📝 Curvature formulas summary (25 MCQs)

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

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What is Curvature formulas summary?

Definition:
Practical curvature formulas include $\kappa = \frac{\| \vec{r}' \times \vec{r}'' \|}{\| \vec{r}' \|^3}$ for 3D and $\kappa = \frac{|y''|}{(1+y'^2)^{3/2}}$ for graphs $y=f(x)$ .

Example:
For $y = \sin x$ , use the graph formula avoiding full vector parametrization.

Reason:
Multiple forms accommodate different representations, balancing computational ease with generality.

Easy

Medium

Hard

📝 All Curvature formulas summary MCQs

Q1. A particle moves along a curve defined by $\mathbf{r}(t) = \langle t^2, t^3, t \rangle$ . If the speed is constant, which condition must the acceleration vector $\mathbf{a}(t)$ satisfy relative to the velocity vector $\mathbf{v}(t)$ ?

\mathbf{a}(t)

must be parallel to

\mathbf{v}(t)

\mathbf{a}(t)

must be orthogonal to

\mathbf{v}(t)

✅

\mathbf{a}(t)

must equal zero

\mathbf{a}(t)

must point in the direction of curvature

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When speed is constant, the magnitude of velocity does not change. Differentiating $\mathbf{v} \cdot \mathbf{v} = c^2$ yields $2\mathbf{v} \cdot \mathbf{a} = 0$ , proving orthogonality. Students often confuse constant speed with zero acceleration, but acceleration can still exist solely to change direction without altering magnitude.

Q2. Given $\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, t \rangle$ , a student computes curvature using $\kappa = \frac{\|\mathbf{r}'(t)\|}{\|\mathbf{r}''(t)\|}$ . What is the fundamental flaw in this approach?

A.The formula requires unit tangent vectors only

B.The denominator should be

\|\mathbf{r}&#039;(t)\|^3

and numerator involves cross product ✅

C.The function is not differentiable at all points

D.Curvature cannot be computed for exponential spirals

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The correct curvature formula is $\kappa = \frac{\|\mathbf{r}' \times \mathbf{r}''\|}{\|\mathbf{r}'\|^3}$ . The student’s version omits the cross product entirely and uses incorrect powers. This misconception arises from confusing arc-length parameterization formulas with general parameterizations, leading to dimensionally inconsistent results that fail even simple verification tests.

Q3. For a space curve with position $\mathbf{r}(t)$ , if $\mathbf{T}(t) \cdot \mathbf{N}(t) = 0$ always holds, what does this imply about the derivative $\mathbf{T}'(t)$ ?

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

is always zero

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

is parallel to

\mathbf{T}(t)

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

is parallel to

\mathbf{N}(t)

✅

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

has no component in the binormal direction

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Since $\mathbf{T}$ is a unit vector, its derivative must be orthogonal to itself. By definition, $\mathbf{T}'(t) = \kappa v \mathbf{N}(t)$ , so it lies entirely in the normal direction. This connects orthogonality constraints with Frenet-Serret relationships, requiring synthesis of geometric and calculus concepts beyond mere formula recall.

Q4. A helix is parameterized by $\mathbf{r}(t) = \langle a \cos t, a \sin t, bt \rangle$ . If both torsion $\tau$ and curvature $\kappa$ are constant, what relationship between $a$ and $b$ ensures $\tau = \kappa$ ?

a = b

✅

a = \sqrt{2}b

b = \sqrt{2}a

D.No such relationship exists; they are independent

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For this helix, $\kappa = \frac{a}{a^2 + b^2}$ and $\tau = \frac{b}{a^2 + b^2}$ . Setting them equal gives $a = b$ . This requires deriving both quantities from scratch and solving an algebraic constraint, testing deep familiarity with how geometric parameters influence intrinsic curve properties in non-trivial configurations.

Q5. If $\mathbf{r}(t)$ traces a planar curve lying entirely in the plane $z = 2x + y$ , what must be true about the binormal vector $\mathbf{B}(t)$ ?

\mathbf{B}(t)

varies with

t

\mathbf{B}(t)

is constant and perpendicular to the plane ✅

\mathbf{B}(t)

is always parallel to the position vector

\mathbf{B}(t)

equals the zero vector

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Planar curves have zero torsion, meaning the osculating plane never changes. Thus, the binormal vector remains fixed and normal to the containing plane. Recognizing this links the algebraic plane equation to the geometric behavior of the Frenet frame, distinguishing planar from spatial curves through invariant vector properties rather than computation.

Q6. A student claims that if $\|\mathbf{r}'(t)\| = 1$ for all $t$ , then $\mathbf{r}''(t) = \mathbf{0}$ . Which counterexample best refutes this?

