A conservative vector field F has potential function f, with f(A) = 3 and f(B) = 11. A particle travels from A to B along path C₁ (a straight line) and from A to B along path C₂ (a long curved detour). What is the work done along each path?
AWork along C₁ equals 8; work along C₂ is larger because the particle travels a greater distance
BBoth equal 8, since the line integral of a conservative field depends only on the endpoints, not the path
CThe work depends on the particle's speed along each path, not just the endpoints
DWork along C₂ equals 0 because the particle loops before reaching B
The Fundamental Theorem of Line Integrals states that ∫_C F·dr = f(endpoint) − f(startpoint) for any conservative field. Both paths share the same endpoints A and B, so both give f(B) − f(A) = 11 − 3 = 8. The path is completely irrelevant — its length, shape, and direction do not affect the integral. This path-independence is the defining property of conservative fields and is exactly why gravity does the same work lifting a book regardless of the route taken.
Question 2 Multiple Choice
A student verifies that ∂P/∂y = ∂Q/∂x everywhere for a field F = ⟨P, Q⟩ defined on the punctured plane ℝ² \ {(0,0)}. What can she conclude?
AF is conservative — the cross-partial condition is both necessary and sufficient for conservativity
BF is conservative on any path that avoids the origin, since the condition holds wherever F is defined
CF might not be conservative — the cross-partial condition is sufficient only on simply connected domains, and removing the origin creates a topological hole that can prevent this
DF is definitely not conservative, because fields with singularities are never conservative
The cross-partial condition (∂P/∂y = ∂Q/∂x) is necessary for conservativity and is sufficient on simply connected domains — domains with no holes. The punctured plane is not simply connected: the missing origin is a hole. The classic counterexample is F = ⟨−y, x⟩/(x²+y²), which satisfies the cross-partial condition everywhere it is defined yet has a nonzero line integral around any closed curve enclosing the origin. The hole allows a 'winding number' obstruction that makes the field non-conservative despite passing the cross-partial test.
Question 3 True / False
For any conservative vector field F, the line integral ∮_C F·dr around any closed curve C equals zero.
TTrue
FFalse
Answer: True
If F = ∇f, then ∮_C F·dr = f(endpoint) − f(startpoint) by the Fundamental Theorem of Line Integrals. For a closed curve, the endpoint is the same as the starting point, so the integral equals f(p) − f(p) = 0. This zero-integral property for closed curves is equivalent to path independence and is often used as the definition of a conservative field. It generalizes the observation that gravity does zero net work over any closed path.
Question 4 True / False
If ∂P/∂y = ∂Q/∂x at most point in a vector field's domain, then the field is very likely to be conservative, regardless of the shape of the domain.
TTrue
FFalse
Answer: False
The cross-partial condition is sufficient for conservativity only on simply connected domains. On domains with holes — like the punctured plane — the condition can hold everywhere yet the field can still be non-conservative. The field F = ⟨−y, x⟩/(x²+y²) is the standard counterexample: its cross-partials are equal on the punctured plane, but its integral around a unit circle centered at the origin equals 2π, not 0. The topological structure of the domain matters as much as the algebraic condition.
Question 5 Short Answer
Explain in your own words why the existence of a potential function f makes line integrals path-independent. What does the Fundamental Theorem of Line Integrals say, and how is it analogous to the single-variable Fundamental Theorem of Calculus?
Think about your answer, then reveal below.
Model answer: If F = ∇f, then moving through F is like moving through a landscape where f measures elevation. The work done equals the change in elevation — f(endpoint) − f(startpoint) — regardless of the route, because all paths between the same two heights change elevation by the same amount. The Fundamental Theorem of Line Integrals formalizes this: ∫_C F·dr = f(B) − f(A). The analogy to single-variable calculus is exact: ∫_a^b f'(x) dx = f(b) − f(a) because f' is the 'gradient' in 1D, and integration telescopes to endpoint evaluations. In both cases, the antiderivative (or potential function) absorbs all information about the interior of the path.
The key insight is that a potential function reduces a path integral — which naively depends on every point along the route — to two evaluations of a scalar function at the endpoints. This is computationally powerful: instead of parametrizing a complicated curve and computing a vector line integral, you simply find f and evaluate it at start and end. The method works only when F is conservative (F = ∇f), which is why identifying conservative fields is the first step in any line integral problem.