Antiderivatives of vector functions: 25 Practice MCQs with Answers and Explanation

Q1. A particle moves with velocity $\mathbf{v}(t) = \langle 2t, e^t, -\sin t \rangle$ . If its position at $t=0$ is $\mathbf{r}(0) = \langle 1, 0, 2 \rangle$ , which expression correctly models the displacement vector from $t=0$ to $t=\pi$ ?

A.

\int_0^\pi \mathbf{v}(t)\,dt + \mathbf{r}(0)

B.

\int_0^\pi \|\mathbf{v}(t)\|\,dt

C.

\mathbf{r}(\pi) - \mathbf{r}(0)

where

\mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C}

and

\mathbf{C}

is solved via initial condition ✅

D.

\int_0^\pi \mathbf{v}&#039;(t)\,dt

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Displacement is defined as the net change in position $\mathbf{r}(\pi) - \mathbf{r}(0)$ . While option A computes the final position, not displacement. Option B calculates arc length (distance traveled), a common misconception. The correct approach requires finding the specific antiderivative satisfying the initial condition, then evaluating the difference, demonstrating multi-step reasoning involving both integration and vector subtraction.

Q2. Given $\mathbf{a}(t) = \langle 6t, 2, 0 \rangle$ and initial conditions $\mathbf{v}(0)=\langle 1,-1,3 \rangle$ , $\mathbf{r}(0)=\langle 0,0,0 \rangle$ , a student claims $\mathbf{r}(t) = \langle t^3+t, t^2-t, 3t \rangle$ . What is the fundamental error in this derivation?

A.The student integrated acceleration directly to position without finding velocity first.

B.The student used scalar integration constants instead of vector constants for each component.

C.The student correctly found velocity but failed to apply the initial position condition when integrating velocity to get position. ✅

D.The student differentiated instead of integrated when moving from acceleration to velocity.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This error analysis question targets the critical two-step process required for second-order vector antiderivatives. The student's velocity $\mathbf{v}(t)=\langle 3t^2+1, 2t-1, 3 \rangle$ is correct. However, integrating this yields $\mathbf{r}(t)=\langle t^3+t+C_1, t^2-t+C_2, 3t+C_3 \rangle$ . Applying $\mathbf{r}(0)=\mathbf{0}$ gives all constants as zero, making the answer coincidentally correct numerically but the reasoning flawed if constants were non-zero. The explanation clarifies that skipping the constant determination step is a systematic failure in modeling physical systems.

Q3. If $\mathbf{F}'(t) = \langle \cos t, \sec^2 t, \frac{1}{1+t^2} \rangle$ and $\mathbf{F}(0) = \langle 0, 0, 0 \rangle$ , what is the geometric significance of the third component of $\mathbf{F}(\pi/4)$ ?

A.It represents the area under

y=\frac{1}{1+t^2}

from 0 to

\pi/4

, equaling

\arctan(\pi/4)

. ✅

B.It represents the slope of the tangent line to the curve at

t=\pi/4

.

C.It represents the total distance traveled along the z-axis, requiring absolute value integration.

D.It represents the curvature of the vector function at that specific instant.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This conceptual question links vector antiderivatives to single-variable calculus geometry. Since $F_z'(t) = \frac{1}{1+t^2}$ , the antiderivative is $\arctan(t)+C$ . With $F_z(0)=0$ , $C=0$ . Evaluating at $\pi/4$ gives $\arctan(\pi/4)$ , which by the Fundamental Theorem of Calculus equals the signed area under the derivative curve. This reinforces that vector antiderivatives are computed component-wise and retain their scalar geometric interpretations, distinguishing them from magnitude-based quantities like distance or curvature.

Q4. A drone’s acceleration is modeled by $\mathbf{a}(t) = \langle -4\cos(2t), -4\sin(2t), 0 \rangle$ . If it starts from rest at the origin, which statement best describes the relationship between the velocity vector $\mathbf{v}(t)$ and position vector $\mathbf{r}(t)$ obtained through successive antidifferentiation?

A.

\mathbf{v}(t)

is always perpendicular to

\mathbf{r}(t)

because the motion is circular.

B.

\mathbf{v}(t)

and

\mathbf{r}(t)

are parallel at all times due to harmonic motion.

C.

