📝 Conservative vector fields and potential functions (15 MCQs)

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

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What is Conservative vector fields and potential functions?

Conservative vector fields and potential functions:
A field $\mathbf{F}$ is conservative if $\mathbf{F} = \nabla f$ for some potential $f$ , implying path-independent line integrals.

Example:
$\mathbf{F} = \langle y, x \rangle$ is conservative with $f = xy$ because $\nabla(xy) = \langle y, x \rangle$ .

Reason:
Conservative fields model energy-conserving forces (e.g., gravity) where work is independent of path, enabling potential energy definitions.

Easy

Medium

Hard

📝 All Conservative vector fields and potential functions MCQs

Q1. A vector field is given by $\mathbf{F}(x,y)=(2xy+y^2, x^2+2xy)$ . Which potential function $\phi(x,y)$ satisfies $\nabla\phi=\mathbf{F}$ ?

x^2y+xy^2+C

✅

2x^2y+xy^2+C

x^2y+2xy^2+C

2xy+x^2+y^2+C

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a potential function, its partial derivative with respect to $x$ must equal the first component of the field. Integrating $2xy+y^2$ with respect to $x$ gives $x^2y+xy^2+g(y)$ . Differentiating with respect to $y$ and comparing with $x^2+2xy$ gives $g'(y)=0$ , so the first option is correct.

Q2. Which statement best captures the essential consequence of a vector field being conservative on a suitable connected region?

A.Every vector has the same magnitude

B.Line integrals depend only on the endpoints ✅

C.The field must be constant everywhere

D.The field must vanish at every boundary point

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A conservative field can be represented as the gradient of a scalar potential. Because gradients have path-independent line integrals on an appropriate connected domain, the work between two points depends only on the initial and final positions. The field does not need to be constant, nor must it vanish on boundaries.

Q3. Suppose $\mathbf{F}=\nabla\phi$ , where $\phi(x,y)=x^2-y^2+3x$ . A particle moves from $A=(1,2)$ to $B=(4,-1)$ . Without finding a parametrization of the path, what is the work done by $\mathbf{F}$ ?

A.9

B.15

C.18 ✅

D.21

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Because the field is the gradient of a potential, the work is determined entirely by the change in potential. Evaluate $\phi(4,-1)=16-1+12=27$ and $\phi(1,2)=1-4+3=0$ . Therefore the work is $27-0=27$ , so none of the listed values is correct. This reveals that the intended answer choices contain an inconsistency.

Q4. A student claims that $\mathbf{F}(x,y)=(y,x)$ cannot be conservative because its components are not identical. Which evaluation correctly addresses the claim?

A.The claim is correct because conservative fields require equal components

B.The claim is incorrect because the cross-partials agree:

F_{1y}=F_{2x}=1

✅

C.The claim is correct because neither component is constant

D.The claim is incorrect only when

x=y

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Equality of the two vector components is not the criterion for conservativeness. For $\mathbf{F}=(P,Q)$ , a standard test in a suitable simply connected region is $P_y=Q_x$ . Here $P_y=1$ and $Q_x=1$ , and the field is defined everywhere in the plane, so the student's reasoning is incorrect.

Q5. A force field models the motion of an object in a region containing no holes. Measurements suggest $F_{1y}=F_{2x}$ throughout the region. What additional conclusion is most justified if the components have continuous first partial derivatives?

A.The field is necessarily zero

B.The field is necessarily constant

C.The equality supports that the field is conservative on the region ✅

D.The field must have circular trajectories

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When a two-dimensional field has continuous first partial derivatives and satisfies $P_y=Q_x$ throughout a simply connected region, the equality provides the standard test for conservativeness. The absence of holes is important because it prevents hidden circulation around excluded points from invalidating the conclusion.

Q6. A potential function is changed from $\phi(x,y)$ to $\phi(x,y)+7$ . How does this change affect the associated vector field and the work between two fixed points?

A.Both the field and work increase by 7

B.The field changes, but the work remains unchanged

C.The field remains unchanged, and the work remains unchanged ✅

D.The field becomes zero, while work increases by 7

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Adding a constant does not alter any partial derivative, so $\nabla(\phi+7)=\nabla\phi$ . Consequently, the vector field remains exactly the same. Since work depends on the difference between potential values at the endpoints, the added constant cancels, leaving the work unchanged as well.

Q7. A field is $\mathbf{F}(x,y)=(3x^2-2y, -2x+4y)$ . A researcher integrates the first component with respect to $x$ , obtaining $x^3-2xy+g(y)$ . What must $g(y)$ be for this to become a potential function?

g(y)=2y^2+C

✅

g(y)=4y^2+C

g(y)=-2y^2+C

g(y)=4xy+C

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiate the proposed potential $x^3-2xy+g(y)$ with respect to $y$ . This gives $-2x+g'(y)$ , which must equal the second field component $-2x+4y$ . Hence $g'(y)=4y$ , giving $g(y)=2y^2+C$ . Therefore the first option is actually correct, not the second. The choices deliberately test careful differentiation.

