📝 Integration a Vector Field Along a Curve (14 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 14 questions available

←Previous Topic:Line integrals with respect to x y z📝 Integration a Vector Field Along a CurveWorking a Line Integral:Next Topic→

What is Integration a Vector Field Along a Curve?

Integration a Vector Field Along a Curve:
This is the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ , summing the tangential component of $\mathbf{F}$ along the curve, representing work done by the field.

Example:
Work by $\mathbf{F} = \langle y, -x \rangle$ along a circle $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ , $0 \le t \le 2\pi$ , is $\int_0^{2\pi} (-1) dt = -2\pi$ .

Reason:
This integral quantifies energy transfer in force fields, a core concept in mechanics and electromagnetism.

Easy

Medium

Hard

📝 All Integration a Vector Field Along a Curve MCQs

Q1. A particle moves along the curve $C$ given by $\mathbf r(t)=(t,t^2)$ , $0\le t\le1$ , in the vector field $\mathbf F(x,y)=\langle y,x\rangle$ . Which expression correctly represents the work done by the field along the motion?

\int_0^1 (t^2+2t^2)\,dt

✅

\int_0^1 (t^2+2t)\,dt

\int_0^1 (t+t^2)\,dt

\int_0^1 (2t+t^2)\,dt

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a vector field, the work integral is $\int_C\mathbf F\cdot d\mathbf r$ . Here $\mathbf r'(t)=\langle1,2t\rangle$ , while $\mathbf F(\mathbf r(t))=\langle t^2,t\rangle$ . Their dot product is $t^2+2t^2=3t^2$ , giving option A. The key step is substituting the curve before taking the dot product.

Q2. Two students parameterize the same directed curve differently. Student 1 uses $\mathbf r_1(t)$ , while Student 2 uses $\mathbf r_2(s)$ with $s$ increasing in the opposite direction. How should their vector-field line integrals compare?

A.They must always be equal

B.They must have opposite signs ✅

C.They are equal only for constant fields

D.Their magnitudes must be different

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A vector-field line integral depends on orientation because $d\mathbf r$ changes sign when the direction of traversal is reversed. Therefore, reversing the direction of the same curve changes $\int_C\mathbf F\cdot d\mathbf r$ into its negative. This is fundamentally different from scalar arc-length integrals, which are unaffected by orientation.

Q3. A force field is $\mathbf F(x,y)=\langle2x,2y\rangle$ . A particle travels from $(1,0)$ to $(0,1)$ along any smooth curve that stays in the plane. Which conclusion is justified without knowing the exact path?

A.The work depends only on the curve length

B.The work is zero because the endpoints are perpendicular

C.The work is the same for every such path ✅

D.The work depends on the speed of traversal

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The field $\mathbf F=\langle2x,2y\rangle$ is the gradient of the potential $f(x,y)=x^2+y^2$ . Hence the work is path independent and equals $f(0,1)-f(1,0)=1-1=0$ . The important reasoning is recognizing a conservative field rather than attempting to parameterize an unspecified path.

Q4. A student evaluates $\int_C\mathbf F\cdot d\mathbf r$ by calculating $\int_C|\mathbf F|\,ds$ . Which statement best identifies the student's error?

A.The student used a scalar integral instead of a vector integral and ignored the directional component of the field ✅

B.The student forgot to reverse the curve orientation

C.The student should always use

\mathbf F\times d\mathbf r

D.The student's method is correct for every vector field

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The expression $\int_C|\mathbf F|\,ds$ measures accumulated field magnitude along the path, whereas work is $\int_C\mathbf F\cdot d\mathbf r$ . The dot product includes the component of the force in the direction of motion. Replacing it with magnitude discards directional information and can produce a completely different result.

Q5. A graph shows a vector field whose arrows are generally tangent to a curve $C$ and point in the same direction as the curve's orientation. Which qualitative prediction is most reasonable for $\int_C\mathbf F\cdot d\mathbf r$ , assuming the field is nonzero along most of $C$ ?

A.It should tend to be positive ✅

B.It must be exactly zero

C.It must be negative

D.Its sign cannot be inferred at all

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The dot product $\mathbf F\cdot d\mathbf r$ is positive when the field and displacement point in generally the same direction. If the graph shows vectors predominantly tangent to the oriented curve and pointing forward, most contributions to the integral are positive. Exact magnitude still requires information about field strength and curve geometry.

Q6. A student claims that if a vector field is perpendicular to the tangent vector at every point of a curve, then the field necessarily has zero magnitude along the curve. What is the best evaluation of this claim?

A.Correct, because perpendicular vectors must both be zero

B.Correct, because the tangent vector determines field magnitude

C.Incorrect, because perpendicularity makes the work contribution zero but does not imply zero field magnitude ✅

D.Incorrect, because perpendicularity makes the work positive

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: If $\mathbf F$ is perpendicular to the tangent direction $d\mathbf r$ , then $\mathbf F\cdot d\mathbf r=0$ , so the work contribution vanishes. However, a nonzero vector can certainly be perpendicular to another nonzero vector. Thus, zero line integral contribution does not imply that the vector field itself is zero along the curve.

Q7. A cyclist moves along a path from $A$ to $B$ . On the first half, the force generally assists the motion; on the second half, it generally opposes the motion. The assisting force is stronger than the opposing force over comparable distances. Which conclusion is most plausible?

A.The total work is necessarily zero

B.The total work is likely positive ✅

C.The total work is necessarily negative

D.The total work cannot depend on field direction

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Work is the accumulated dot product between force and displacement. Positive contributions occur where the force assists the motion, while negative contributions occur where it opposes the motion. If the positive component is stronger over comparable portions of the trajectory, the net integral is likely positive. Exact equality cannot be concluded without quantitative information.