\mathbf{v}(t) \cdot \mathbf{r}(t) = 0

only at discrete points, despite uniform circular acceleration. ✅

D.The dot product

\mathbf{v}(t) \cdot \mathbf{r}(t)

is constantly negative, indicating inward spiral motion.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This challenging problem requires computing $\mathbf{v}(t) = \langle -2\sin(2t), 2\cos(2t)-2, 0 \rangle$ and $\mathbf{r}(t) = \langle \cos(2t)-1, \sin(2t)-2t, 0 \rangle$ using initial conditions. Their dot product simplifies to $-2\sin(2t)(\cos(2t)-1) + (2\cos(2t)-2)(\sin(2t)-2t)$ , which is not identically zero. Students often assume circular acceleration implies circular position, but the ‘starts from rest’ condition introduces phase shifts and linear drift terms. This tests deep understanding that antiderivatives encode initial state information that fundamentally alters geometric relationships beyond the derivative’s form.

Q5. Consider the graph of $\|\mathbf{v}(t)\|$ versus $t$ showing a symmetric bell-shaped curve from $t=0$ to $t=2$ . Without knowing $\mathbf{v}(t)$ ’s direction, what can be definitively concluded about the antiderivative $\mathbf{r}(t)$ ?

A.The magnitude

\|\mathbf{r}(2) - \mathbf{r}(0)\|

equals the area under the

\|\mathbf{v}(t)\|

curve.

B.The total path length from

t=0

to

t=2

equals the area under the

\|\mathbf{v}(t)\|

curve. ✅

C.The displacement vector

\mathbf{r}(2) - \mathbf{r}(0)

must be zero due to symmetry.

D.The average velocity vector over [0,2] has magnitude equal to half the peak speed.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This graph-based question distinguishes between scalar and vector accumulation. The area under $\|\mathbf{v}(t)\|$ always gives total distance traveled (path length), regardless of directional changes. Displacement magnitude (option A) requires knowledge of direction; symmetry in speed doesn’t imply zero displacement (option C) unless direction also reverses symmetrically. Average velocity magnitude isn’t simply half the peak (option D) without knowing the exact functional form. Only option B holds universally, testing conceptual understanding that antiderivatives of speed yield arc length, while antiderivatives of velocity vectors yield displacement—two fundamentally different quantities often confused in graphical interpretation.

Q6. When computing $\int \langle t e^t, \ln t, \sqrt{1-t^2} \rangle \, dt$ , which component presents the greatest risk of domain-related errors during antidifferentiation, and why?

A.The first component, because integration by parts may introduce extraneous constants.

B.The second component, because

\int \ln t \, dt = t\ln t - t + C

is undefined for

t \leq 0

, restricting the entire vector function’s domain.

C.The third component, because trigonometric substitution assumes

t \in [-1,1]

, but the antiderivative’s domain must match the original integrand’s domain.

D.All components equally, since vector antiderivatives require consistent domains across all entries. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: This error analysis question emphasizes that a vector-valued antiderivative exists only where all components are simultaneously defined. While individual antiderivatives have known forms, the vector function’s domain is the intersection of all component domains: $t>0$ (from ln t) AND $t \in [-1,1]$ (from sqrt), yielding $t \in (0,1]$ . Students often compute antiderivatives component-wise without checking domain compatibility, leading to invalid expressions outside this interval. This mixed-concept problem integrates calculus techniques with set theory, highlighting that vector antidifferentiation imposes stricter constraints than scalar cases.

Q7. A student computes $\int \langle \sin t, \cos t, t \rangle \, dt = \langle -\cos t, \sin t, t^2/2 \rangle + \mathbf{C}$ and applies $\mathbf{F}(0) = \langle 1, 0, 0 \rangle$ to get $\mathbf{C} = \langle 2, 0, 0 \rangle$ . Is this solution valid for modeling physical position?

A.Yes, because the constant vector satisfies the initial condition algebraically.

B.No, because

\mathbf{C}

must be determined component-wise; here

C_y

should satisfy

\sin(0) + C_y = 0 \Rightarrow C_y=0

, which is correct, but

C_z

should satisfy

0 + C_z = 0 \Rightarrow C_z=0

, so

\mathbf{C}=\langle 2,0,0\rangle

is actually correct. ✅

C.No, because the antiderivative of

\cos t

is

\sin t + C

, and at

t=0

,

\sin(0)+C_y=0

implies

C_y=0

, but the student’s method is sound; the issue is elsewhere.

