Smooth parametrizations of curves: 24 Practice MCQs with Answers and Explanation

Q1. A particle moves along a curve defined by $\mathbf{r}(t) = \langle t^3, t^2 \rangle$ . At $t=0$ , the velocity vector is zero. Which statement best characterizes the smoothness of this parametrization at the origin?

A.The curve is smooth because both component functions are infinitely differentiable polynomials.

B.The parametrization is not smooth at

t=0

because

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

, creating a cusp despite the curve's geometric continuity. ✅

C.The curve lacks smoothness because the second derivative

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

does not exist.

D.The parametrization is smooth everywhere because the speed

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

is continuous for all real

t

.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Smoothness requires a non-zero continuous tangent vector. While components are differentiable, $\mathbf{r}'(0)=\mathbf{0}$ violates the regularity condition. This distinguishes geometric smoothness from parametric smoothness; a cusp exists at the origin even though the path is continuous, making standard arc-length formulas inapplicable directly at that point.

Q2. Consider two parametrizations of the same semicircle: $\mathbf{r}_1(t) = \langle \cos t, \sin t \rangle$ for $t \in [0, \pi]$ and $\mathbf{r}_2(u) = \langle 1-2u, \sqrt{2u-u^2} \rangle$ for $u \in [0,1]$ . Why might $\mathbf{r}_2$ be problematic for computing line integrals compared to $\mathbf{r}_1$ ?

A.

\mathbf{r}_2

traverses the curve in the opposite orientation, negating the integral result.

B.

\mathbf{r}_2

has an undefined derivative at the endpoints

u=0

and

u=1

, violating smoothness on the closed interval. ✅

C.

\mathbf{r}_2

covers only half the semicircle due to domain restrictions.

D.

\mathbf{r}_2

produces a negative speed function which is physically impossible.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Although $\mathbf{r}_2$ traces the correct geometric path, the square root term yields infinite derivatives at endpoints. Smooth parametrizations require continuous non-zero tangents on the entire interval including endpoints for proper integration. This highlights that geometric equivalence does not guarantee parametric suitability for calculus operations requiring regularity.

Q3. Given a smooth curve $C$ with parametrization $\mathbf{r}(t)$ , a student reparametrizes using $t = s^3$ . If the original curve had non-zero velocity everywhere, what is the effect on the new parametrization $\mathbf{R}(s) = \mathbf{r}(s^3)$ at $s=0$ ?

A.The new parametrization remains smooth because

s^3

is a bijective smooth function.

B.The new parametrization loses smoothness at

s=0

because

\frac{dt}{ds} = 3s^2

vanishes there. ✅

C.The curvature of the curve changes at

s=0

due to the nonlinear time scaling.

D.The orientation of the curve reverses for

s < 0

, breaking smoothness.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Chain rule gives $\mathbf{R}'(s) = \mathbf{r}'(s^3) \cdot 3s^2$ . Even if $\mathbf{r}' \neq \mathbf{0}$ , the factor $3s^2$ creates a zero velocity at $s=0$ . This demonstrates that not all smooth coordinate changes preserve parametric regularity; the transition function must have a non-vanishing derivative to maintain smoothness.

Q4. An engineer models a roller coaster track using piecewise cubic splines. At a junction point, position and first derivative match, but the second derivative is discontinuous. From the perspective of smooth parametrizations, why is this insufficient for passenger comfort?

A.Passenger comfort depends only on positional continuity, so the design is adequate.

B.Velocity continuity ensures no sudden jolts, making acceleration discontinuity irrelevant.

C.Acceleration corresponds to force; a discontinuous second derivative implies an instantaneous jerk, violating physical smoothness requirements. ✅

D.The parametrization fails because cubic splines cannot represent circular loops accurately.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Smooth parametrizations in physics require at least $C^2$ continuity for realistic motion. While mathematically the curve is $C^1$ , the abrupt change in acceleration (jerk) causes discomfort. This applies mathematical smoothness concepts to engineering constraints, showing that higher-order differentiability matters beyond mere tangent continuity in applied modeling scenarios.

