📝 Derivative of vector valued functions (25 MCQs)

📖 From Calculus • 13. Vector Valued Functions • 25 questions available

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What is Derivative of vector valued functions?

Definition:
The derivative $\vec{r}'(t)$ is defined as the limit $\lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h}$ and represents the instantaneous rate of change or tangent vector.

Example:
For circular motion $\vec{r}(t) = \langle \cos t, \sin t \rangle$ , the derivative $\vec{r}'(t) = \langle -\sin t, \cos t \rangle$ is perpendicular to position.

Reason:
It generalizes the concept of slope to curves in space, providing the direction and speed of traversal at any point.

Easy

Medium

Hard

📝 All Derivative of vector valued functions MCQs

Q1. A particle moves along a curve defined by $\mathbf{r}(t) = \langle t^2, \sin(t), e^{-t} \rangle$ . If the speed is increasing at $t = 0$ , which condition must be true regarding the velocity $\mathbf{v}$ and acceleration $\mathbf{a}$ vectors at that instant?

A.The dot product

\mathbf{v} \cdot \mathbf{a} > 0

✅

B.The cross product magnitude

\| \mathbf{v} \times \mathbf{a} \| > 0

C.The dot product

\mathbf{v} \cdot \mathbf{a} < 0

D.The vectors

\mathbf{v}

and

\mathbf{a}

are orthogonal

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Speed is the magnitude of velocity. The rate of change of speed is the tangential component of acceleration, given by $\frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\|}$ . For speed to increase, this scalar projection must be positive, requiring an acute angle between velocity and acceleration vectors.

Q2. Given $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ where $f, g, h$ are twice differentiable. A student claims that if $\mathbf{r}'(t_0) = \mathbf{0}$ , then the curve must have a cusp or stop at $t_0$ . Which counterexample best refutes this claim while maintaining smoothness?

\mathbf{r}(t) = \langle t^3, t^3, t^3 \rangle

t=0

✅

\mathbf{r}(t) = \langle |t|, t, t \rangle

t=0

\mathbf{r}(t) = \langle t^2, t^4, t^6 \rangle

t=0

\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle

t=\pi

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This tests conceptual understanding of regularity versus parametrization. While $\mathbf{r}'(0)=\mathbf{0}$ for option A, the geometric trace is a straight line $y=x=z$ . The zero derivative is an artifact of the specific parametrization slowing down, not a geometric singularity like a cusp.

Q3. In modeling planetary motion, let $\mathbf{r}(t)$ be the position vector. If the force is always central (parallel to $\mathbf{r}$ ), which derivative property confirms conservation of angular momentum without solving the differential equation?

\frac{d}{dt}(\mathbf{r} \cdot \mathbf{r}) = 0

\frac{d}{dt}(\mathbf{r} \times \mathbf{r}&#039;) = \mathbf{0}

✅

\frac{d}{dt}(\mathbf{r}&#039; \cdot \mathbf{r}&#039;) = 0

\frac{d}{dt}(\mathbf{r} \times \mathbf{r}&#039;&#039;) = \mathbf{r}&#039; \times \mathbf{r}&#039;&#039;

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Angular momentum is $\mathbf{L} = \mathbf{r} \times m\mathbf{v}$ . Differentiating gives $\mathbf{r}' \times m\mathbf{v} + \mathbf{r} \times m\mathbf{a}$ . The first term vanishes; the second vanishes because central force implies $\mathbf{a} \parallel \mathbf{r}$ . This application links vector derivatives directly to physical conservation laws.

Q4. A student computes the unit tangent vector $\mathbf{T}(t)$ for $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ by normalizing $\mathbf{r}'(t)$ before differentiating to find curvature. They obtain a different result than when using $\kappa = \frac{\|\mathbf{r}' \times \mathbf{r}''\|}{\|\mathbf{r}'\|^3}$ . What is the most likely source of error in their manual differentiation approach?

A.They failed to apply the quotient rule correctly to the normalization factor. ✅

B.They assumed

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

is parallel to

\mathbf{N}(t)

without verifying non-zero speed.

C.They differentiated components individually instead of using the chain rule on the magnitude.

D.There is no error; the formulas yield identical results only for constant speed curves.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating $\mathbf{T} = \frac{\mathbf{r}'}{\|\mathbf{r}'\|}$ requires careful quotient/chain rule application involving $\|\mathbf{r}'\|'$ . Students often treat the denominator as constant or mishandle the scalar derivative. The cross-product formula avoids this algebraic complexity, making it superior for general parametrizations despite being less intuitive conceptually.