D.The solution is mathematically correct but physically meaningless without units; however, within pure mathematics, it is valid.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This question tests meticulous application of initial conditions to vector antiderivatives. The student’s work is actually correct: $-\cos(0)+C_x=1 \Rightarrow C_x=2$ ; $\sin(0)+C_y=0 \Rightarrow C_y=0$ ; $0+C_z=0 \Rightarrow C_z=0$ . Distractors exploit common fears about vector constants. The explanation confirms validity while reinforcing that each component’s constant is independent and must satisfy its own initial value equation. This prevents overcomplication and builds confidence in systematic component-wise solving, addressing misconceptions that vector constants behave differently than scalar ones.

Q8. Suppose $\mathbf{r}''(t) = \langle 0, 0, -g \rangle$ with $g>0$ , $\mathbf{r}'(0)=\langle v_0\cos\theta, 0, v_0\sin\theta \rangle$ , and $\mathbf{r}(0)=\langle 0,0,h \rangle$ . Which modification to the antiderivative process would model air resistance proportional to velocity?

A.Replace

-g

with

-g - k\|\mathbf{r}&#039;(t)\|

in the z-component before integrating.

B.Change the problem to a differential equation

\mathbf{r}&#039;&#039;(t) = \langle 0,0,-g \rangle - k\mathbf{r}&#039;(t)

, making direct antidifferentiation impossible. ✅

C.Add a term

-k\mathbf{r}(t)

to the acceleration vector before taking antiderivatives.

D.Integrate

\langle 0,0,-g \rangle

normally, then subtract

k\int \mathbf{r}&#039;(t)\,dt

as a correction factor.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This application question bridges antiderivatives and differential equations. Air resistance makes acceleration depend on velocity, creating $\mathbf{r}'' + k\mathbf{r}' = \langle 0,0,-g \rangle$ , which cannot be solved by simple antidifferentiation since $\mathbf{r}'$ is unknown. Options A, C, and D incorrectly treat resistance as an explicit function of time or position, ignoring its velocity dependence. Only B recognizes that the problem transcends basic antiderivative techniques, requiring ODE methods. This tests higher-order thinking by identifying when standard calculus tools fail, a crucial skill in mathematical modeling where idealized antiderivative problems meet real-world complexity.

Q9. Given $\mathbf{F}'(t) = \langle f(t), g(t), h(t) \rangle$ where $f,g,h$ are continuous on [a,b], and $\mathbf{G}(t)$ is another antiderivative of the same function, which statement must be true?

A.

\mathbf{F}(t) - \mathbf{G}(t) = \mathbf{0}

for all t in [a,b].

B.

\mathbf{F}(t) - \mathbf{G}(t) = \mathbf{C}

, a constant vector, for all t in [a,b]. ✅

C.

\|\mathbf{F}(t) - \mathbf{G}(t)\|

is constant, but the vector difference may vary in direction.

D.

\mathbf{F}(b) - \mathbf{G}(b) = \mathbf{F}(a) - \mathbf{G}(a)

, but intermediate differences may fluctuate.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This foundational recall question establishes the vector analogue of the scalar antiderivative uniqueness theorem. Since differentiation is linear and component-wise, if two vector functions have identical derivatives on an interval, their difference has zero derivative everywhere, implying each component difference is constant. Thus, the entire vector difference is a constant vector. Options C and D confuse vector constancy with magnitude constancy or endpoint equality, common misconceptions when transitioning from scalar to vector calculus. Mastery of this principle is essential for correctly applying initial conditions and understanding solution spaces in vector differential equations.

Q10. A spacecraft’s velocity is $\mathbf{v}(t) = \langle e^{-t}\cos t, e^{-t}\sin t, 0 \rangle$ . To find total displacement from $t=0$ to $\infty$ , a student writes $\lim_{b\to\infty} \int_0^b \mathbf{v}(t)\,dt$ . What hidden assumption makes this improper integral well-defined?

A.That

\mathbf{v}(t)

is absolutely integrable, ensuring convergence of each component’s improper integral.

