📝 Independence of Path: Conservative Vector Fields (16 MCQs)

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

←Previous Topic:Line integrals over piecewise smooth curves📝 Independence of Path: Conservative Vector FieldsWork integrals in vector fields:Next Topic→

What is Independence of Path: Conservative Vector Fields?

Independence of Path: Conservative Vector Fields:
A line integral is path-independent if $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{C_2} \mathbf{F} \cdot d\mathbf{r}$ for any two paths between the same endpoints, which occurs iff $\mathbf{F}$ is conservative.

Example:
For $\mathbf{F} = \langle y, x \rangle$ , integral from $(0,0)$ to $(1,1)$ equals 1 for any path.

Reason:
Path independence simplifies calculations and defines potential energy, a fundamental principle in conservative force systems.

Easy

Medium

Hard

📝 All Independence of Path: Conservative Vector Fields MCQs

Q1. A vector field $F$ has the property that every line integral between two fixed points gives the same value, regardless of the smooth path chosen. Which conclusion is most directly justified?

A.The field must have constant magnitude everywhere

B.The field is conservative on the relevant region ✅

C.Every path between the points must have the same length

D.The field must be perpendicular to every possible path

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Path independence means that the integral depends only on the initial and terminal points, not on the route taken. This is precisely the defining behavior associated with a conservative vector field on the region under consideration. Constant magnitude, equal path lengths, and universal perpendicularity are neither necessary nor implied.

Q2. For a conservative vector field $F$ , suppose $F=\nabla \phi$ . A particle moves from $A$ to $B$ along two different smooth paths. What feature of the resulting line integrals is guaranteed?

A.They are equal because both depend only on

\phi(B)-\phi(A)

✅

B.They are equal only when the two paths have equal lengths

C.They are equal only if the paths have identical parameterizations

D.They may differ whenever the paths curve in different directions

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If $F$ is the gradient of a scalar potential $\phi$ , then the line integral from $A$ to $B$ equals $\phi(B)-\phi(A)$ . Since the endpoints are identical for both paths, the potential difference is identical. Therefore, the shape, length, and parameterization of the paths do not change the integral.

Q3. Suppose a field has zero circulation around every closed curve contained in a region. What does this property strongly suggest about line integrals between two points in that region?

A.The integral must depend on the speed used to traverse the curve

B.The integral is path independent within the region ✅

C.Only straight-line paths can produce nonzero integrals

D.The field must have zero magnitude everywhere

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If the integral around every closed curve is zero, two different paths joining the same endpoints can be combined into a closed loop by reversing one path. The total circulation being zero then forces the two path integrals to be equal. Thus, zero circulation around closed curves is closely tied to path independence.

Q4. A field is conservative on a region, and a scalar potential is known to satisfy $\phi(2,1)=7$ and $\phi(5,4)=19$ . A particle moves from $(2,1)$ to $(5,4)$ along a complicated curved path. What is the work done by the field?

A.-12 ✅

B.-26

C.The answer depends on the length of the path

D.The answer depends on the curvature of the path

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a conservative field, the work between two endpoints is determined by the change in potential, $W=\phi(B)-\phi(A)$ . Here the change is $19-7=12$ . The complicated geometry of the path is irrelevant because the field's line integral is independent of the particular route connecting the endpoints.

Q5. Consider $F(x,y)=(2xy+y^2, x^2+2xy)$ . A student claims that the field is not conservative because both components depend on two variables. Which assessment is correct?

A.The student is correct because conservative fields can depend on only one variable per component

B.The student is correct because mixed-variable terms prevent path independence

C.The student is incorrect because the cross-partial derivatives agree, so a potential can exist on a suitable simply connected region ✅

D.The student is incorrect only if the field has constant magnitude

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Dependence on multiple variables does not prevent a field from being conservative. For this field, the relevant cross-partial derivatives are equal: the derivative of the first component with respect to $y$ matches the derivative of the second component with respect to $x$ . On a suitable simply connected region, this supports conservativity.

Q6. A student evaluates $\int_C F\cdot dr$ along one convenient path and obtains $8$ . The student then concludes that the integral along every other path with the same endpoints is also $8$ , without checking any property of $F$ . What is the logical flaw?

A.A convenient path can never be used for a line integral

B.One path determines all other path integrals only when path independence has been established ✅

C.Line integrals always require numerical approximation

D.The endpoints have no role in a line integral

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Computing one path integral does not automatically establish that every other path gives the same value. That conclusion requires a valid reason, such as demonstrating that the field is conservative on the relevant region or otherwise proving path independence. Without such justification, the result applies only to the path actually evaluated.

Q7. A field $F=(P,Q)$ satisfies $P_y=Q_x$ throughout a region containing the origin except that the origin itself is removed. A student concludes immediately that $F$ must be conservative everywhere in that region. Which issue should be investigated first?

A.Whether the field has constant magnitude

B.Whether the region has a hole that could prevent a global potential from existing ✅

C.Whether every curve is a straight line

D.Whether

P

and

Q

have the same sign

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Equality of the cross-partial derivatives is an important local test, but global conservativity can also depend on the topology of the domain. Removing the origin can create a puncture or hole. In such a region, closed-loop circulation may remain nonzero even when the local derivative condition is satisfied, so the domain must be examined.

Q8. A field $F=(y,x)$ is used to calculate work from $A=(0,0)$ to $B=(2,3)$ . Which approach is most efficient if the field's structure suggests a potential function?