Q5. Examine the graph of a parametric curve that appears visually smooth but has a parametrization where $|\mathbf{r}'(t)| \to 0$ as $t \to a$ . If you attempt to compute the unit tangent vector $\mathbf{T}(t)$ near $t=a$ , what fundamental issue arises?

A.The unit tangent vector becomes undefined because division by zero speed occurs. ✅

B.The unit tangent vector oscillates wildly due to numerical instability.

C.The curve self-intersects at

t=a

, making the tangent ambiguous.

D.The limit of

\mathbf{T}(t)

exists but differs from the geometric tangent direction.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The unit tangent is defined as $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$ . When speed approaches zero, this ratio becomes indeterminate regardless of geometric appearance. This tests understanding that visual smoothness doesn't guarantee parametric regularity; the algebraic definition breaks down at singular points even when the traced path looks perfectly smooth.

Q6. A student claims that $\mathbf{r}(t) = \langle t^2, t^4 \rangle$ is not a smooth parametrization of $y=x^2$ because it traces the parabola twice. Identify the primary flaw in this reasoning.

A.The curve is actually smooth; tracing twice doesn't violate smoothness, but

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

does. ✅

B.The student is correct; double-tracing inherently violates the definition of smooth parametrizations.

C.The parametrization is invalid because

t^4

grows faster than

t^2

, causing asymptotic behavior.

D.Smoothness requires injectivity, so any non-one-to-one parametrization is automatically non-smooth.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The error conflates injectivity with smoothness. The actual problem is $\mathbf{r}'(0)=\mathbf{0}$ , not double-tracing. Many valid smooth parametrizations aren't injective (e.g., periodic curves). This tests precise understanding: smoothness concerns derivative regularity, not mapping uniqueness. The student misidentified the singularity source while correctly sensing something was wrong.

Q7. When converting the implicit curve $x^{2/3} + y^{2/3} = 1$ (an astroid) to parametric form, why is $\mathbf{r}(t) = \langle \cos^3 t, \sin^3 t \rangle$ preferred over solving for $y(x)$ explicitly?

A.Explicit solutions avoid trigonometric complexity and are always smoother.

B.The parametric form naturally handles vertical tangents and cusps without domain splitting, maintaining global smoothness except at isolated points. ✅

C.Implicit differentiation always produces discontinuous derivatives for closed curves.

D.The parametric version has constant speed, simplifying arc length calculations.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Solving explicitly requires four separate branches with domain restrictions and undefined derivatives at axes. The trigonometric parametrization unifies all branches into one expression where derivatives exist except at cusp points. This demonstrates how clever parametrization resolves geometric singularities that plague explicit representations, showcasing the power of parametric methods for non-functional curves.

Q8. Suppose $\mathbf{r}(t)$ is smooth on $[a,b]$ and $f: [c,d] \to [a,b]$ is a smooth bijection with $f'(u) > 0$ . Which property is NOT necessarily preserved under reparametrization $\mathbf{R}(u) = \mathbf{r}(f(u))$ ?

A.Regularity (non-zero velocity)

B.Orientation of traversal

C.Arc length of the curve

D.Curvature as a function of parameter value ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: While geometric curvature is invariant, its functional dependence on the parameter changes because $\kappa(u) = \kappa(f(u))$ composed with chain rule factors. Regularity, orientation, and total arc length are preserved under positive-diffeomorphism reparametrizations. This subtle distinction between geometric invariants and parametric expressions tests deep conceptual understanding of what 'smooth equivalence' truly means.

Q9. A computer graphics algorithm generates a curve via $\mathbf{r}(t) = \langle t, |t| \rangle$ . A developer argues it's smooth because both coordinates are continuous and the graph is a straight line for $t>0$ and $t<0$ . What is the most rigorous rebuttal?

A.Continuity alone doesn't imply differentiability;

|t|

lacks a derivative at

t=0

, so

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

doesn't exist. ✅

B.The curve has a corner at the origin, which visually disproves smoothness.

C.Straight lines are trivially smooth, so the issue must be computational precision.

