Questions: Conservative Vector Fields and Potential Functions

Question 1

For a conservative vector field F, you compute ∫_{C₁} F · dr = 7 along path C₁ from point A to point B. What is ∫_{C₂} F · dr along a completely different path C₂ from A to B?

• A. It cannot be determined without knowing the specific field and path C₂.
• B. −7, because the field reverses orientation when a different path is taken.
• C. 7, because path independence means the line integral depends only on the endpoints.
• D. 0, because conservative fields produce zero work for any path between distinct points.

Answer: (C)

Path independence is the defining property of conservative fields: the line integral between two fixed endpoints is the same regardless of which path connects them. Since both C₁ and C₂ go from A to B, both equal f(B) − f(A) = 7. Option D confuses path independence with the zero-circulation property — a round trip (A to B back to A) gives zero, but a one-way trip from A to B does not.

Question 2

For F = ⟨P, Q⟩ on a simply connected region, which condition is necessary and sufficient to confirm F is conservative?

• A. P and Q are both continuous and positive throughout the region.
• B. The line integral of F along every straight-line path equals zero.
• C. ∂P/∂y = ∂Q/∂x — the cross-partial derivatives of the two components are equal.
• D. F has constant magnitude at every point in the region.

Answer: (C)

The curl-free condition ∂P/∂y = ∂Q/∂x is both necessary and sufficient on a simply connected domain. It is necessary because if F = ∇f, then P = ∂f/∂x and Q = ∂f/∂y, and Clairaut's theorem forces the mixed partials of f to be equal. It is sufficient because on a simply connected region, this condition guarantees a potential function f can be constructed by integration. On non-simply connected regions (like a punctured plane), the condition is necessary but not sufficient.

Question 3

For a conservative field F = ∇f, the line integral from A to B equals f(B) − f(A), regardless of the path taken.

• T. True
• F. False

Answer: True

True — this is the fundamental theorem for line integrals. If F = ∇f, then ∫_C F · dr = ∫_C ∇f · dr = f(endpoint) − f(startpoint). The path cancels entirely; only the values of the potential function at the two endpoints matter. This mirrors the single-variable result ∫_a^b f'(x) dx = f(b) − f(a).

Question 4

If ∮_C F · dr = 0 for one specific closed loop C, then F should be conservative.

• T. True
• F. False

Answer: False

False. Zero circulation around one particular loop does not prove conservatism. A conservative field has zero circulation around every closed loop, not just some. A non-conservative field could produce zero net circulation around a specific path by cancellation. To confirm conservatism, you need either the curl test (∂P/∂y = ∂Q/∂x on a simply connected domain) or verified path independence for all pairs of endpoints.

Question 5

Explain why a conservative vector field has path-independent line integrals — why do only the endpoints matter?

Model answer: A conservative field F is the gradient of a potential function f, so F = ∇f. The fundamental theorem for line integrals then gives ∫_C ∇f · dr = f(endpoint) − f(startpoint) for any path C. This holds because the integral telescopes: intermediate values of f along the path cancel in the telescoping sum, leaving only the difference between the terminal values of f. The path's shape is irrelevant — only where f starts and ends matters.

The physical analogy is gravitational potential energy: moving an object between two heights always requires the same energy regardless of route, because energy is a function of position alone. Conservative vector fields are exactly the force fields with this property — they are gradients of potential energy functions, and line integrals reduce to potential differences.