📝 Derivative rules for vector functions (25 MCQs)

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

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What is Derivative rules for vector functions?

Definition:
Derivative rules include product rules for dot and cross products, such as $\frac{d}{dt}[\vec{u} \cdot \vec{v}] = \vec{u}' \cdot \vec{v} + \vec{u} \cdot \vec{v}'$ .

Example:
Differentiating $\vec{r}(t) \times \vec{r}'(t)$ yields $\vec{r}(t) \times \vec{r}''(t)$ since $\vec{r}' \times \vec{r}' = \vec{0}$ .

Reason:
These rules maintain algebraic consistency while respecting the non-commutative nature of vector multiplication.

Easy

Medium

Hard

📝 All Derivative rules for vector functions MCQs

Q1. A particle moves along a space curve defined by $\mathbf{r}(t)$ . If $\mathbf{r}'(t) \cdot \mathbf{r}''(t) = 0$ for all $t$ , which physical interpretation is necessarily true regarding the motion?

A.The acceleration vector is always zero.

B.The speed of the particle is constant. ✅

C.The velocity vector is always perpendicular to the position vector.

D.The tangential component of acceleration is non-zero.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative of the squared speed is $\frac{d}{dt}(\mathbf{v} \cdot \mathbf{v}) = 2\mathbf{v} \cdot \mathbf{a}$ . If this dot product vanishes identically, the rate of change of speed is zero. Students often confuse orthogonality of velocity and acceleration with geometric properties of the curve itself rather than kinematic constraints.

Q2. Given $\mathbf{u}(t)$ and $\mathbf{v}(t)$ are differentiable vector functions, a student computes $\frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)]$ as $\mathbf{u}'(t) \times \mathbf{v}'(t)$ . What is the fundamental error in this reasoning?

A.They applied the scalar product rule instead of the cross product rule.

B.They failed to account for the non-commutative nature of the cross product in the differentiation sequence.

C.They assumed derivatives distribute over cross products without maintaining operand order. ✅

D.They confused the cross product with the dot product derivative.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The correct rule is $\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$ . The student’s answer omits the original functions entirely. This misconception arises from incorrectly generalizing scalar multiplication rules to vector operations, ignoring that differentiation acts on each factor sequentially while preserving the cross product structure and order.

Q3. If $\mathbf{r}(t)$ describes a smooth curve and $\|\mathbf{r}'(t)\| = c$ where $c > 0$ is constant, what can be deduced about $\mathbf{r}''(t)$ relative to $\mathbf{r}'(t)$ ?

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

is parallel to

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

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

is orthogonal to

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

. ✅

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

must be the zero vector.

D.No specific geometric relationship exists.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Differentiating $\mathbf{r}' \cdot \mathbf{r}' = c^2$ yields $2\mathbf{r}' \cdot \mathbf{r}'' = 0$ . Thus acceleration is purely normal when speed is constant. This applies directly to uniform circular motion but holds generally. Students selecting parallel confuse constant speed with straight-line motion or misapply chain rule concepts.

Q4. Consider two vector functions where $\mathbf{F}(t) = f(t)\mathbf{u}(t)$ . If $f'(t_0) = 0$ and $\mathbf{u}'(t_0) \neq \mathbf{0}$ , what determines the direction of $\mathbf{F}'(t_0)$ ?

A.The direction depends solely on

\mathbf{u}(t_0)

B.The direction depends solely on

\mathbf{u}&#039;(t_0)

C.The direction depends on both

f(t_0)

and

\mathbf{u}&#039;(t_0)

. ✅

D.The derivative is undefined at

t_0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Using the product rule: $\mathbf{F}' = f'\mathbf{u} + f\mathbf{u}'$ . At $t_0$ , the first term vanishes, leaving $f(t_0)\mathbf{u}'(t_0)$ . Unless $f(t_0)=0$ , the direction aligns with $\mathbf{u}'(t_0)$ scaled by $f(t_0)$ . This tests understanding of how scalar zeros affect vector derivative directionality beyond simple computation.

Q5. A graph shows $\|\mathbf{r}'(t)\|$ increasing while the angle between $\mathbf{r}'(t)$ and $\mathbf{r}''(t)$ remains acute. Which statement best characterizes the tangential acceleration?