Q8. A force field is $\mathbf F(x,y)=\langle y,-x\rangle$ , and a particle traverses the unit circle counterclockwise once. A student argues that the work is zero because the particle returns to its starting point. Which response is correct?

A.The student is correct because every closed curve has zero work

B.The student is correct because the field has constant magnitude

C.The student is incorrect because returning to the starting point does not guarantee path-independent work ✅

D.The student is incorrect because every vector field gives positive work on a circle

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Returning to the starting point guarantees zero net displacement, but a line integral of a vector field is not generally determined solely by endpoint displacement. The field $\langle y,-x\rangle$ circulates around the origin, and its direction has a consistent relationship with the counterclockwise tangent direction. Therefore, the closed-curve work need not vanish.

Q9. A curve is parameterized by $\mathbf r(t)=\langle 2\cos t,2\sin t\rangle$ , $0\le t\le\pi$ . If a vector field is always directed radially outward from the origin, what can be concluded about $\mathbf F\cdot\mathbf r'(t)$ along this curve?

A.It is generally zero ✅

B.It is always negative

C.It is always positive

D.It alternates sign exactly once

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The tangent vector to a circle is perpendicular to the radial direction. A radially outward field is parallel to the position vector, while $\mathbf r'(t)$ is tangent to the circle. Hence the two vectors are perpendicular, making their dot product zero at every point. Consequently, the vector field contributes no work along this circular path.

Q10. Two possible routes connect the same warehouse locations. Route 1 is short but moves mostly against a prevailing force field. Route 2 is longer but follows the field direction. Which statement best explains why Route 2 could require less external work against the field?

A.Line integrals depend only on distance traveled

B.The dot product can make field work more favorable when motion aligns with the field ✅

C.Longer paths always produce smaller work

D.A vector field has no effect on route selection

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The work of a vector field depends on the dot product $\mathbf F\cdot d\mathbf r$ , not merely on path length. Motion aligned with the field produces positive field work, while motion against it produces negative field work. Therefore, a longer path can still be energetically more favorable depending on the direction of motion relative to the field.

Q11. Suppose a graph shows a vector field with arrows pointing upward on the left side of a closed curve and downward on the right side. The curve is traversed counterclockwise. Which reasoning is most appropriate when estimating the sign of the total work?

A.Only arrow magnitude matters

B.Only curve length matters

C.Compare the field direction with the tangent direction on each segment before combining contributions ✅

D.A closed curve always has zero work

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The sign of each contribution comes from the dot product between the field vector and the curve's directed tangent. Along different portions of a closed curve, the tangent direction changes, so the same field orientation may assist motion on one segment and oppose it on another. A qualitative graph must therefore be analyzed segment by segment.

Q12. For a vector field $\mathbf F=\langle P,Q\rangle$ , a student writes the line integral along $x=g(t)$ , $y=h(t)$ as $\int P(x,y)\,dt+\int Q(x,y)\,dt$ . Which correction is essential?

A.Replace both terms with

P+Q

multiplied by arc length

B.Use

\int[P(g(t),h(t))g&#039;(t)+Q(g(t),h(t))h&#039;(t)]\,dt

✅

C.Use

\int[P(g(t),h(t))+Q(g(t),h(t))]\,ds

D.Integrate only the first component because

x

is the independent variable

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a vector-field line integral, the differential displacement is $d\mathbf r=\langle dx,dy\rangle$ . Under the parameterization, $dx=g'(t)dt$ and $dy=h'(t)dt$ . Therefore, $\mathbf F\cdot d\mathbf r=P\,dx+Q\,dy$ , producing the stated expression. Omitting $g'(t)$ and $h'(t)$ loses the directional geometry of the curve.

Q13. A vector field gives zero line integral around every closed curve contained in a simply connected region. A student concludes that the field must be zero everywhere in that region. Which conclusion is more accurate?

A.The student is correct

B.The field may be nonzero but can still have zero circulation because it may be conservative ✅

C.The field must have constant magnitude

D.The field must be perpendicular to every curve

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Zero circulation around every closed curve in an appropriate simply connected region is consistent with a conservative field, not necessarily a zero field. For example, a gradient field can be nonzero while its integral around every closed path is zero. The crucial distinction is between having zero circulation and having zero vector magnitude.

Q14. A path is split into three oriented pieces $C_1,C_2,C_3$ . A numerical model estimates their work contributions as $4.2$ , $-7.5$ , and $3.8$ units. What is the total work, and what does its sign indicate?

15.5

units; the field strongly assists the motion

0.5

units; the assisting contributions slightly exceed the opposing contribution ✅

-0.5

units; the opposing contribution slightly exceeds the assisting contributions

-15.5

units; the field strongly opposes the motion

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The total line integral is obtained by adding the contributions from each oriented segment: $4.2+(-7.5)+3.8=0.5$ . The positive result means the field performs a small net amount of positive work along the complete path. This illustrates why segment-by-segment analysis is useful when field direction changes along a trajectory.

🔗 Related Topics (MCQs)

📝 Vector Fields in Calculus MCQs 📝 Graphical Representations of Vector Fields MCQs 📝 Vector field notation and components MCQs 📝 Inverse square vector fields MCQs 📝 Gradient vector fields MCQs

←Previous Topic:Line integrals with respect to x y z📝 Integration a Vector Field Along a CurveWorking a Line Integral:Next Topic→