📝 Flow fields in vector calculus (14 MCQs)

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

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What is Flow fields in vector calculus?

Flow fields in vector calculus:
A flow field $\mathbf{F}$ describes velocity of a fluid; flux through a surface gives net volume flow rate, and divergence indicates sources/sinks.

Example:
For $\mathbf{F} = \langle x, y, z \rangle$ through a sphere, flux = $\nabla \cdot \mathbf{F} \cdot V = 3 \cdot (4\pi/3) = 4\pi$ , matching surface integral.

Reason:
Flow field analysis uses divergence and flux to understand fluid dynamics, aerodynamics, and weather patterns.

Easy

Medium

Hard

📝 All Flow fields in vector calculus MCQs

Q1. A flow field in the plane is represented by $\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle$ . What does the vector $\mathbf{F}(2,-1)$ most directly describe at the point $(2,-1)$ ?

A.The total distance traveled from the origin to

(2,-1)

B.The instantaneous direction and magnitude assigned to a particle located at

(2,-1)

✅

C.The average velocity of every particle in the entire field

D.The acceleration of a particle regardless of its location

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A flow field assigns a vector to each point in space. Therefore, evaluating $\mathbf{F}(2,-1)$ gives the local vector governing motion at that particular location, including both its direction and magnitude. It does not automatically represent accumulated distance, a global average velocity, or acceleration unless the model specifically defines it that way.

Q2. Two vector fields are $\mathbf{F}_1=\langle y,-x\rangle$ and $\mathbf{F}_2=\langle -y,x\rangle$ . Which relationship between their local motions is correct?

A.They always produce identical vectors at every point

B.They produce vectors in opposite directions with equal magnitudes at every point ✅

C.They have equal components but different magnitudes

D.They are perpendicular everywhere except at the origin

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For every point $(x,y)$ , $\mathbf{F}_2=-\mathbf{F}_1$ . Multiplying a vector by $-1$ reverses its direction while preserving its magnitude. Thus, a particle subjected to one field would move instantaneously opposite to the direction predicted by the other, except that both vectors become zero at the origin.

Q3. A fluid model assigns $\mathbf{F}(x,y)=\langle x,2y\rangle$ . At $(1,2)$ , the vector is $\langle1,4\rangle$ , while at $(-1,2)$ it is $\langle-1,4\rangle$ . What is the most meaningful interpretation of this comparison?

A.The fluid moves with the same direction at both points

B.The vertical component is unchanged, while the horizontal component reverses because the

x

-coordinate changes sign ✅

C.The magnitude must be identical because only one coordinate changes

D.The field is constant because the vertical component is unchanged

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The field depends on both coordinates, so changing $x$ from $1$ to $-1$ changes only the first component. The vertical component remains $4$ because $y=2$ is unchanged. This demonstrates how a flow field can produce related but distinct local motions at nearby or symmetric points.

Q4. A student claims that if a flow field has $\mathbf{F}(x,y)=\langle 3,4\rangle$ everywhere, particles must follow curved paths because the field is a two-dimensional vector field. Which evaluation is most accurate?

A.The claim is correct because all two-dimensional fields generate curves

B.The claim is correct because the vector has two nonzero components

C.The claim is incorrect because constant velocity vectors produce straight-line motion ✅

D.The claim is incorrect because particles cannot move in two dimensions under a vector field

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A constant vector field assigns exactly the same velocity vector at every location. A particle therefore maintains both its horizontal and vertical velocity components, giving a trajectory of the form $x=x_0+3t$ , $y=y_0+4t$ . Eliminating $t$ produces a straight line rather than a curved path.

Q5. Consider $\mathbf{F}(x,y)=\langle -y,x\rangle$ . A particle starts at $(2,0)$ . Without solving the full differential equation, which qualitative motion is most consistent with the field?