D.The absolute value function is smooth almost everywhere, making this acceptable for rendering.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This targets the misconception that visual linearity implies smoothness. Rigorously, smoothness requires existence and continuity of derivatives. Since $d|t|/dt$ is undefined at zero, the vector derivative doesn't exist there. This basic recall question anchors HOTS items by ensuring foundational definitions are solid before tackling complex scenarios involving parametrization subtleties.

Q10. Compare the arc length computation for $\mathbf{r}(t) = \langle t, t^2 \rangle$ on $[0,1]$ versus its arc-length parametrization $\mathbf{r}(s)$ . Why is finding $\mathbf{r}(s)$ analytically often impractical despite being theoretically ideal?

A.Arc-length parametrization eliminates the need for integration entirely.

B.Inverting

s(t) = \int_0^t |\mathbf{r}&#039;(u)| du

typically involves non-elementary integrals or transcendental equations. ✅

C.The arc-length parametrization always introduces artificial singularities.

D.Numerical methods for arc-length parametrization are less accurate than direct integration.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: While $\mathbf{r}(s)$ simplifies theoretical proofs (unit speed), practical computation faces inversion barriers. For $\langle t,t^2\rangle$ , $s(t)$ involves $\int \sqrt{1+4u^2} du$ , whose inverse isn't elementary. This application question bridges theory and practice, explaining why textbooks use arbitrary parameters despite arc-length being the 'natural' choice.

Q11. A student computes the tangent line to $\mathbf{r}(t) = \langle t^3, t^6 \rangle$ at $t=0$ by taking $\lim_{t\to 0} \mathbf{r}'(t)/|\mathbf{r}'(t)|$ and gets $\langle 1, 0 \rangle$ . However, direct substitution gives $\mathbf{0}/0$ . Is their limiting approach valid here?

A.Yes, because the limit of the unit tangent exists and defines the geometric tangent despite parametric singularity. ✅

B.No, because L'Hôpital's rule cannot be applied to vector-valued functions.

C.No, because the curve

y=x^2

has no well-defined tangent at the origin.

D.Yes, but only because the exponent ratio is exactly 2; otherwise the limit wouldn't exist.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Although $\mathbf{r}'(0)=\mathbf{0}$ , the direction stabilizes as $t\to 0$ . Computing $\mathbf{r}'(t)/|\mathbf{r}'(t)| = \langle 3t^2, 6t^5 \rangle / \sqrt{9t^4+36t^{10}} \to \langle 1,0 \rangle$ . This shows geometric tangents can exist at parametric singularities through limits. This challenging problem separates parametric regularity from geometric tangent existence, requiring careful limit analysis.

Q12. In modeling fluid flow along a wire shaped as a helix, why would a non-uniform parametrization $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2), t^2 \rangle$ produce incorrect flux calculations if treated as uniform?

A.The parametrization reverses orientation for negative

t

, flipping flux sign.

B.Speed varies with

t

, so equal parameter intervals don't correspond to equal arc lengths, distorting density-weighted integrals. ✅

C.Helices require cylindrical coordinates; Cartesian parametrizations are inherently invalid.

D.The second derivative discontinuity introduces artificial turbulence in the model.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Flux integrals depend on arc length element $ds = |\mathbf{r}'(t)| dt$ . Non-uniform speed means sampling equally in $t$ oversamples slow regions and undersamples fast ones. Without correcting by $|\mathbf{r}'|$ , physical quantities like mass flux become erroneous. This application links parametrization choice to conservation laws in physics modeling.

Q13. Which modification transforms the non-smooth parametrization $\mathbf{r}(t) = \langle t, t^{1/3} \rangle$ at $t=0$ into a smooth one representing the same curve $x=y^3$ ?

A.Replace

t

with

t^3

to get

\langle t^3, t \rangle

, eliminating fractional exponents. ✅

B.Take absolute value:

\langle |t|, |t|^{1/3} \rangle

to ensure positivity.

C.Differentiate component-wise and integrate back to remove singularities.