B.That the limit exists component-wise, which holds here since

\int_0^\infty e^{-t}\cos t\,dt

and

\int_0^\infty e^{-t}\sin t\,dt

both converge.

C.That

\|\mathbf{v}(t)\|

decays faster than

1/t

, guaranteeing path length finiteness.

D.That the antiderivative approaches a finite limit vector as

t\to\infty

, which is equivalent to component-wise convergence. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: This conceptual question probes understanding of improper integrals for vector functions. Convergence is defined component-wise: the vector integral converges iff each scalar component integral converges. Option D correctly identifies that existence of $\lim_{t\to\infty} \mathbf{R}(t)$ (where $\mathbf{R}'=\mathbf{v}$ ) is equivalent to this. Option A’s absolute integrability is sufficient but not necessary. Option C confuses displacement with arc length. Option B states a true fact but doesn’t articulate the defining criterion. The explanation emphasizes that vector limits inherit scalar definitions component-wise, preventing misapplication of norm-based convergence criteria inappropriate for displacement calculations.

Q11. In modeling planetary motion, $\mathbf{r}''(t) = -\frac{k}{\|\mathbf{r}(t)\|^3}\mathbf{r}(t)$ . Why can’t we find $\mathbf{r}(t)$ by direct antidifferentiation as done in constant-acceleration problems?

A.Because the right-hand side depends on

\mathbf{r}(t)

itself, making it a nonlinear ODE rather than an explicit function of t. ✅

B.Because gravitational constant k varies with time in realistic models.

C.Because vector antiderivatives require Cartesian coordinates, but planetary motion is naturally polar.

D.Because the antiderivative of

1/\|\mathbf{r}\|^3

involves elliptic integrals with no closed form.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This challenging question contrasts solvable antiderivative problems with genuine dynamical systems. Direct antidifferentiation requires the integrand to be an explicit function of the independent variable t alone. Here, acceleration depends on the unknown position vector $\mathbf{r}(t)$ , creating a coupled nonlinear system. Options B and C cite irrelevant complications. Option D incorrectly focuses on integration difficulty rather than structural impossibility. Recognizing this distinction is crucial: many physics problems appear similar to basic antiderivative exercises but belong to a fundamentally different class requiring advanced methods. This tests metacognitive awareness of problem taxonomy beyond computational technique.

Q12. Two students compute $\int \langle 2t, 3t^2, 4t^3 \rangle \, dt$ . Student A gets $\langle t^2, t^3, t^4 \rangle + \mathbf{C}$ ; Student B gets $\langle t^2+1, t^3-2, t^4+\pi \rangle + \mathbf{D}$ . Given no initial conditions, whose answer is more general?

A.Student A, because it uses the simplest constant vector

\mathbf{C}

.

B.Student B, because specific constants demonstrate deeper understanding.

C.Neither; both represent the same family of solutions since

\mathbf{C}

and

\mathbf{D}

are arbitrary constant vectors. ✅

D.Student A, because adding specific numbers unnecessarily restricts the solution space.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: This conceptual question addresses the nature of indefinite vector integrals. Both answers are equally valid general antiderivatives because the constant vector absorbs any specific values. Student B’s constants are just particular instances of the arbitrary $\mathbf{D}$ ; renaming $\mathbf{D}' = \mathbf{D} + \langle 1,-2,\pi \rangle$ recovers Student A’s form. The key insight is that ‘arbitrary constant vector’ encompasses all possible constant offsets, making specific numerical choices neither more nor less general. This prevents students from mistakenly believing simplified constants are preferable or that added numbers indicate error, reinforcing abstraction in vector calculus.

Q13. A robot arm’s angular velocity leads to tip velocity $\mathbf{v}(t) = \langle -\sin t, \cos t, 0 \rangle$ . If the arm starts at $\mathbf{r}(0) = \langle 1, 0, 0 \rangle$ , what does $\mathbf{r}(\pi)$ reveal about the limitations of antiderivatives in capturing rotational constraints?

A.

\mathbf{r}(\pi) = \langle -1, 0, 0 \rangle

correctly shows half-circle motion, proving antiderivatives fully capture rotation.

B.