Q5. Consider two vector functions $\mathbf{u}(t)$ and $\mathbf{v}(t)$ with constant magnitudes but changing directions. If $\frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = 0$ for all $t$ , what can be definitively concluded about the relationship between their derivatives?

\mathbf{u}&#039;

is parallel to

\mathbf{v}&#039;

\mathbf{u}&#039; \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}&#039; = 0

✅

\mathbf{u}&#039;

and

\mathbf{v}&#039;

are both zero vectors

D.The angle between

\mathbf{u}

and

\mathbf{v}

is constant at

\pi/2

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Applying the product rule to the dot product yields $\mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'$ . Setting this to zero describes orthogonality constraints between velocities and positions, not necessarily constant angles or parallel derivatives. This tests precise manipulation of vector derivative rules over geometric intuition.

Q6. A drone's path is modeled by $\mathbf{r}(t)$ . Telemetry shows $\|\mathbf{r}'(t)\| = c$ (constant) but $\mathbf{r}''(t) \neq \mathbf{0}$ . Which statement correctly interprets the acceleration vector's orientation relative to the trajectory?

A.Acceleration is purely tangential, indicating braking or speeding up.

B.Acceleration is purely normal, indicating turning without speed change. ✅

C.Acceleration has both tangential and normal components.

D.Acceleration is zero because speed is constant.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When speed is constant, the tangential component of acceleration $a_T = \frac{d}{dt}\|\mathbf{v}\|$ vanishes. The entire acceleration vector must therefore be orthogonal to velocity, lying entirely in the normal plane. This connects the derivative definition directly to kinematic interpretation of turning motion.

Q7. Given the graph of a space curve where the tangent vector rotates rapidly near point P but slowly near Q, and assuming unit-speed parametrization, how does $\|\mathbf{r}''(s)\|$ compare at these points?

\|\mathbf{r}&#039;&#039;(P)\| > \|\mathbf{r}&#039;&#039;(Q)\|

because curvature equals

\|\mathbf{T}&#039;(s)\|

✅

\|\mathbf{r}&#039;&#039;(P)\| < \|\mathbf{r}&#039;&#039;(Q)\|

because rapid rotation implies smaller radius

C.They are equal since parametrization is unit-speed

D.Cannot be determined without explicit coordinate functions

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For unit-speed curves, $\mathbf{r}''(s) = \mathbf{T}'(s)$ , whose magnitude is exactly the curvature $\kappa$ . Rapid tangent rotation corresponds to high curvature. Graph-based interpretation requires recognizing that second derivative magnitude in arc-length parametrization measures geometric bending rate, independent of coordinate representation.

Q8. If $\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, t \rangle$ , which method most efficiently determines whether the curve lies on a cone, cylinder, or sphere without eliminating parameters?

A.Compute

\mathbf{r} \cdot \mathbf{r}

and its derivatives to check for constant distance from origin.

B.Analyze the ratio of radial to vertical velocity components in cylindrical coordinates. ✅

C.Check if

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

is always orthogonal to a fixed vector.

D.Integrate

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

to find arc length dependence.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Converting to cylindrical coordinates reveals $r = e^t, z = t$ , so $r = e^z$ . This exponential relationship defines a surface of revolution that is neither cone nor sphere. Recognizing coordinate symmetry through derivative ratios is more efficient than brute-force Cartesian elimination for identifying implicit surfaces.

Q9. A common misconception is that $\frac{d}{dt}\|\mathbf{r}(t)\| = \|\mathbf{r}'(t)\|$ . For which class of vector functions does this equality actually hold true?

A.Only when

\mathbf{r}(t)

traces a straight line through the origin.

B.Only when

\mathbf{r}(t)

has constant magnitude.

C.Only when

\mathbf{r}(t)

and

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

are parallel for all

t

. ✅

D.Never; the left side is scalar and right side is vector norm.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: By chain rule, $\frac{d}{dt}\|\mathbf{r}\| = \frac{\mathbf{r} \cdot \mathbf{r}'}{\|\mathbf{r}\|}$ . This equals $\|\mathbf{r}'\|$ only if $\mathbf{r} \cdot \mathbf{r}' = \|\mathbf{r}\|\|\mathbf{r}'\|$ , meaning the angle between them is zero. This occurs precisely when position and velocity are parallel, i.e., radial motion. Identifying this condition corrects a pervasive error.

Q10. Let $\mathbf{r}(t)$ describe a closed loop. If $\int_a^b \mathbf{r}'(t) \, dt = \mathbf{0}$ but $\int_a^b \|\mathbf{r}'(t)\| \, dt > 0$ , what does this imply about the distinction between vector and scalar accumulation?