D.No reparametrization can make

x=y^3

smooth because it has a vertical tangent.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Original has $y'(0)$ undefined. Substituting $t=u^3$ yields $\mathbf{R}(u)=\langle u^3, u \rangle$ , where $\mathbf{R}'(u)=\langle 3u^2, 1 \rangle \neq \mathbf{0}$ everywhere. Vertical tangents aren't inherently non-smooth; poor parametrization causes apparent singularities. This tests ability to diagnose whether non-smoothness is geometric or merely parametric, a crucial skill in curve analysis.

Q14. Two students debate whether $\mathbf{r}(t) = \langle e^{-1/t^2}, 0 \rangle$ for $t \neq 0$ and $\mathbf{r}(0)=\langle 0,0 \rangle$ is smooth. Student A says yes (all derivatives vanish at 0); Student B says no (function is flat at origin). Who is correct regarding smooth parametrization?

A.Student A, because infinite differentiability with continuous derivatives satisfies smoothness definition.

B.Student B, because vanishing derivatives mean the curve stops moving, violating regularity.

C.Both are partially correct; it's smooth but not regular at

t=0

. ✅

D.Neither; the function isn't defined for negative

t

, so smoothness is meaningless.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The function is $C^\infty$ (classic flat function example), satisfying smoothness. However, $\mathbf{r}'(0)=\mathbf{0}$ , failing regularity. Smooth parametrization typically requires both. This nuanced distinction trips many learners who conflate differentiability class with regularity. Recognizing that smooth ≠ regular is essential for advanced differential geometry applications.

Q15. When animating a camera path along $\mathbf{r}(t)$ , why might constant-speed reparametrization cause undesirable visual effects near high-curvature regions despite being mathematically smooth?

A.Constant speed forces the camera to rotate too quickly in tight turns, causing motion blur or disorientation. ✅

B.High curvature regions inherently lack smooth parametrizations.

C.Animation software cannot handle arc-length parametrization numerically.

D.Constant speed violates the Frenet-Serret equations, introducing torsion errors.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Mathematical smoothness doesn't guarantee perceptual quality. In high-curvature zones, constant linear speed demands rapid angular acceleration for orientation tracking, exceeding human visual comfort. This mixed-concept question connects differential geometry to animation principles, showing that optimal parametrization balances mathematical regularity with application-specific constraints like viewer physiology and rendering limitations.

Q16. A curve is given by $\mathbf{r}(t) = \langle t^2 \sin(1/t), t^2 \cos(1/t) \rangle$ for $t \neq 0$ and $\mathbf{r}(0)=\mathbf{0}$ . Despite wild oscillations near zero, why is this actually a smooth parametrization?

A.The amplitude decay

t^2

dominates oscillation frequency, making all derivatives vanish continuously at origin. ✅

B.Oscillatory functions are never smooth, so this must be a trick question.

C.Smoothness only requires continuity, not differentiability, near singular points.

D.The curve spirals infinitely but maintains constant speed throughout.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Computing derivatives shows $\mathbf{r}'(t) \to \mathbf{0}$ as $t\to 0$ because $t^2$ suppresses $1/t$ growth. Higher derivatives similarly vanish via product rule dominance. This Olympiad-style problem tests ability to analyze competing asymptotic behaviors and recognize that rapid oscillation doesn't preclude smoothness when amplitude decays sufficiently fast.

Q17. If a smooth parametrization $\mathbf{r}(t)$ satisfies $\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{0}$ for all $t$ , what can be concluded about the curve's geometry?

A.The curve lies entirely in a plane passing through the origin.

B.The curve is a straight line (or part thereof). ✅

C.The curvature is constant but nonzero.

D.The torsion is undefined everywhere.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Cross product zero implies $\mathbf{r}'$ and $\mathbf{r}''$ are parallel, meaning acceleration has no normal component. Thus, the curve doesn't bend; velocity direction is constant. Integrating shows $\mathbf{r}(t) = \mathbf{a} f(t) + \mathbf{b}$ , a line. This tests linking vector calculus conditions to geometric classification through multi-step deduction.