\mathbf{r}(\pi) = \langle -1, 0, 0 \rangle

, but the actual path length is

\pi

, while

\|\mathbf{r}(\pi)-\mathbf{r}(0)\|=2

; antiderivatives give displacement, not trajectory details. ✅

C.Antiderivatives fail here because rotational motion requires quaternion representation, not vectors.

D.The result is incorrect because

\int \cos t \, dt = \sin t

misses the radius factor in circular motion.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This application question highlights what antiderivatives do and don’t provide. Computing $\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle$ satisfies initial conditions and gives correct displacement $\langle -2,0,0 \rangle$ . However, the antiderivative doesn’t encode path length ( $\pi$ ) or confirm circularity—it only gives endpoints. Students might think vector antiderivatives describe full trajectories, but they merely accumulate infinitesimal displacements. This distinction is vital in robotics where path planning requires more than endpoint positions. The explanation reinforces that antiderivatives solve kinematic integration but not geometric characterization, preventing overinterpretation of results.

Q14. When solving $\mathbf{r}'(t) = \mathbf{v}(t)$ with $\mathbf{r}(t_0) = \mathbf{r}_0$ , which formulation avoids ambiguity in the constant of integration?

A.

\mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C}

, then solve

\mathbf{C} = \mathbf{r}_0 - \int \mathbf{v}(t_0)\,dt

.

B.

\mathbf{r}(t) = \mathbf{r}_0 + \int_{t_0}^t \mathbf{v}(s)\,ds

. ✅

C.

\mathbf{r}(t) = \int_{0}^t \mathbf{v}(s)\,ds + \mathbf{r}_0

, assuming

t_0=0

.

D.

\mathbf{r}(t) = \int \mathbf{v}(t)\,dt

evaluated between

t_0

and

t

, plus

\mathbf{r}_0

.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This conceptual question compares equivalent but pedagogically distinct formulations. Option B uses the definite integral with variable upper limit, which automatically satisfies $\mathbf{r}(t_0)=\mathbf{r}_0$ without solving for $\mathbf{C}$ , eliminating ambiguity. Option A works but risks errors in evaluating indefinite integrals at $t_0$ . Option C assumes $t_0=0$ , losing generality. Option D is syntactically incorrect (definite integrals don’t need ‘plus r0’ after evaluation). The definite integral form is preferred in applied contexts because it embeds initial conditions intrinsically, reducing algebraic steps and potential sign errors. This promotes robust problem-solving habits beyond rote antidifferentiation.

Q15. A student argues: ‘Since $\frac{d}{dt}\|\mathbf{r}(t)\| = \frac{\mathbf{r}(t)\cdot\mathbf{r}'(t)}{\|\mathbf{r}(t)\|}$ , then $\int \|\mathbf{r}'(t)\|\,dt = \|\mathbf{r}(t)\| + C$ .’ What is the flaw in this reasoning?

A.The formula for derivative of magnitude is correct, but

\|\mathbf{r}&#039;(t)\| \neq \frac{d}{dt}\|\mathbf{r}(t)\|

unless motion is radial.

B.The student confused

\|\mathbf{r}&#039;(t)\|

with

\|\mathbf{r}(t)\|&#039;

; the former is speed, the latter is rate of change of distance from origin.

C.Both A and B identify the same core error: equating speed with radial velocity. ✅

D.The reasoning is actually valid for any smooth curve passing through the origin.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This error analysis question exposes a pervasive misconception. Speed $\|\mathbf{r}'(t)\|$ measures instantaneous path traversal rate, while $\frac{d}{dt}\|\mathbf{r}(t)\|$ measures how fast the object moves toward/away from the origin. They’re equal only when velocity is parallel to position (radial motion). The student’s integral claims arc length equals distance from origin, which fails for circular orbits (arc length grows, distance stays constant). Options A and B are equivalent restatements of this error. Recognizing this distinction is fundamental to understanding that vector antiderivatives of velocity give displacement, while scalar antiderivatives of speed give path length—two unrelated quantities except in special cases.

Q16. Given $\mathbf{F}'(t) = \langle t^2, \sin t, e^t \rangle$ and $\mathbf{F}(1) = \langle 0, 0, 0 \rangle$ , which computational strategy minimizes error when finding $\mathbf{F}(2)$ ?

