📝 Conservative vector field test (15 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 15 questions available

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What is Conservative vector field test?

Conservative vector field test:
In 2D, $\mathbf{F} = \langle P, Q \rangle$ is conservative iff $P_y = Q_x$ ; in 3D, iff $\nabla \times \mathbf{F} = \mathbf{0}$ on a simply connected domain.

Example:
For $\mathbf{F} = \langle xy, x^2 \rangle$ , $P_y = x$ , $Q_x = 2x$ , not equal, so not conservative.

Reason:
This test provides a quick check for path independence and potential existence, essential for solving physical problems efficiently.

Easy

Medium

Hard

📝 All Conservative vector field test MCQs

Q1. For a continuously differentiable vector field $F(x,y)=\langle P(x,y),Q(x,y)\rangle$ on a simply connected region, which condition provides the standard local test for conservativeness?

P_x=Q_y

throughout the region

P_y=Q_x

throughout the region ✅

P_x=Q_x

throughout the region

P_y=Q_y

throughout the region

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a continuously differentiable field on a simply connected region, equality of the cross-partials $P_y$ and $Q_x$ is the key test. This condition indicates that the field can be associated with a scalar potential, so line integrals depend only on endpoints rather than the particular path.

Q2. A student claims that $P_y=Q_x$ automatically proves a vector field is conservative everywhere. Which missing condition is most important when the domain has holes or excluded points?

A.The field must have constant magnitude

B.The domain must be simply connected or otherwise satisfy an appropriate global condition ✅

C.The field must be zero at the origin

D.The components must both be linear

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Equality of the cross-partials is a local condition and does not by itself guarantee global path independence when the domain contains holes. A suitable domain condition, such as simple connectivity, prevents circulation around excluded regions from invalidating the conclusion.

Q3. Consider $F(x,y)=\langle 2xy+x^2, x^2+3y^2\rangle$ . A student checks the cross-partials and concludes that the field is conservative on $\mathbb R^2$ . What is the best evaluation of this reasoning?

A.It is correct because

P_y=Q_x=2x

, and the domain has no holes ✅

B.It is incorrect because

P_x

must equal

Q_y

C.It is incorrect because

P

and

Q

must be constants

D.It is correct only when

x=y

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Here $P_y=2x$ and $Q_x=2x$ , so the required equality holds. The domain $\mathbb R^2$ is simply connected, meaning there are no holes that could create global path-dependence problems. Therefore the student's conclusion is justified.

Q4. A force field is modeled by $F(x,y)=\langle 3x^2y, x^3+4y\rangle$ . A particle moves between two fixed points along different smooth paths. What conclusion follows most efficiently after applying the conservative-field test?

A.The work must be computed separately along every path

B.The work is path independent because

P_y=Q_x

\mathbb R^2

✅

C.The work is zero for every path

D.The work depends only on the length of each path

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $P=3x^2y$ , we obtain $P_y=3x^2$ . For $Q=x^3+4y$ , we obtain $Q_x=3x^2$ , so the cross-partials agree. Since the field is defined on all of $\mathbb R^2$ , the domain is simply connected, giving path independence.

Q5. Which field should raise the strongest concern that the cross-partial test alone may be insufficient to establish conservativeness on its entire stated domain?

F=\langle y,x\rangle

\mathbb R^2

F=\langle 2x,2y\rangle

\mathbb R^2

F=\langle -y/(x^2+y^2),x/(x^2+y^2)\rangle

\mathbb R^2\setminus{(0,0)}

✅

F=\langle x^2,y^2\rangle

\mathbb R^2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The third field has matching cross-partials away from the origin, but its domain excludes the origin and therefore contains a hole. This is a classic situation where a local derivative test can pass while global circulation around the missing point prevents the field from being conservative.

Q6. Suppose $F=\langle P,Q\rangle$ is defined on a disk-shaped region. You calculate $P_y-Q_x=0$ at every sampled point on a computer grid, but not analytically. What is the strongest mathematical conclusion?

A.The field is definitely conservative

B.The field is definitely nonconservative

C.The numerical evidence suggests equality but does not constitute a rigorous proof ✅

D.The field must have zero magnitude

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Checking a finite collection of grid points cannot establish an identity throughout a continuous region. Numerical evidence may strongly suggest that $P_y=Q_x$ , but a rigorous conservativeness argument requires an analytic verification of the condition together with appropriate assumptions about differentiability and the domain.

Q7. A field $F=\langle P,Q\rangle$ satisfies $P_y=Q_x$ everywhere in a rectangular region. A second field has the same property only inside a region containing a circular hole. Which comparison is most accurate?

A.Both fields are guaranteed conservative because the derivative condition is identical

B.The rectangular-region field has a stronger global guarantee because the region is simply connected ✅

C.The field with the hole is automatically more conservative

D.Neither field can be conservative because derivatives are involved

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both fields satisfy the local derivative condition, but the domains matter. A rectangle is simply connected, so the local condition extends to a global conservative conclusion under the usual smoothness assumptions. A hole can allow nonzero circulation and therefore requires additional analysis.

Q8. A student tests $P_x=Q_y$ instead of $P_y=Q_x$ and obtains equality for a particular field. Why is this reasoning unreliable?