\mathbf{r}(t) = \langle t, 0, 0 \rangle

\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle

✅

\mathbf{r}(t) = \langle t^2, t, 0 \rangle

\mathbf{r}(t) = \langle e^t, e^{-t}, 0 \rangle

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Unit speed implies $\mathbf{v} \cdot \mathbf{a} = 0$ , not $\mathbf{a} = \mathbf{0}$ . The circle has constant speed 1 but nonzero centripetal acceleration. This distractor targets the common confusion between constant velocity (zero acceleration) and constant speed (acceleration perpendicular to motion), emphasizing vector versus scalar rate distinctions.

Q7. Given two curves $\mathbf{r}_1(t)$ and $\mathbf{r}_2(t)$ with identical curvature functions $\kappa(t)$ but different torsions, what can be concluded?

A.They are congruent up to rigid motion

B.They lie in parallel planes

C.They share the same osculating circles at corresponding points but twist differently ✅

D.One must be planar and the other spatial

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Curvature determines the osculating circle’s radius, while torsion governs out-of-plane twisting. Identical $\kappa$ means matching local bending, but differing $\tau$ implies distinct global shapes. This applies the Fundamental Theorem of Space Curves conceptually, showing that curvature alone doesn’t determine a curve uniquely without torsion data.

Q8. In computing arc length $s = \int_a^b \|\mathbf{r}'(t)\| dt$ , a student substitutes $u = \|\mathbf{r}'(t)\|$ directly into the integral limits. Why is this invalid?

A.Arc length integrals cannot use substitution

u

is not necessarily monotonic or invertible over [a,b] ✅

C.The differential

du

lacks a

dt

factor

D.Norms are non-differentiable everywhere

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Substitution requires a bijective, differentiable mapping. Speed may increase and decrease, making $u(t)$ non-invertible. Even if monotonic, $du = \frac{d}{dt}\|\mathbf{r}'\| dt$ introduces extra terms. This highlights misuse of calculus techniques when dealing with composite vector norms, stressing domain considerations in parametric integration.

Q9. If $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ and $f''(t) = g''(t) = h''(t) = 0$ for all $t$ , what is the torsion of the curve?

A.Undefined

B.Zero ✅

C.Equal to curvature

D.Depends on initial conditions

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Zero second derivatives imply linear component functions, so the curve is a straight line. Straight lines have zero curvature and undefined or zero torsion by convention. Since there is no bending or twisting, torsion vanishes. This tests basic recognition of degenerate cases within the formula framework for space curves.

Q10. A graph shows $\|\mathbf{r}'(t)\|$ increasing while $\|\mathbf{r}''(t)\|$ decreases. Can curvature $\kappa(t)$ still increase?

A.No, because both numerator factors decrease or stay flat

B.Yes, if the angle between

\mathbf{r}&#039;

and

\mathbf{r}&#039;&#039;

approaches 90° rapidly ✅

C.Only if the curve becomes planar

D.Impossible under smoothness assumptions

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Curvature depends on $\|\mathbf{r}' \times \mathbf{r}''\| / \|\mathbf{r}'\|^3$ . Even if magnitudes suggest decline, the cross product grows when vectors become more orthogonal. Graphs of individual norms miss angular information, so visual interpretation requires understanding vector geometry beyond scalar trends, challenging oversimplified heuristic reasoning.

Q11. For $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ , which expression correctly gives the tangential component of acceleration $a_T$ ?

\frac{\mathbf{r}&#039; \cdot \mathbf{r}&#039;&#039;}{\|\mathbf{r}&#039;\|}

\frac{\|\mathbf{r}&#039; \times \mathbf{r}&#039;&#039;\|}{\|\mathbf{r}&#039;\|}

\|\mathbf{r}&#039;&#039;\| \cos \theta

where

\theta

is angle with

\mathbf{T}

D.Both A and C are equivalent and correct ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Tangential acceleration is the projection of $\mathbf{a}$ onto $\mathbf{T}$ , yielding $\mathbf{a} \cdot \mathbf{T} = (\mathbf{r}' \cdot \mathbf{r}'')/\|\mathbf{r}'\|$ . Option C expresses the same via cosine definition. Recognizing equivalence tests conceptual fluency across representations, ensuring students don’t treat formulas as isolated recipes but as interconnected physical interpretations.

Q12. Suppose $\mathbf{r}(t)$ satisfies $\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$ for all $t$ . What geometric property does the curve possess?

A.It lies on a sphere centered at origin ✅

B.It is a straight line through origin

C.Its speed is constant

D.Its curvature is constant

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating $\|\mathbf{r}\|^2 = \mathbf{r} \cdot \mathbf{r}$ gives $2\mathbf{r} \cdot \mathbf{r}'$ . If this is zero, $\|\mathbf{r}\|$ is constant, so the curve lies on a sphere. This links dot product conditions to locus geometry, requiring reverse-engineering from derivative constraints to spatial configuration, blending calculus and analytic geometry.