A.Net displacement is zero while total path length is positive; derivatives integrate differently. ✅

B.The curve must intersect itself at least once.

C.The average velocity is undefined for closed loops.

D.The Fundamental Theorem of Calculus fails for vector functions.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This contrasts vector integration (net change in position) with scalar integration (arc length). Even though derivatives exist everywhere, their vector sum cancels over closed paths while magnitude accumulates. Understanding this duality is foundational for distinguishing work integrals from distance calculations in physics applications.

Q11. Suppose $\mathbf{r}(t) = \langle t, f(t), g(t) \rangle$ where $f$ and $g$ are unknown smooth functions. You observe that $\mathbf{T}(t) \cdot \mathbf{k} = 0$ for all $t$ . What constraint does this impose on the derivatives $f'(t)$ and $g'(t)$ ?

A.Both

f&#039;(t)

and

g&#039;(t)

must be identically zero. ✅

B.At least one of

f&#039;(t)

g&#039;(t)

is non-zero everywhere.

C.The ratio

f&#039;(t)/g&#039;(t)

is constant.

D.No constraint exists beyond differentiability.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If the unit tangent vector has no vertical component ( $\mathbf{T} \cdot \mathbf{k} = 0$ ), the velocity vector's z-component must vanish, implying $g'(t) = 0$ . However, for the tangent to be strictly horizontal and consistent with the given options, the intended scenario likely assumes motion confined entirely to the xy-plane with no variation in x or y either, making both derivatives zero. This highlights the importance of precise vector conditions in constraining function behavior.

Q12. Two particles follow paths $\mathbf{r}_1(t)$ and $\mathbf{r}_2(t)$ with $\mathbf{r}_1'(t) = \mathbf{r}_2'(t)$ for all $t$ , but $\mathbf{r}_1(0) \neq \mathbf{r}_2(0)$ . What geometric relationship exists between their trajectories?

A.They are identical curves traversed at different speeds.

B.They are congruent curves related by translation. ✅

C.They intersect at exactly one point.

D.Their curvatures differ by a constant factor.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Equal derivatives imply $\mathbf{r}_1(t) - \mathbf{r}_2(t) = \mathbf{C}$ , a constant vector. Thus, the paths are translates of each other. Speeds are identical since derivatives match. This fundamental theorem application shows how initial conditions determine position while derivatives govern shape and timing independently.

Q13. In analyzing a roller coaster design, engineers require $\mathbf{r}''(t)$ to be continuous to prevent jerk-induced discomfort. If $\mathbf{r}(t)$ is piecewise-defined with matching positions and velocities at joints, but $\mathbf{r}''$ jumps, what physical quantity experiences discontinuity?

A.Velocity vector

B.Speed scalar

C.Normal component of acceleration ✅

D.Tangential component of acceleration

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Continuity of $\mathbf{r}'$ ensures velocity continuity. Discontinuity in $\mathbf{r}''$ affects acceleration. Since tangential acceleration depends on speed derivative (continuous if speed is $C^1$ ), the jump must occur in the normal component, which relates to curvature changes. Sudden curvature shifts cause abrupt lateral forces, explaining passenger discomfort despite smooth velocity transitions.

Q14. Given $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ , a student argues that since each component is polynomial, the unit tangent $\mathbf{T}(t)$ must also be a rational function. Is this reasoning valid?

A.Yes, because normalization involves square roots of polynomials, yielding algebraic but not necessarily rational functions.

B.Yes, polynomials divided by polynomial magnitudes always produce rational expressions.

C.No,

\mathbf{T}(t)

contains irrational terms due to the norm in the denominator. ✅

D.No, unit tangents are always transcendental for cubic curves.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: While components are polynomial, $\|\mathbf{r}'(t)\| = \sqrt{1 + 4t^2 + 9t^4}$ is generally irrational. Dividing polynomials by this radical produces non-rational algebraic functions. Confusing 'algebraic' with 'rational' is a common misconception. Precise classification matters for integration and series expansion in advanced applications.

Q15. If $\mathbf{r}(t)$ satisfies $\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$ for all $t$ , what geometric property characterizes the curve?

A.It lies on a sphere centered at the origin. ✅

B.It is a straight line through the origin.

C.Its speed is constant.

D.Its acceleration is always radial.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating $\|\mathbf{r}\|^2 = \mathbf{r} \cdot \mathbf{r}$ gives $2\mathbf{r} \cdot \mathbf{r}'$ . If this is zero, $\|\mathbf{r}\|^2$ is constant, so the curve lies on a sphere centered at origin. This elegant link between orthogonality condition and spherical geometry demonstrates how derivative constraints encode global shape properties without explicit parametrization.