A.Compute separate integrals along many possible paths and compare them

B.Find a scalar function

\phi

with

\nabla\phi=F

, then use the endpoint values ✅

C.Approximate the field by a constant vector along the straight segment

D.Multiply the endpoint coordinates and use that as the work

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The components $y$ and $x$ are consistent with the gradient of $\phi(x,y)=xy$ , since $\nabla\phi=(y,x)$ . Once this potential is identified, the work is obtained directly from $\phi(B)-\phi(A)$ , avoiding unnecessary parameterization and integration along a particular path.

Q9. A delivery robot moves from $A$ to $B$ through a force field. Route 1 is a direct road, while Route 2 contains several loops and detours. Measurements show that the field is conservative throughout the accessible region. Which statement best models the work?

A.Route 2 always requires more work because it is longer

B.Route 1 and Route 2 produce the same work because they have the same endpoints ✅

C.The work on Route 2 must be zero because it contains loops

D.The work depends only on the maximum distance from

A

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a conservative force field, work is determined by the initial and final positions. Detours, loops, and route length do not alter the net work between the same endpoints. Thus, provided both routes remain in the region where the field is conservative, the robot experiences identical total work on both routes.

Q10. A student finds a potential candidate $\phi(x,y)=x^2y+y^3$ for $F=(2xy, x^2+3y^2)$ . Another student argues that the candidate cannot be valid because the second component contains both $x$ and $y$ . Which response is best?

A.The argument is correct because a potential's derivatives cannot contain mixed variables

B.The argument is correct because potentials must be linear

C.The argument is incorrect; differentiating

\phi

with respect to

x

and

y

produces exactly the two components of

F

✅

D.The argument is incorrect only when

x=y

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A scalar potential may contain products and powers of several variables. Differentiating $\phi=x^2y+y^3$ gives $\phi_x=2xy$ and $\phi_y=x^2+3y^2$ , exactly matching the field components. Therefore, the presence of mixed variables is not evidence against conservativity or against a valid potential.

Q11. Imagine a diagram showing arrows of a vector field that point generally outward from a central point, with arrow lengths increasing smoothly with distance. Two curves connect the same endpoints, one nearly straight and one highly curved. If the field is known to be $F=\nabla\phi$ , what should the diagram lead you to expect?

A.The curved path must always have greater work

B.The straight path must always have greater work

C.Both paths should give the same work, although the local contributions along them may differ ✅

D.Both paths must have zero work

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A gradient field can produce different local values of $F\cdot dr$ along different routes because the direction of travel changes. However, the total line integral between fixed endpoints depends only on the potential difference. Therefore, the two curves should produce the same total work even though their intermediate contributions differ.

Q12. A graph shows several curves joining points $A$ and $B$ . One student selects the shortest curve to evaluate a line integral, claiming that the shortest path always minimizes the integral. If the field is conservative, how should this claim be evaluated?

A.It is correct because line integrals are proportional to path length

B.It is correct only for curved paths

C.It is incorrect because all paths between

A

and

B

have the same total integral ✅

D.It is incorrect because conservative fields always give zero integrals

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a conservative field, the total line integral between fixed endpoints is independent of the chosen path. Therefore, selecting the shortest geometric curve does not minimize or maximize the integral; every admissible path produces the same total value. The integral is controlled by the endpoints through the potential difference.

Q13. A field is tested on a rectangular loop. The calculated line integral around the loop is $0$ . A student concludes that the field is therefore conservative throughout the entire plane. Which conclusion is most defensible?

A.The conclusion is guaranteed because one closed loop is sufficient

B.The conclusion is guaranteed if the rectangle has positive area

C.The result provides evidence but is insufficient by itself to establish global conservativity ✅

D.The result proves that the field is identically zero

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A single closed-loop integral equal to zero does not establish that every possible closed loop has zero circulation. Global conservativity requires stronger evidence, such as path independence for all relevant paths, existence of a global potential, or suitable derivative and domain conditions. The field also need not be zero.

Q14. Suppose a field has potential $\phi(x,y)=x^2+xy-y^2$ . A student computes the work from $A=(1,2)$ to $B=(3,-1)$ by integrating separately along a horizontal segment and then a vertical segment. Another student evaluates only $\phi(B)-\phi(A)$ . Which comparison is correct?

A.Only the first method can be valid

B.Only the second method can be valid

C.Both methods can be valid, but the potential method is generally more efficient because path independence eliminates the need to parameterize the route ✅

D.Neither method is valid because the field is not constant

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The gradient of the given potential defines a conservative field, so any path connecting the endpoints gives the same work. The piecewise horizontal-then-vertical calculation is valid if performed correctly, but it requires parameterization and integration. Evaluating $\phi(B)-\phi(A)$ reaches the same result more directly.

Q15. A researcher claims that a vector field is conservative because numerical calculations along two randomly chosen paths between $A$ and $B$ agree to six decimal places. What is the strongest criticism of this reasoning?

A.Numerical calculations can never be used in mathematics

B.Agreement for two paths is suggestive but does not prove independence for all admissible paths ✅

C.Two paths can never share the same endpoints

D.Conservative fields cannot be tested computationally

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Numerical agreement along two paths provides useful evidence but cannot establish a universal property by itself. Path independence concerns every admissible path in the relevant domain. A rigorous argument would require structural information about the field, such as a potential function or appropriate conditions that guarantee conservativity.

Q16. Let $F=(P,Q)$ be smooth on a simply connected region, and suppose $P_y=Q_x$ everywhere in that region. Which conclusion follows most strongly from these conditions?

A.Every line integral is zero

B.The field is conservative and line integrals are independent of path within the region ✅

C.The field has constant magnitude

D.Every path between two points has identical geometric length

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: In a simply connected region, the absence of holes removes the main global obstruction that can separate local derivative information from global path independence. If the cross-partial derivatives agree throughout the region and the field is sufficiently smooth, the field is conservative there. Consequently, its line integrals depend only on endpoints.

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