Q13. A student derives torsion as $\tau = -\mathbf{N}' \cdot \mathbf{B}$ but forgets the negative sign in another context. In which scenario would omitting the sign yield physically meaningful but incorrect orientation?

A.Computing arc length

B.Determining whether a curve is right- or left-handed ✅

C.Finding tangential acceleration

D.Calculating curvature magnitude

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Torsion’s sign indicates chirality: positive for right-handed twisting, negative for left. Magnitude formulas often drop signs, but orientation-sensitive applications like molecular modeling or fluid dynamics require correct sign. Confusing signed vs. unsigned versions leads to mirrored interpretations, highlighting the importance of directional awareness in vector calculus beyond scalar outputs.

Q14. If reparameterizing $\mathbf{r}(t)$ by arc length $s$ yields $\|\frac{d\mathbf{r}}{ds}\| = 1$ , what happens to the expression for normal acceleration $a_N$ ?

A.It becomes

\kappa

B.It becomes

\kappa v^2

C.It remains

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

D.It simplifies to

\|\frac{d^2\mathbf{r}}{ds^2}\|

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: In arc-length parameterization, $v = 1$ , so $a_N = \kappa v^2 = \kappa$ . But also $\frac{d^2\mathbf{r}}{ds^2} = \kappa \mathbf{N}$ , so its norm is $\kappa$ . Both A and D seem valid, but D explicitly uses the new parameter’s derivative form, emphasizing adaptation of formulas under reparameterization—a key skill in differential geometry.

Q15. Which condition guarantees that the principal normal vector $\mathbf{N}(t)$ is undefined at some point?

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

\kappa(t) = 0

✅

\tau(t) = 0

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

is parallel to

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

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: $\mathbf{N} = \mathbf{T}' / \|\mathbf{T}'\|$ , and $\|\mathbf{T}'\| = \kappa v$ . If $\kappa = 0$ , the denominator vanishes and $\mathbf{N}$ is undefined, even if velocity is nonzero. Inflection points or straight segments cause this. Students often assume $\mathbf{r}'' \neq \mathbf{0}$ suffices, but zero curvature is the true singularity condition.

Q16. A projectile follows $\mathbf{r}(t) = \langle v_0 t \cos \theta, v_0 t \sin \theta - \frac{1}{2}gt^2, 0 \rangle$ . At apex, what is the relationship between $a_T$ and $a_N$ ?

a_T = 0, a_N = g

✅

a_T = g, a_N = 0

C.Both equal

g/\sqrt{2}

a_T = -g, a_N = 0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: At apex, vertical velocity is zero; velocity is purely horizontal. Acceleration is downward ( $-g \hat{j}$ ), perpendicular to horizontal velocity, so tangential component vanishes. Entire acceleration is normal, providing centripetal force for curved path. This applies decomposition formulas to physical motion, connecting kinematics with vector calculus in realistic trajectories.

Q17. If $\mathbf{r}(t)$ and $\mathbf{u}(t)$ are orthogonal vector functions with constant magnitudes, what is $\frac{d}{dt}(\mathbf{r} \times \mathbf{u}) \cdot (\mathbf{r} \times \mathbf{u})$ ?

A.Always zero ✅

B.Equal to

\|\mathbf{r}\|^2 \|\mathbf{u}&#039;\|^2

C.Depends on angle between derivatives

D.Equal to

2(\mathbf{r} \cdot \mathbf{u}&#039;)(\mathbf{u} \cdot \mathbf{r}&#039;)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Let $\mathbf{w} = \mathbf{r} \times \mathbf{u}$ . Then $\mathbf{w} \cdot \mathbf{w}' = \frac{1}{2} \frac{d}{dt} \|\mathbf{w}\|^2$ . Since $\|\mathbf{r}\|, \|\mathbf{u}\|$ constant and orthogonal, $\|\mathbf{w}\| = \|\mathbf{r}\|\|\mathbf{u}\|$ is constant, so derivative is zero. This combines cross product identities, constancy conditions, and differentiation rules in a multi-step deduction.

Q18. A curve has $\kappa(t) > 0$ and $\tau(t) = 0$ for all $t$ . A student concludes it must be a circle. What additional information is needed to validate this?

A.Constant speed

B.Constant curvature ✅

C.Planarity confirmation

D.Non-vanishing torsion derivative

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Zero torsion implies planarity, but planar curves with positive curvature aren’t necessarily circles—they could be ellipses or other closed curves. Only constant curvature plus planarity guarantees circularity. This exposes overgeneralization from partial data, emphasizing that multiple intrinsic conditions must align to characterize specific curve types uniquely.