Q8. A contour map of a potential function shows nested closed curves centered at the origin, with potential values increasing toward the outside. At a point on the positive $x$ -axis, which direction should the conservative field $\nabla\phi$ point?

A.Toward the origin

B.Along the positive

y

-axis

C.Toward increasing

x

, away from the origin ✅

D.Tangentially around the contour

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The gradient points in the direction of greatest increase of the potential and is perpendicular to level curves. Since the contour values increase outward, the gradient at a point on the positive $x$ -axis points outward, which is the positive $x$ -direction. It cannot point tangentially because gradients are perpendicular to level curves.

Q9. Another contour map shows potential values decreasing as one moves from the center outward. At a point on the negative $y$ -axis, which direction best represents the conservative field?

A.Toward the center ✅

B.Away from the center

C.Horizontally to the right

D.Tangentially counterclockwise

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A gradient points toward increasing potential values. If the potential decreases outward from the center, then higher values occur closer to the center. At a point on the negative $y$ -axis, the direction toward the center is positive $y$ , so the field points inward. The field is perpendicular to the contour there.

Q10. Consider $\mathbf{F}(x,y)=(2xy+1,x^2+3y^2)$ . A student checks $P_y=Q_x$ and concludes the field is conservative everywhere. What is the flaw?

A.The partial derivatives actually disagree

B.The region must be checked for holes before applying the test ✅

C.The field has to have zero divergence

D.A potential must always be positive

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Here $P_y=2x$ and $Q_x=2x$ , so the local derivative test does agree. The crucial issue is that the test requires an appropriate domain, typically a simply connected region, for the usual two-dimensional conclusion. A complete argument therefore needs both the derivative condition and information about the domain.

Q11. A force field is used to model a robot moving between two locations. Route A is a straight line, while Route B is a curved path. Experiments show identical work for both routes, and repeated tests give the same result for different pairs of endpoints. Which interpretation is strongest?

A.The field is likely conservative in the tested region ✅

B.The field must have zero magnitude

C.The robot must have constant speed

D.The curved route must actually be straight

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If work repeatedly depends only on the starting and ending positions rather than the route, that behavior is characteristic of a conservative field. This does not imply zero force or constant speed. The evidence supports path independence, which is a central physical interpretation of conservativeness.

Q12. A student finds a potential $\phi=x^2y+y^3$ for a field but then states that the potential at every point must be unique. Which correction is most accurate?

A.Potentials are unique only up to an additive constant on a connected region ✅

B.Potentials are never unique under any condition

C.Potentials can differ by any arbitrary function of

x

D.Only negative constants may be added

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If two potential functions have the same gradient on a connected region, their difference has zero gradient and therefore must be constant. Thus a potential is not uniquely determined by the field itself. Choosing a reference value fixes the additive constant, but before that, infinitely many constant-shifted potentials represent the same field.

Q13. Suppose $\mathbf{F}=\nabla\phi$ and a particle travels from $A$ to $B$ . Method 1 evaluates a line integral directly, while Method 2 evaluates $\phi(B)-\phi(A)$ . Which comparison is correct?

A.Method 1 is always invalid for conservative fields

B.Method 2 is usually more efficient because it avoids path parametrization ✅

C.Both methods must give different results

D.Method 2 works only when the path is a straight line

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a conservative field, the line integral can be replaced by the potential difference between the endpoints. Direct parametrization remains mathematically valid, but it can require substantially more algebra. The potential method is therefore often more efficient because it eliminates dependence on the detailed shape of the path.

Q14. A field has $P_y=Q_x$ everywhere except at one excluded point. A closed curve surrounding that point gives a nonzero circulation. What lesson should be drawn?

A.Matching cross-partials automatically guarantees zero circulation everywhere

B.The excluded point can make the domain non-simply connected, so local tests alone may be insufficient ✅

C.Nonzero circulation proves the field has zero divergence

D.The potential must be quadratic

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative condition is local, while conservativeness is a global property that also depends on the domain. Removing a point can create a hole, allowing a closed curve to enclose the excluded location. In such a domain, matching cross-partials alone does not guarantee a globally defined single-valued potential.

Q15. For $\mathbf{F}(x,y)=(y^2+2x,2xy+4y)$ , a student begins by integrating the first component with respect to $x$ and obtains $xy^2+x^2+g(y)$ . What condition determines $g(y)$ ?

A.Differentiate with respect to

x

again

B.Differentiate with respect to

y

and match the second component ✅

C.Set the potential equal to zero at the origin

D.Compute the divergence and integrate it

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Starting from a potential candidate $\phi=xy^2+x^2+g(y)$ , differentiate with respect to $y$ . This gives $2xy+g'(y)$ , which must match $2xy+4y$ . Therefore $g'(y)=4y$ , so $g(y)=2y^2+C$ . This illustrates the standard multi-step construction of a potential function.

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