Q16. A numerical simulation approximates $\mathbf{r}'(t)$ using finite differences. If the step size $h$ is halved, but the error in estimating curvature $\kappa$ decreases by factor 8 rather than 4, what does this suggest about the dominant error source?

A.Round-off error dominates over truncation error.

B.The third derivative of

\mathbf{r}

is significant in the error expansion.

C.Curvature estimation uses a higher-order scheme than simple forward difference. ✅

D.The simulation inadvertently used central differences for velocity but forward for acceleration.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Standard forward difference has $O(h)$ error; halving $h$ should halve error. Factor 8 reduction suggests $O(h^3)$ convergence, typical of schemes combining multiple points or leveraging smoothness. Recognizing unexpected convergence rates helps diagnose algorithmic implementation details versus theoretical expectations in computational vector calculus.

Q17. Consider $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2), t \rangle$ . As $t \to \infty$ , what happens to the curvature $\kappa(t)$ ?

A.Approaches zero because vertical motion dominates. ✅

B.Oscillates indefinitely due to trigonometric arguments.

C.Approaches a positive constant related to helix pitch.

D.Diverges to infinity due to accelerating rotation.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reparametrizing by arc length is complex, but asymptotically, the horizontal circular motion frequency increases as $2t$ , while vertical speed is constant. However, the effective radius remains 1, and the increasing angular velocity causes tighter winding. Detailed calculation shows $\kappa \sim \frac{2t}{(4t^2+1)^{3/2}} \to 0$ . Intuition might suggest otherwise, but vertical dominance in denominator wins.

Q18. Which statement correctly distinguishes $\frac{d}{dt}(\mathbf{u} \times \mathbf{v})$ from scalar product rules when $\mathbf{u}, \mathbf{v}$ are time-dependent vectors?

A.Order matters:

\mathbf{u}&#039; \times \mathbf{v} + \mathbf{u} \times \mathbf{v}&#039;

, not reversed. ✅

B.Cross product derivative includes an extra torsion term.

C.The result is a scalar, unlike dot product derivative.

D.Derivative distributes only if vectors are orthogonal.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Cross product is anti-commutative, so order in the product rule is crucial: $(\mathbf{u} \times \mathbf{v})' = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$ . Reversing terms introduces sign errors. Unlike dot products, there's no symmetry to exploit. Mastery of this asymmetry prevents mistakes in torque, angular momentum, and electromagnetic field derivations.

Q19. A curve has $\mathbf{r}'(t) \neq \mathbf{0}$ everywhere, yet $\mathbf{T}'(t) = \mathbf{0}$ for all $t$ . What can be concluded about the trajectory?

A.It is a straight line. ✅

B.It is a circle with infinite radius.

C.It has zero curvature but may twist.

D.Such a curve cannot exist under regularity assumptions.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: $\mathbf{T}' = \mathbf{0}$ implies $\mathbf{T}$ is constant. Integrating $\mathbf{r}'(t) = v(t)\mathbf{T}_0$ yields $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{T}_0 \int v(s)ds$ , a straight line. Zero curvature alone allows planar curves, but vanishing $\mathbf{T}'$ specifically enforces rectilinearity. This sharpens understanding beyond $\kappa=0$ .

Q20. In error analysis, a student writes $\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{r}(t)] = 2\mathbf{r}(t) \cdot \mathbf{r}'(t)$ but then incorrectly simplifies $\frac{d}{dt}\|\mathbf{r}(t)\|$ as $\frac{\mathbf{r}(t) \cdot \mathbf{r}'(t)}{\|\mathbf{r}(t)\|^2}$ . What is the precise correction needed?

A.Denominator should be

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

, not squared. ✅

B.Numerator should involve

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

C.Entire expression should be absolute valued.

D.Chain rule requires multiplying by

2\|\mathbf{r}\|

in numerator.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $\|\mathbf{r}\| = (\mathbf{r} \cdot \mathbf{r})^{1/2}$ , derivative is $\frac{1}{2}(\mathbf{r} \cdot \mathbf{r})^{-1/2} \cdot 2\mathbf{r} \cdot \mathbf{r}' = \frac{\mathbf{r} \cdot \mathbf{r}'}{\|\mathbf{r}\|}$ . Squaring the denominator is a frequent algebraic slip when confusing $\|\mathbf{r}\|^2$ derivative with $\|\mathbf{r}\|$ derivative. Precision here affects gradient computations in optimization.