A.Tangential acceleration is negative.

B.Tangential acceleration is positive. ✅

C.Tangential acceleration is zero.

D.Tangential acceleration oscillates.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Tangential acceleration equals $\mathbf{a} \cdot \mathbf{T} = \|\mathbf{a}\|\cos\theta$ . An acute angle means $\cos\theta > 0$ , so tangential acceleration is positive, consistent with increasing speed shown in the magnitude graph. Students choosing zero confuse orthogonal acceleration with general motion; those choosing negative misinterpret the angle-speed relationship.

Q6. Let $\mathbf{r}(t) = \langle t^2, t^3, t^4 \rangle$ . Without computing components explicitly, determine the value of $\mathbf{r}'(0) \times \mathbf{r}''(0)$ .

\mathbf{0}

✅

\langle 0, 0, 2 \rangle

\langle 0, -2, 0 \rangle

D.Undefined

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At $t=0$ , $\mathbf{r}'(0) = \langle 0,0,0 \rangle$ since all components have powers ≥2. Any cross product involving the zero vector yields zero. This avoids tedious calculation and tests recognition of singular points where standard curvature formulas fail due to vanishing velocity, a subtle but critical concept.

Q7. Which condition ensures that $\frac{d}{dt}(\mathbf{u} \cdot \mathbf{u}) = 0$ implies $\mathbf{u}(t)$ has constant magnitude but not necessarily constant direction?

\mathbf{u}&#039;(t) = \mathbf{0}

\mathbf{u}(t) \cdot \mathbf{u}&#039;(t) = 0

✅

\|\mathbf{u}&#039;(t)\| = 0

\mathbf{u}&#039;&#039;(t) \cdot \mathbf{u}(t) = 0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The derivative of squared magnitude is $2\mathbf{u} \cdot \mathbf{u}'$ . Setting this to zero gives orthogonality between vector and its derivative, guaranteeing constant length. Option A is sufficient but overly restrictive; C is equivalent to A; D relates to second derivative. This foundational identity underpins unit tangent vector derivations.

Q8. In modeling planetary orbits, if $\mathbf{r}(t) \times \mathbf{r}'(t) = \mathbf{h}$ (constant vector), what does differentiating this expression reveal about the force field?

A.The force is central and lies in the orbital plane. ✅

B.The force is perpendicular to the orbital plane.

C.The force varies inversely with distance cubed.

D.The angular momentum is not conserved.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Differentiating gives $\mathbf{r}' \times \mathbf{r}' + \mathbf{r} \times \mathbf{r}'' = \mathbf{0}$ , so $\mathbf{r} \times \mathbf{a} = \mathbf{0}$ . Thus acceleration is parallel to position, implying central force. This connects derivative rules to physics conservation laws. Misconceptions arise from confusing cross product constancy with planar motion alone without recognizing force implications.

Q9. A student claims that if $\mathbf{r}'(t) \neq \mathbf{0}$ , then $\frac{d}{dt}\left( \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \right)$ is always parallel to $\mathbf{r}'(t)$ . Why is this incorrect?

A.Because the unit tangent derivative is always orthogonal to the unit tangent itself. ✅

B.Because the denominator cannot be differentiated.

C.Because the result depends on the parametrization being arc-length.

D.Because the derivative is always zero.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Let $\mathbf{T} = \mathbf{r}'/\|\mathbf{r}'\|$ . Since $\mathbf{T} \cdot \mathbf{T} = 1$ , differentiating gives $2\mathbf{T} \cdot \mathbf{T}' = 0$ . Thus $\mathbf{T}' \perp \mathbf{T}$ , never parallel unless zero. The student confuses direction of velocity with direction of its rate of change, missing the geometric constraint of unit vectors.

Q10. Suppose $\mathbf{r}(t)$ satisfies $\mathbf{r}'''(t) = \mathbf{0}$ for all $t$ . What is the maximum possible degree of the polynomial components of $\mathbf{r}(t)$ ?

A.Three

B.Two ✅

C.One

D.Four

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Integrating $\mathbf{r}''' = \mathbf{0}$ thrice yields quadratic polynomials at most. Each integration increases degree by one starting from zero. Students selecting three forget that third derivative being zero eliminates cubic terms; those choosing four misunderstand integration constants. This links derivative orders to polynomial structure in vector calculus contexts.