A.It compares the wrong pair of partial derivatives for the standard two-dimensional test ✅

B.It differentiates both functions with respect to variables they cannot contain

C.It proves the field is always zero

D.It can only be used for three-dimensional fields

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For $F=\langle P,Q\rangle$ , the relevant condition compares the cross-partials $P_y$ and $Q_x$ . Comparing $P_x$ with $Q_y$ examines derivatives in the same respective variables and does not test the required compatibility condition for a scalar potential.

Q9. A graph of a two-dimensional vector field shows arrows that appear tangent to concentric circles centered at the origin, with arrows circulating counterclockwise. The field is defined everywhere except at the center. Which interpretation is most appropriate?

A.The field is likely conservative because the arrows are smooth

B.The missing center is irrelevant because the arrows are continuous elsewhere

C.The circulation pattern and excluded center suggest that the field may fail to be globally conservative ✅

D.The field must have zero circulation because all arrows have equal length

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A circulating vector field around an excluded center is a strong warning sign. Even if the cross-partial condition holds away from the center, a closed curve surrounding the missing point can have nonzero circulation. Thus the topology of the domain must be considered before declaring the field conservative.

Q10. A contour-style sketch shows a vector field whose arrows consistently point perpendicular to nested level curves of a scalar-looking surface, and the arrows reverse direction when moving across certain contours. What does this visual evidence most strongly suggest?

A.The field may be related to a gradient field, but a derivative and domain test is still needed ✅

B.The field cannot be conservative because it crosses contours

C.The field must have constant magnitude

D.The field is necessarily tangent to every contour

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Gradient fields are perpendicular to level curves of their potential functions, so the sketch provides useful qualitative evidence of a possible conservative structure. However, visual appearance alone cannot prove conservativeness. The cross-partial condition and domain properties should be checked mathematically.

Q11. Two methods are proposed for testing a smooth planar field on a simply connected region. Method I compares $P_y$ and $Q_x$ . Method II attempts to find a potential function directly by integrating $P$ with respect to $x$ . Which statement best compares them?

A.Method I is always invalid; only Method II can prove conservativeness

B.Method II is always faster for every field

C.Method I provides an efficient test, while Method II can additionally construct the potential when successful ✅

D.The two methods are unrelated

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: On a simply connected region, comparing $P_y$ and $Q_x$ provides an efficient test for conservativeness. Directly constructing a potential function can require more algebra, but it supplies additional information because the resulting function explicitly describes the scalar potential associated with the field.

Q12. A field is defined on an annular region. Its cross-partials agree everywhere in the annulus. A student says that any two paths with the same endpoints must therefore produce the same line integral. What should an instructor say?

A.Correct, because cross-partial equality always implies path independence

B.Incorrect, because the annulus is not simply connected and closed-loop circulation must be investigated ✅

C.Correct, because annuli have smooth boundaries

D.Incorrect, because line integrals can never be path independent on curved regions

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: An annulus contains a hole, so it is not simply connected. Matching cross-partials establishes a local compatibility condition but does not automatically eliminate circulation around the hole. The student's conclusion is therefore premature; the global topology or circulation around representative closed curves must be examined.

Q13. A modeling team modifies a potential-based force field by adding a rotational component. After the modification, measurements show that $P_y-Q_x$ is nonzero in part of the domain. What is the most defensible conclusion?

A.The modified field remains conservative because the original field was conservative

B.The modified field cannot be conservative on any region containing points where

P_y-Q_x\neq0

✅

C.The modification only changes the potential by a constant

D.The field becomes conservative if the particle moves slowly

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For a sufficiently smooth conservative planar field, the cross-partials must agree throughout the relevant region. If $P_y-Q_x$ is nonzero at any point, this necessary condition fails there. Consequently, no scalar potential can represent the field throughout a region containing that point.

Q14. A field has $P_y=Q_x$ everywhere except along a single curve where the formula for one component changes discontinuously. A student ignores the curve because it has zero area. Why can this be a serious mistake?

A.Zero-area sets never affect line integrals

B.A line path can cross or follow the exceptional curve, so continuity and differentiability assumptions may fail ✅

C.The field must become constant on the curve

D.Only three-dimensional fields can have exceptional curves

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The conservative-field test relies on suitable smoothness assumptions, not merely on area considerations. A curve of discontinuity can directly affect a path because a one-dimensional trajectory may cross it or travel along it. Therefore, ignoring such a curve can invalidate the assumptions needed for the test.

Q15. Consider $F=\langle 2xy+y^2, x^2+2xy\rangle$ on $\mathbb R^2$ . A student notices $P_y=2x+2y$ and $Q_x=2x+2y$ , then claims that the work around every closed curve is zero. Which additional reasoning makes the conclusion valid?

A.The field has equal component values at every point

B.The domain is simply connected, so the matching cross-partials give global conservativeness and hence zero circulation around closed curves ✅

C.The vector field is symmetric in

x

and

y

, which alone guarantees zero work

D.Every closed curve has zero length

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative calculation gives $P_y=Q_x=2x+2y$ , establishing the local condition. The crucial global step is that the domain $\mathbb R^2$ is simply connected. Therefore the field is conservative, and any line integral around a closed curve must vanish, regardless of the curve's shape.

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