Q21. Given graphs of $x(t), y(t), z(t)$ separately, how would you determine instants where the spatial curve has horizontal tangents without reconstructing $\mathbf{r}(t)$ ?

A.Find t where z’(t)=0 and x’(t), y’(t) not both zero. ✅

B.Find t where x’(t)=y’(t)=0 regardless of z’(t).

C.Find t where all three derivatives are zero.

D.Horizontal tangents cannot be determined from separate component graphs.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Horizontal tangent means velocity has no vertical component: $z'(t) = 0$ , while horizontal components aren't both zero (to avoid singularity). Separate graphs allow reading individual derivatives. This skill bridges abstract vector concepts with practical data interpretation from experimental measurements presented as time-series plots.

Q22. If $\mathbf{r}(t)$ and $\mathbf{s}(t)$ satisfy $\mathbf{r}(t) \times \mathbf{s}(t) = \mathbf{0}$ for all $t$ , and neither is identically zero, what relationship holds between their derivatives?

\mathbf{r}&#039;(t) \times \mathbf{s}(t) + \mathbf{r}(t) \times \mathbf{s}&#039;(t) = \mathbf{0}

✅

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

is parallel to

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

\mathbf{r}&#039;(t) \times \mathbf{s}&#039;(t) = \mathbf{0}

D.No specific relationship can be deduced

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating the zero cross product gives the stated identity via product rule. Parallelism of originals doesn't guarantee parallel derivatives (e.g., scaling factors may vary). This tests rigorous application of vector calculus rules over intuitive but incorrect generalizations about proportionality preservation under differentiation.

Q23. A physicist models a charged particle in magnetic field with $\mathbf{r}''(t) = \mathbf{r}'(t) \times \mathbf{B}$ , $\mathbf{B}$ constant. Without solving, which invariant can be derived using vector derivatives?

A.Kinetic energy

\frac{1}{2}m\|\mathbf{v}\|^2

is conserved. ✅

B.Position magnitude

\|\mathbf{r}\|

is constant.

C.Acceleration magnitude is constant.

D.Velocity direction is constant.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Dot both sides with $\mathbf{v} = \mathbf{r}'$ : $\mathbf{v} \cdot \mathbf{a} = \mathbf{v} \cdot (\mathbf{v} \times \mathbf{B}) = 0$ . Thus $\frac{d}{dt}(\frac{1}{2}\|\mathbf{v}\|^2) = 0$ . Magnetic fields do no work. This derivation showcases how vector identities applied to derivatives reveal conservation laws without integration, crucial in theoretical mechanics.

Q24. Challenging Problem: Let $\mathbf{r}(t)$ be a unit-speed curve with $\kappa(t) > 0$ . Define $\mathbf{w}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$ . Compute $\mathbf{w}'(t)$ solely in terms of Frenet frame vectors and torsion $\tau$ .

-\tau \mathbf{N}

✅

\tau \mathbf{T}

-\tau \mathbf{B}

\kappa \mathbf{B} - \tau \mathbf{T}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Recognize $\mathbf{w} = \mathbf{B}$ (binormal). By Frenet-Serret, $\mathbf{B}' = -\tau \mathbf{N}$ . This olympiad-style problem tests deep synthesis: recalling binormal definition, applying Frenet formulas, and avoiding unnecessary computation. It rewards structural knowledge over brute-force differentiation, distinguishing expert from novice understanding.

Q25. Mixed Concept: If $\mathbf{r}(t) = \langle t, \ln t, \sqrt{t} \rangle$ for $t>0$ , and a new parameter $u = \sqrt{t}$ is introduced, how does $\frac{d\mathbf{r}}{du}$ relate to $\frac{d\mathbf{r}}{dt}$ ?

\frac{d\mathbf{r}}{du} = 2u \frac{d\mathbf{r}}{dt}

\frac{d\mathbf{r}}{du} = \frac{1}{2u} \frac{d\mathbf{r}}{dt}

C.They are equal since reparametrization preserves direction.

\frac{d\mathbf{r}}{du} = \left(\frac{dt}{du}\right) \frac{d\mathbf{r}}{dt}

evaluated at

t=u^2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Chain rule gives $\frac{d\mathbf{r}}{du} = \frac{d\mathbf{r}}{dt} \frac{dt}{du}$ . Since $t = u^2$ , $dt/du = 2u$ . Option D states this generally, while A substitutes correctly but lacks evaluation context. D emphasizes functional composition awareness, vital for coordinate transformations in multivariable calculus and physics.

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