A.Compute indefinite integral

\langle t^3/3, -\cos t, e^t \rangle + \mathbf{C}

, solve for

\mathbf{C}

using

t=1

, then evaluate at

t=2

.

B.Use definite integral

\mathbf{F}(2) = \mathbf{F}(1) + \int_1^2 \mathbf{F}&#039;(t)\,dt = \int_1^2 \langle t^2, \sin t, e^t \rangle \, dt

. ✅

C.Approximate numerically using Simpson’s rule on each component separately.

D.Differentiate

\mathbf{F}&#039;(t)

to verify continuity before integrating.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This application question evaluates methodological efficiency. Since $\mathbf{F}(1)=\mathbf{0}$ , the definite integral from 1 to 2 directly gives $\mathbf{F}(2)$ without solving for constants, avoiding two extra algebraic steps where sign or arithmetic errors commonly occur. Option A is valid but prone to mistakes in constant determination. Option C sacrifices exactness unnecessarily. Option D is irrelevant verification. The definite integral approach leverages the given condition optimally, demonstrating strategic thinking beyond mechanical computation. This aligns with best practices in applied mathematics where problem structure should guide method selection to enhance accuracy and efficiency.

Q17. Consider $\mathbf{v}(t) = \langle \cos(t^2), \sin(t^2), 0 \rangle$ . Although each component lacks elementary antiderivative, what can still be asserted about $\mathbf{r}(t) = \int_0^t \mathbf{v}(s)\,ds$ ?

A.

\mathbf{r}(t)

does not exist because the components aren’t integrable.

B.

\mathbf{r}(t)

exists and is smooth, but cannot be expressed in closed form using elementary functions. ✅

C.

\mathbf{r}(t)

traces a circle because

\cos^2+\sin^2=1

.

D.

\mathbf{r}(t)

has constant speed 1, so

\|\mathbf{r}(t)\| = t

.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging question separates existence from expressibility. Continuity of $\mathbf{v}(t)$ guarantees $\mathbf{r}(t)$ exists and is differentiable (by FTC), even without elementary antiderivatives. Option A falsely conflates non-elementary with non-integrable. Option C confuses velocity direction with position trajectory; $\mathbf{r}(t)$ is a Cornu spiral, not a circle. Option D mistakes speed of $\mathbf{v}$ for growth rate of $\|\mathbf{r}\|$ ; $\|\mathbf{r}(t)\| \neq t$ generally. This tests sophisticated understanding that calculus operates beyond symbolic computation, emphasizing analytical properties over closed-form requirements—a key mindset for advanced mathematics and scientific computing.

Q18. In comparing $\int_a^b \mathbf{r}'(t)\,dt$ and $\int_a^b \|\mathbf{r}'(t)\|\,dt$ , which scenario makes these two quantities most nearly equal in magnitude?

A.When

\mathbf{r}(t)

traces a closed loop returning to start.

B.When

\mathbf{r}(t)

moves in a straight line without reversing direction. ✅

C.When

\mathbf{r}&#039;(t)

is orthogonal to

\mathbf{r}(t)

throughout [a,b].

D.When

\|\mathbf{r}&#039;(t)\|

is constant over [a,b].

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This mixed-concept question connects vector and scalar integrals geometrically. The vector integral gives displacement $\mathbf{r}(b)-\mathbf{r}(a)$ ; its magnitude equals the scalar integral (path length) only when motion is unidirectional along a straight line, so no cancellation occurs and path coincides with displacement. Closed loops (A) make displacement zero but path length positive. Orthogonality (C) relates to energy, not path efficiency. Constant speed (D) doesn’t prevent directional changes that increase path length relative to displacement. This reinforces that equality of magnitudes is exceptional, occurring precisely when the curve is a monotonic line segment—a critical insight for optimization and physics applications.

Q19. A physicist models fluid flow with $\mathbf{V}(x,y,z,t)$ , but seeks particle paths via $d\mathbf{r}/dt = \mathbf{V}(\mathbf{r}(t),t)$ . Why is this not a standard vector antiderivative problem?

A.Because

\mathbf{V}

depends on spatial coordinates that are themselves functions of t, creating implicit dependence. ✅

B.Because fluid velocity fields are typically discontinuous in real applications.

C.Because particle tracking requires Lagrangian coordinates, making Eulerian

\mathbf{V}

irrelevant.