A.It moves directly outward along the positive

x

-axis

B.It moves directly inward toward the origin

C.It initially moves upward and is consistent with counterclockwise circular motion ✅

D.It initially moves downward and is consistent with clockwise circular motion

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At $(2,0)$ , the field gives $\mathbf{F}(2,0)=\langle0,2\rangle$ , so the particle initially moves upward. More generally, $\langle-y,x\rangle$ is tangent to circles centered at the origin because its dot product with the radial vector $\langle x,y\rangle$ is zero. Thus circular counterclockwise motion is the natural qualitative interpretation.

Q6. A wind field is modeled by $\mathbf{W}(x,y)=\langle 6-y,2x\rangle$ . A drone travels from $(0,1)$ to $(3,1)$ . At which location does the wind have a stronger horizontal component, and why does this matter for path planning?

A.At

(0,1)

, because the horizontal component is

6

B.At

(3,1)

, because the horizontal component is

5

C.At

(0,1)

, because the horizontal component is

7

✅

D.Both locations have the same horizontal component because only

x

controls horizontal motion

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At $(0,1)$ , the horizontal component is $6-1=5$ , whereas at $(3,1)$ it is also $6-1=5$ . Therefore neither location has a stronger horizontal component; the horizontal wind contribution is equal at both points. The tempting value $7$ comes from incorrectly substituting $x=1$ into $6-y$ , while $x=3$ does not affect that component.

Q7. A particle moves according to $\mathbf{F}(x,y)=\langle x,-y\rangle$ . It starts at $(2,3)$ . Which statement best predicts the initial tendency of its trajectory?

A.It moves right and downward, so both coordinates initially move away from zero ✅

B.It moves left and upward, so both coordinates initially move toward zero

C.It moves right and upward, so both coordinates increase

D.It moves left and downward, so both coordinates decrease

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: At $(2,3)$ , the field equals $\langle2,-3\rangle$ . Hence the instantaneous horizontal motion is positive and the vertical motion is negative. The particle initially moves right and downward. Notice that this does not mean both coordinates move away from zero: $x$ increases away from zero while $y$ decreases toward zero.

Q8. A researcher models water velocity by $\mathbf{F}(x,y)=\langle y, -x\rangle$ . At a sampling station located at $(4,-3)$ , the measured velocity is predicted to be $\langle-3,-4\rangle$ . What should the researcher conclude about the direction of motion relative to the origin?

A.The velocity points directly away from the origin

B.The velocity points directly toward the origin

C.The velocity is tangent to the circle centered at the origin through the station ✅

D.The velocity has no relation to the position vector

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The position vector is $\langle4,-3\rangle$ , while the velocity is $\langle-3,-4\rangle$ . Their dot product is $4(-3)+(-3)(-4)=0$ , so the vectors are perpendicular. A velocity perpendicular to the radial direction is tangent to the circle centered at the origin. Thus the flow initially changes angular position rather than radial distance.

Q9. A student analyzes $\mathbf{F}(x,y)=\langle -x,-y\rangle$ at $(3,-4)$ and writes: 'The field points upward because the $y$ -coordinate is negative, so the second component must be positive.' What is the correct diagnosis?

A.The student is correct because the field always points upward when

y<0

B.The student is wrong because the field equals

\langle-3,4\rangle

, which points up and left ✅

C.The student is wrong because the field equals

\langle3,-4\rangle

, which points down and right

D.The student is correct because both components are determined by the coordinates themselves

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Substituting $(3,-4)$ gives $\mathbf{F}(3,-4)=\langle-3,4\rangle$ . The negative sign in the second component changes the negative $y$ -coordinate into a positive component. Therefore the vector points upward and left. The error is not in identifying the sign of $y$ , but in failing to apply the negative coefficient in the field.

Q10. A flow diagram shows vectors along the positive $x$ -axis pointing right, along the negative $x$ -axis pointing left, and vectors above and below the axis pointing approximately toward the $x$ -axis. Which qualitative field best matches this diagram?