Q11. When computing $\frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))]$ , which expansion correctly applies derivative rules while preserving scalar triple product properties?

\mathbf{u}&#039; \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v}&#039; \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}&#039;)

✅

(\mathbf{u}&#039; \cdot \mathbf{v}) \times \mathbf{w} + \mathbf{u} \cdot (\mathbf{v}&#039; \times \mathbf{w}&#039;)

\mathbf{u}&#039; \cdot (\mathbf{v}&#039; \times \mathbf{w}&#039;)

\frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) \times \mathbf{w} + \mathbf{u} \cdot \frac{d}{dt}(\mathbf{v} \times \mathbf{w})

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The scalar triple product is a scalar-valued function of three vectors. The product rule extends linearly: differentiate each vector factor once while holding others fixed, summing results. Order within cross products must be preserved. Incorrect options violate associativity or omit terms, reflecting poor grasp of multilinear operator differentiation.

Q12. A curve has $\mathbf{r}'(t) = \langle \cos t, \sin t, t \rangle$ . Without integrating, what is $\frac{d}{dt}\|\mathbf{r}'(t)\|^2$ at $t = \pi/2$ ?

\pi

✅

2\pi

0

1

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Compute $\|\mathbf{r}'\|^2 = \cos^2 t + \sin^2 t + t^2 = 1 + t^2$ . Derivative is $2t$ ; at $\pi/2$ this equals $\pi$ . This avoids finding $\mathbf{r}(t)$ and tests efficient use of norm-squared identities. Students integrating unnecessarily waste time; those miscalculating trig identities miss simplification opportunities.

Q13. If $\mathbf{r}(t)$ is twice differentiable and $\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{0}$ for all $t$ in an interval, what geometric property must the curve possess?

A.It lies entirely in a plane.

B.It is a straight line segment.

C.Its curvature is identically zero. ✅

D.Its torsion is undefined.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Vanishing cross product implies $\mathbf{r}''$ is parallel to $\mathbf{r}'$ , meaning no normal acceleration component. Curvature $\kappa = \|\mathbf{r}' \times \mathbf{r}''\| / \|\mathbf{r}'\|^3 = 0$ . Planarity requires torsion zero, not curvature zero. Straight lines are special cases; the general conclusion is zero curvature everywhere, allowing reparametrized lines.

Q14. In error analysis of numerical differentiation for vector functions, why might $\|\mathbf{r}'(t)\|$ computed via finite differences exhibit larger relative error than individual components?

A.Vector norms amplify component errors through squaring and square root operations. ✅

B.Finite difference schemes are invalid for vector magnitudes.

C.Component derivatives cancel out in norm calculations.

D.Norms are insensitive to directional changes.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Each component error propagates nonlinearly into the norm. Squaring magnifies absolute errors, and the square root doesn’t fully compensate. Additionally, cancellation in components may reduce their individual errors while norm aggregates them. Students overlooking error propagation in composite functions miss why vector calculus demands careful numerical treatment beyond scalar analogs.

Q15. Given $\mathbf{r}(t) = e^{at}\mathbf{u} + e^{bt}\mathbf{v}$ with constant vectors $\mathbf{u}, \mathbf{v}$ and scalars $a \neq b$ , find $\mathbf{r}'(t) \times \mathbf{r}''(t)$ in simplest form.

ab(b-a)e^{(a+b)t}(\mathbf{u} \times \mathbf{v})

✅

(a^2b - ab^2)e^{(a+b)t}(\mathbf{u} \times \mathbf{v})

\mathbf{0}

(a-b)^2 e^{(a+b)t}(\mathbf{u} \times \mathbf{v})

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Compute derivatives: $\mathbf{r}' = ae^{at}\mathbf{u} + be^{bt}\mathbf{v}$ , $\mathbf{r}'' = a^2e^{at}\mathbf{u} + b^2e^{bt}\mathbf{v}$ . Cross product yields $(ab^2 - a^2b)e^{(a+b)t}(\mathbf{u} \times \mathbf{v}) = ab(b-a)e^{(a+b)t}(\mathbf{u} \times \mathbf{v})$ . Factoring reveals elegant structure testing algebraic manipulation alongside vector calculus rules.