D.Because the antiderivative would require integrating over four variables simultaneously.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This Olympiad-style question distinguishes autonomous/non-autonomous systems from standard antiderivatives. Standard problems have $\mathbf{r}'(t) = \mathbf{f}(t)$ with explicit t-dependence only. Here, $\mathbf{V}$ depends on $\mathbf{r}(t)$ , making the ODE implicit and generally unsolvable by direct integration. Options B and C cite practical issues, not mathematical structure. Option D misrepresents dimensionality. Recognizing this structural difference is essential for identifying when calculus tools apply versus when dynamical systems theory is needed. This elevates understanding beyond computation to problem classification, a hallmark of expert mathematical thinking.

Q20. Given $\mathbf{a}(t) = \langle 0, -9.8, 0 \rangle$ , $\mathbf{v}(0) = \langle 20, 15, 0 \rangle$ , $\mathbf{r}(0) = \langle 0, 2, 0 \rangle$ , at what time t does the vertical component of position reach maximum, and how is this found via antiderivatives?

A.At t=15/9.8, found by setting

v_y(t) = -9.8t + 15 = 0

after integrating acceleration.

B.At t=2/9.8, found by setting

r_y(t) = -4.9t^2 + 15t + 2 = 0

.

C.At t=15/9.8, found by maximizing

r_y(t)

directly via derivative test on the position antiderivative.

D.Both A and C are valid approaches yielding the same result through different stages of antidifferentiation. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: This multi-step application question integrates physics and calculus. Integrating $a_y=-9.8$ gives $v_y=-9.8t+15$ ; setting to zero yields t=15/9.8. Alternatively, integrating again gives $r_y=-4.9t^2+15t+2$ ; its derivative is $v_y$ , so maximizing position uses the same critical point. Both methods rely on antiderivatives: A uses velocity antiderivative, C uses position antiderivative’s derivative. Option B solves for ground impact, not max height. The equivalence demonstrates flexibility in applying calculus concepts, reinforcing that antiderivatives create interconnected solution pathways rather than isolated procedures.

Q21. If $\mathbf{F}'(t) = \mathbf{0}$ for all t in an open interval I, what can be concluded about $\mathbf{F}(t)$ on I?

A.

\mathbf{F}(t) = \mathbf{0}

for all t in I.

B.

\mathbf{F}(t)

is a constant vector on I, but not necessarily zero. ✅

C.

\|\mathbf{F}(t)\|

is constant, but direction may vary.

D.

\mathbf{F}(t)

is linear in t with zero slope.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This direct recall question establishes the vector zero-derivative theorem. Component-wise, $F_i'(t)=0$ implies $F_i(t)=c_i$ , so $\mathbf{F}(t)=\langle c_1,c_2,c_3 \rangle$ , a constant vector. It need not be zero (e.g., $\mathbf{F}(t)=\langle 1,2,3 \rangle$ ). Option C describes constant magnitude with varying direction, which would require nonzero derivative. Option D contradicts zero derivative. This foundational result underpins uniqueness of antiderivatives up to constants and is prerequisite for solving initial value problems. Mastery ensures students don’t conflate ‘zero derivative’ with ‘zero function,’ a subtle but critical distinction in vector analysis.

Q22. A student computes $\int \langle \sec t \tan t, \csc^2 t, \sinh t \rangle \, dt = \langle \sec t, -\cot t, \cosh t \rangle + \mathbf{C}$ and checks by differentiating. What additional validation step is essential for correctness in applied contexts?

A.Verify that the domain of the antiderivative matches the domain of the original vector function. ✅

B.Check that

\mathbf{C}

has units consistent with the integrand.

C.Confirm that

\sec t > 0

to avoid sign errors in the first component.

D.Ensure that

\cosh t \geq 1

to maintain physical plausibility.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This error analysis question emphasizes domain consistency in applied antidifferentiation. While differentiation verifies algebraic correctness, it doesn’t ensure the antiderivative is valid where the original function is defined. For example, $\sec t \tan t$ is undefined at $t=\pi/2 + n\pi$ , and $\sec t$ shares these discontinuities—but if a student wrote $|\sec t|$ , differentiation might still work piecewise while violating domain alignment. In modeling, using an antiderivative outside the integrand’s domain produces nonsensical results. This step bridges pure calculus and application, ensuring mathematical objects remain meaningful in context—a crucial HOTS skill often overlooked in computational drills.