\mathbf{F}(x,y)=\langle x,y\rangle

\mathbf{F}(x,y)=\langle -x,-y\rangle

\mathbf{F}(x,y)=\langle x,-y\rangle

✅

\mathbf{F}(x,y)=\langle -y,x\rangle

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: On the positive $x$ -axis, $y=0$ , so $\langle x,-y\rangle=\langle x,0\rangle$ points right. On the negative $x$ -axis it points left. Above the axis, the second component is negative, directing vectors downward; below it, the second component is positive, directing vectors upward. Thus the field tends toward the $x$ -axis, matching the diagram.

Q11. An engineer compares two candidate velocity models for airflow: $\mathbf{F}_1=\langle x,y\rangle$ and $\mathbf{F}_2=\langle -y,x\rangle$ . At a point on the circle $x^2+y^2=25$ , which reasoning correctly distinguishes their local behavior?

A.Both fields are tangent to the circle because both have magnitude

5

\mathbf{F}_1

is radial while

\mathbf{F}_2

is tangent to the circle ✅

\mathbf{F}_1

is tangent while

\mathbf{F}_2

is radial

D.Neither field has a predictable relation to the circle because the point is unspecified

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $\mathbf{F}_1=\langle x,y\rangle$ , the vector is exactly the position vector, so it points radially outward. For $\mathbf{F}_2=\langle-y,x\rangle$ , its dot product with $\langle x,y\rangle$ is zero, making it perpendicular to the radius and therefore tangent to the circle. The distinction is geometric rather than based only on magnitude.

Q12. Suppose $\mathbf{F}(x,y)=\langle 2x,-2y\rangle$ . A particle begins at $(1,1)$ , while another begins at $(-1,1)$ . Which comparison is most accurate about their initial velocities and subsequent qualitative behavior?

A.Their initial velocities are identical and their paths coincide

B.Their initial velocities are opposites, producing mirror-image motion across the

y

-axis ✅

C.Their initial velocities differ only vertically, so their paths must remain parallel

D.Their initial velocities are perpendicular, so the particles immediately move along circles

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: At $(1,1)$ , the velocity is $\langle2,-2\rangle$ ; at $(-1,1)$ , it is $\langle-2,-2\rangle$ . The horizontal components are opposite while the vertical components are equal. This symmetry about the $y$ -axis means the resulting motions exhibit mirror-image behavior rather than identical or perpendicular trajectories.

Q13. A flow field is $\mathbf{F}(x,y)=\langle y,-x\rangle$ . A student argues that because $\mathbf{F}\cdot\langle x,y\rangle=0$ , a particle starting at any point must remain on a circle forever. Which response is mathematically strongest?

A.The conclusion is valid because perpendicular vectors always imply constant speed

B.The conclusion is invalid because perpendicularity only gives zero instantaneous radial component; additional analysis is needed to establish the complete trajectory ✅

C.The conclusion is invalid because the dot product is never zero for this field

D.The conclusion is valid only when

x=y

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The zero dot product shows that the velocity is perpendicular to the position vector at each point, so the instantaneous radial component is zero. For this particular field, further analysis can indeed establish circular motion, but the stated reasoning alone does not prove that conclusion merely from one local orthogonality observation. A complete argument must connect the condition throughout the motion.

Q14. Let $\mathbf{F}(x,y)=\langle x^2-y^2,2xy\rangle$ . At points on the unit circle, the field magnitude is observed to be $1$ . A student concludes that the field therefore has constant direction on the entire circle. Which evaluation is correct?

A.Correct, because constant magnitude always implies constant direction

B.Correct, because every point on the unit circle has the same coordinates

C.Incorrect, because constant magnitude does not prevent direction from changing; for this field the direction varies with position ✅

D.Incorrect, because the magnitude is actually

2

on the unit circle

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For $x^2+y^2=1$ , the squared magnitude is $(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2=1$ , so the magnitude is indeed constant. However, the components depend on $x$ and $y$ , causing the direction to rotate as the point moves around the circle. Constant speed or magnitude does not imply constant direction.

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