dx + C acknowledges the same ambiguity.

Question 5

Explain why the mixed partial test (∂P/∂y = ∂Q/∂x) can fail to identify a non-conservative field, and what condition on the domain ensures the test is sufficient.

Model answer: The test can fail on domains with holes — regions that are not simply connected. On such domains, a field can satisfy ∂P/∂y = ∂Q/∂x everywhere yet still have a non-zero loop integral around a hole, because the hole prevents certain loops from being contracted to a point. The condition that makes the test sufficient is simple connectedness: every closed loop in the domain can be continuously shrunk to a point. On simply connected domains (like all of ℝ² or a convex region), equal mixed partials guarantee a potential function exists and the field is conservative.

The deeper reason is Stokes' theorem: on a simply connected domain, a field with zero curl (∂P/∂y = ∂Q/∂x) integrates to zero around every loop because every loop bounds a region over which the curl integrates. When the domain has a hole, some loops don't bound any region in the domain — the integral around them is not constrained to zero by the curl condition alone.