Q23. Suppose $\mathbf{r}(t)$ satisfies $\mathbf{r}'(t) = \mathbf{A}\mathbf{r}(t)$ for constant matrix $\mathbf{A}$ . Why can’t we write $\mathbf{r}(t) = \int \mathbf{A}\mathbf{r}(t)\,dt + \mathbf{C}$ as a solution?

A.Because

\mathbf{r}(t)

appears on both sides, making the expression circular and non-constructive. ✅

B.Because matrix multiplication doesn’t commute with integration.

C.Because

\mathbf{A}

must be diagonalizable for antiderivatives to exist.

D.Because the integral of a matrix-vector product requires tensor calculus.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This challenging question identifies structural barriers to naive antidifferentiation. The equation $\mathbf{r}' = \mathbf{A}\mathbf{r}$ defines $\mathbf{r}$ implicitly; writing $\mathbf{r} = \int \mathbf{A}\mathbf{r}\,dt + \mathbf{C}$ doesn’t solve for $\mathbf{r}$ since it remains inside the integral. This is unlike $\mathbf{r}' = \mathbf{f}(t)$ where $\mathbf{f}$ is known. Options B-D cite technicalities irrelevant to the core issue of self-reference. Recognizing this circularity distinguishes differential equations from integration problems, guiding students toward appropriate methods (eigenvalues, exponentials). This meta-level understanding prevents futile attempts to force antiderivative techniques onto incompatible problem structures.

Q24. When graphing $\mathbf{r}(t) = \int_0^t \langle \cos s, \sin s, s/10 \rangle \, ds$ , which feature of the resulting helix is directly determined by the antiderivative process rather than the integrand’s form?

A.The radius of the helix, fixed by

\cos^2 s + \sin^2 s = 1

.

B.The vertical pitch, determined by integrating

s/10

to

t^2/20

.

C.The starting point at origin, enforced by the lower limit 0 in the definite integral. ✅

D.The constant speed

\sqrt{1 + (t/10)^2}

, derived from the integrand’s magnitude.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This graph-based question isolates the role of integration bounds in shaping curves. The integrand determines local geometry (radius via trig identity, speed via magnitude), but the definite integral from 0 fixes $\mathbf{r}(0)=\mathbf{0}$ , anchoring the helix at origin. An indefinite integral would allow vertical/horizontal translation via $\mathbf{C}$ . Option B misattributes pitch to integration; pitch comes from integrand’s z-component form. Option D confuses speed with position. This highlights that antiderivatives contribute global positioning through initial/boundary conditions, while integrands govern local behavior—a nuanced understanding essential for interpreting parametric curves generated by integration.

Q25. In a navigation system, position is updated via $\mathbf{r}_{n+1} = \mathbf{r}_n + \int_{t_n}^{t_{n+1}} \mathbf{v}(t)\,dt$ . If $\mathbf{v}(t)$ is measured with error $\epsilon(t)$ , how does the antiderivative propagate this error in $\mathbf{r}(T)$ ?

A.Error accumulates as

\|\int_0^T \epsilon(t)\,dt\| \leq \int_0^T \|\epsilon(t)\|\,dt

, bounded by total error magnitude over time. ✅

B.Error cancels out due to vector averaging over intervals.

C.Error grows quadratically because integration is a smoothing operator.

D.Error depends only on

\epsilon(T)

, not historical values, due to the Fundamental Theorem.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This mixed-concept application question links numerical analysis with vector calculus. Measurement error $\epsilon(t)$ in velocity integrates to position error $\int \epsilon(t)\,dt$ . The triangle inequality gives the bound in A, showing worst-case accumulation proportional to total error exposure. Option B falsely assumes cancellation; errors typically add constructively. Option C mischaracterizes integration as amplifying error; it averages but doesn’t square. Option D violates causality; position depends on entire history. This demonstrates that antiderivatives transform instantaneous uncertainties into cumulative ones, a critical consideration in sensor fusion and control systems where understanding error propagation determines system reliability.

📝 Antiderivatives of vector functions (25 MCQs)

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