Q16. Which scenario guarantees that $\frac{d}{dt}(\mathbf{u} \times \mathbf{v})$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ at a specific instant?

\mathbf{u}&#039;

and

\mathbf{v}&#039;

are both parallel to

\mathbf{u} \times \mathbf{v}

. ✅

\mathbf{u}

and

\mathbf{v}

are orthogonal.

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

happens to be orthogonal by coincidence.

D.Never guaranteed unless

\mathbf{u}

and

\mathbf{v}

are constant.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For $\mathbf{w} = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$ to be orthogonal to $\mathbf{u}$ , need $\mathbf{u} \cdot \mathbf{w} = 0$ . If $\mathbf{u}', \mathbf{v}' \parallel \mathbf{u} \times \mathbf{v}$ , then each cross term involves $(\mathbf{u} \times \mathbf{v}) \times \mathbf{v}$ etc., which lie in the $\mathbf{u}$ - $\mathbf{v}$ plane, making $\mathbf{w}$ orthogonal to both. Tests deep vector identity knowledge.

Q17. A student differentiates $\|\mathbf{r}(t)\|$ as $\|\mathbf{r}'(t)\|$ . Under what precise condition is this erroneous formula accidentally correct?

A.Only when

\mathbf{r}(t)

and

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

are parallel and

\|\mathbf{r}(t)\| > 0

. ✅

B.When

\mathbf{r}(t)

is constant.

C.Never correct.

D.When

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

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Correct derivative is $(\mathbf{r} \cdot \mathbf{r}') / \|\mathbf{r}\|$ . Equality with $\|\mathbf{r}'\|$ requires $\mathbf{r} \cdot \mathbf{r}' = \|\mathbf{r}\| \|\mathbf{r}'\|$ , i.e., Cauchy-Schwarz equality holding with positive proportionality. This occurs iff vectors are parallel and same direction. Recognizing accidental correctness prevents blind formula application and reinforces geometric meaning of dot products.

Q18. If a vector function $\mathbf{r}(t)$ satisfies $\mathbf{r}(t) \cdot \mathbf{r}'(t) = k$ (nonzero constant), what can be said about $\|\mathbf{r}(t)\|^2$ ?

A.It grows linearly with

t

. ✅

B.It is constant.

C.It grows quadratically.

D.It oscillates periodically.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since $\frac{d}{dt}\|\mathbf{r}\|^2 = 2\mathbf{r} \cdot \mathbf{r}' = 2k$ , integrating gives $\|\mathbf{r}\|^2 = 2kt + C$ . Linear growth follows directly. Students selecting constant confuse with orthogonality case ( $k=0$ ); quadratic assumes $\mathbf{r} \cdot \mathbf{r}' \propto t$ . This models radial expansion under constant power input in physics.

Q19. Consider $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ where $f,g,h$ are smooth. If $\mathbf{r}'(t_0) = \mathbf{0}$ but $\mathbf{r}''(t_0) \neq \mathbf{0}$ , why is the standard unit tangent vector undefined at $t_0$ ?

A.Because division by zero occurs in normalization. ✅

B.Because the curve has a cusp.

C.Because higher derivatives don't exist.

D.Because the curve leaves the domain.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Unit tangent is $\mathbf{T} = \mathbf{r}' / \|\mathbf{r}'\|$ . Vanishing velocity makes denominator zero regardless of acceleration. Even if the curve is smooth geometrically, the parametrization singularity prevents defining $\mathbf{T}$ via this formula. Reparametrization may resolve it, but the given expression fails fundamentally at stationary points.

Q20. In comparing analytical vs. computational approaches, why might symbolic differentiation of $\mathbf{r}(t) \times \mathbf{r}'(t)$ be preferred over numerical differentiation for detecting conserved quantities?

A.Symbolic methods preserve exact algebraic structures revealing hidden symmetries.

B.Numerical methods introduce truncation errors masking exact zeros. ✅

C.Both are equally effective.

D.Symbolic methods are faster computationally.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Conserved quantities like angular momentum require exact zero derivatives. Numerical noise produces small nonzero values indistinguishable from true variation, especially near machine precision. Symbolic manipulation maintains identities exactly, enabling definitive verification. Graphs of numerical derivatives show spurious fluctuations obscuring conservation, while symbolic results yield clean zero lines confirming invariants reliably.