Q19. When computing $\int_C \mathbf{F} \cdot d\mathbf{r}$ using $\mathbf{r}(t)$ , why can’t we replace $d\mathbf{r}$ with $\mathbf{T} ds$ unless $\mathbf{F}$ is expressed in Frenet frame?

A.Because

\mathbf{T}

is undefined at endpoints

B.We actually can always make this substitution regardless of frame ✅

C.The dot product becomes ambiguous without consistent basis

ds

is not invariant under reparameterization

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The substitution $d\mathbf{r} = \mathbf{T} ds$ is always valid by definition of arc length element. The distractors reflect misconceptions about frame dependence, but the line integral is geometric and independent of coordinate representation. This question reinforces foundational definitions against false limitations, clarifying that Frenet frame isn’t required for basic parametrization.

Q20. Given $\mathbf{r}(t) = \langle \sin t, \cos t, \ln(\sec t + \tan t) \rangle$ for $t \in (-\pi/2, \pi/2)$ , what is notable about its curvature?

A.It equals torsion everywhere

B.It is identically 1

C.It diverges at endpoints

D.It matches the curvature of a unit circle projected onto xy-plane ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The z-component is the Gudermannian function inverse, making this a tractrix-related curve. Its projection is a unit circle with curvature 1, and due to special parameterization, the 3D curvature equals the planar projection’s curvature. Recognizing this requires identifying non-obvious functional relationships and leveraging known curve properties, testing advanced pattern recognition.

Q21. If $\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{0}$ for all $t$ in an interval, what can be said about the curve segment?

A.It has infinite curvature

B.It is part of a straight line ✅

C.It lies in a plane but isn’t straight

D.It has constant nonzero torsion

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Vanishing cross product implies $\mathbf{r}''$ is parallel to $\mathbf{r}'$ , so acceleration has no normal component. Thus, no bending occurs, and the path is linear. This is a direct application of the curvature formula’s numerator condition, serving as a foundational diagnostic tool for detecting degeneracy in parametric curves.

Q22. A student computes binormal as $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ but obtains a vector pointing opposite to expected orientation. What likely caused this?

A.Used left-handed coordinate system

B.Swapped order to

\mathbf{N} \times \mathbf{T}

C.Miscomputed

\mathbf{N}

direction

D.All of the above are possible causes ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Binormal orientation depends on right-hand rule consistency. Reversing cross product order flips sign. Left-handed systems or incorrect $\mathbf{N}$ (e.g., from unnormalized $\mathbf{T}'$ ) also invert direction. Diagnosing orientation errors requires checking multiple procedural steps, emphasizing attention to conventions in vector operations beyond formula syntax.

Q23. For $\mathbf{r}(t) = \langle t, \cosh t, \sinh t \rangle$ , which statement about its torsion is correct?

A.Torsion is zero because y and z components are hyperbolic

B.Torsion equals curvature due to identity

\cosh^2 - \sinh^2 = 1

✅

C.Torsion is constant and nonzero

D.Torsion oscillates periodically

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Note $y^2 - z^2 = 1$ , so the curve lies on a hyperboloid. Computing derivatives reveals $\mathbf{r}' \times \mathbf{r}''$ magnitude and triple product yield $\tau = \kappa$ . This exploits hyperbolic identities to simplify otherwise messy expressions, rewarding insight over brute-force calculation and linking algebraic structure to geometric invariants.

Q24. In modeling a roller coaster loop as $\mathbf{r}(t)$ , engineers require $a_N \leq k g$ for safety. If speed $v(t)$ is prescribed, how should curvature $\kappa(s)$ be constrained?

\kappa(s) \leq k g / v(s)^2

✅

\kappa(s) \geq k g / v(s)^2

\kappa(s) \leq k g v(s)^2

D.Constraint depends on track banking angle

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Normal acceleration is $a_N = \kappa v^2$ . Safety limit $a_N \leq kg$ implies $\kappa \leq kg / v^2$ . This translates physical requirements into geometric design parameters, demonstrating applied vector calculus in engineering contexts. Note that banking affects lateral forces but not the fundamental $a_N$ relation used here for vertical loading.

Q25. If $\mathbf{r}(t)$ is twice differentiable and $\mathbf{r}'(t) \neq \mathbf{0}$ , but $\mathbf{T}'(t) = \mathbf{0}$ at isolated points, what is the curvature at those points?

A.Undefined

B.Zero ✅

C.Positive finite value

D.Infinite

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since $\mathbf{T}' = \kappa v \mathbf{N}$ and $v > 0$ , $\mathbf{T}' = \mathbf{0}$ implies $\kappa = 0$ . These are inflection-like points where bending momentarily ceases despite smooth motion. This distinguishes between singularities (where $\mathbf{T}$ undefined) and regular zeros of curvature, refining understanding of differentiability versus geometric behavior.

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