Q21. Let $\mathbf{r}(t)$ be a unit-speed curve. What simplification occurs in $\frac{d}{dt}(\mathbf{r}'(t) \times \mathbf{r}''(t))$ compared to arbitrary parametrization?

A.The expression reduces to

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

. ✅

B.All terms vanish identically.

C.It becomes

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

D.No simplification occurs.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Unit speed implies $\mathbf{r}' \cdot \mathbf{r}' = 1$ , so $\mathbf{r}' \cdot \mathbf{r}'' = 0$ . Differentiating the cross product: $\mathbf{r}'' \times \mathbf{r}'' + \mathbf{r}' \times \mathbf{r}''' = \mathbf{r}' \times \mathbf{r}'''$ since self-cross product vanishes. Arbitrary parametrization retains extra terms involving speed derivatives. This highlights advantages of natural parametrization in simplifying higher-order expressions.

Q22. A model uses $\mathbf{v}(t) = \omega(t) \times \mathbf{r}(t)$ for rotating frames. If $\omega(t)$ is time-varying, what additional term appears in acceleration beyond Coriolis and centripetal?

A.Euler acceleration:

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

✅

B.Transverse acceleration:

\omega(t) \times \mathbf{v}(t)

C.Radial jerk:

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

D.No additional term needed.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Differentiate $\mathbf{v} = \omega \times \mathbf{r}$ : $\mathbf{a} = \omega' \times \mathbf{r} + \omega \times \mathbf{r}' = \omega' \times \mathbf{r} + \omega \times (\omega \times \mathbf{r})$ . The first term is Euler acceleration from angular acceleration. Students omitting it assume constant rotation rate, failing to model realistic spinning systems like precessing gyroscopes or variable-speed turbines accurately.

Q23. Why does $\frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'$ hold even when $\mathbf{u}, \mathbf{v}$ map to $\mathbb{R}^n$ for $n > 3$ ?

A.Dot product definition and limit laws generalize dimension-independently. ✅

B.Cross product analogues exist in higher dimensions.

C.It only holds for

n=3

by convention.

D.Higher-dimensional spaces lack inner products.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The dot product is defined as sum of component products in any dimension. Differentiation acts component-wise, and the product rule applies to each scalar product independently. Summation preserves the rule universally. This abstraction is crucial for functional analysis and machine learning where vector-valued functions inhabit infinite-dimensional spaces, yet derivative rules remain structurally identical.

Q24. If $\mathbf{r}(t)$ traces a helix with constant pitch and radius, which combination of derivatives remains invariant under reparametrization?

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

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

C.Curvature

\kappa

and torsion

\tau

✅

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

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Geometric invariants like curvature and torsion depend only on the curve’s shape, not parametrization. Velocity magnitude and cross products scale with parameter speed. While $\mathbf{r}' \times \mathbf{r}''$ relates to curvature, its magnitude isn’t invariant alone. Only properly normalized combinations yield true geometric descriptors. This distinguishes parametric artifacts from intrinsic properties essential in differential geometry.

Q25. Suppose $\mathbf{F}(t) = \int_a^t \mathbf{G}(s) ds$ where $\mathbf{G}$ is continuous. What is $\frac{d}{dt}(\mathbf{F}(t) \cdot \mathbf{F}(t))$ evaluated using fundamental theorem and derivative rules?

2\mathbf{F}(t) \cdot \mathbf{G}(t)

✅

\mathbf{G}(t) \cdot \mathbf{G}(t)

2\mathbf{G}(t) \cdot \mathbf{G}(t)

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By FTC, $\mathbf{F}'(t) = \mathbf{G}(t)$ . Then $\frac{d}{dt}(\mathbf{F} \cdot \mathbf{F}) = 2\mathbf{F} \cdot \mathbf{F}' = 2\mathbf{F} \cdot \mathbf{G}$ . This combines integral calculus with vector derivative rules seamlessly. Students selecting B confuse with derivative of $\|\mathbf{F}\|$ ; C squares G incorrectly; D differentiates G unnecessarily. Tests synthesis of multiple calculus pillars in vector context.

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