Q18. A student attempts to verify smoothness of $\mathbf{r}(t) = \langle t, \sqrt[3]{t} \rangle$ by checking if $x(t)$ and $y(t)$ are individually smooth. They conclude it's smooth since cube roots are defined everywhere. What critical aspect did they miss?

A.Cube root function has unbounded derivative at

t=0

, so

y&#039;(t)

doesn't exist there. ✅

B.Domain of cube root excludes negative numbers, breaking smoothness.

C.Smoothness requires boundedness, not just differentiability.

D.Individual component smoothness guarantees vector smoothness, so their conclusion is correct.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: While $\sqrt[3]{t}$ is continuous everywhere, $d/dt(t^{1/3}) = \frac{1}{3}t^{-2/3}$ blows up at zero. Component-wise differentiability fails at origin. This error analysis question targets the common mistake of assuming algebraic definability implies differentiability. Students must distinguish between function domain and derivative existence, especially for fractional powers.

Q19. Given a smooth closed curve $\mathbf{r}(t)$ with period $T$ , why can't we generally find a global arc-length parametrization that is also periodic with the same period $T$ ?

A.Arc-length parametrization always has period equal to total length

L

, not original parameter period. ✅

B.Periodicity is incompatible with smoothness for closed curves.

C.The speed function must be constant for periodicity, which rarely holds.

D.Reparametrization destroys the closure property of the curve.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Arc-length parameter $s$ ranges over $[0,L]$ per cycle, so natural period is $L$ . Original period $T$ equals $L$ only if speed is identically 1. Generally $T \neq L$ , so periods differ. This subtle point reveals that 'natural' parametrization changes temporal structure, important in Fourier analysis of curves and signal processing on manifolds.

Q20. In robotics, a manipulator follows path $\mathbf{r}(t)$ . If joint velocities saturate at certain $t$ , engineers reparametrize to $\mathbf{R}(u)$ respecting actuator limits. How does this affect the mathematical smoothness of the executed trajectory?

A.Reparametrization for hardware constraints may introduce

C^1

breakpoints where velocity profiles switch, reducing smoothness class. ✅

B.Actuator saturation doesn't affect mathematical smoothness since paths remain identical.

C.Smoothness improves because constrained motion avoids high-frequency oscillations.

D.Only acceleration smoothness degrades; velocity remains perfectly smooth.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Physical constraints often necessitate piecewise-defined time scalings with matching conditions. Even if individual segments are smooth, junctions may only achieve $C^1$ or $C^0$ continuity depending on constraint transitions. This real-world scenario shows idealized smooth parametrizations must be compromised for feasibility, balancing mathematical purity with engineering reality.

Q21. Consider $\mathbf{r}(t) = \langle t^3, t^3 \sin(1/t) \rangle$ for $t \neq 0$ , $\mathbf{r}(0)=\mathbf{0}$ . Unlike the $t^2$ case, why does this fail to be smooth at origin despite similar structure?

A.The derivative

y&#039;(t)

contains

t^2 \sin(1/t) - t \cos(1/t)

, whose

\cos(1/t)

term doesn't vanish as

t \to 0

.

B.Sine oscillations prevent any limit from existing at zero.

C.The

t^3

factor is insufficient to bound the derivative of

\sin(1/t)

, unlike

t^2

which provides extra damping. ✅

D.Both components must have matching decay rates for smoothness.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Differentiating: $y' = 3t^2 \sin(1/t) + t^3 \cos(1/t)(-1/t^2) = 3t^2 \sin(1/t) - t \cos(1/t)$ . As $t\to 0$ , first term vanishes but second oscillates without converging. Thus $y'(0)$ doesn't exist. This contrasts with $t^2$ case where derivative had $t \cos(1/t) \to 0$ . Tests precise asymptotic analysis.

Q22. A graph shows a parametric curve with a visible loop. The parametrization is smooth everywhere. What must be true about the velocity vector at the self-intersection point?

A.Velocity must be zero at intersection to allow direction change.

B.Velocity vectors at different passage times through the intersection point must be distinct and non-zero. ✅

C.Velocity must be perpendicular to itself, implying zero magnitude.

D.Self-intersections are impossible for smooth parametrizations.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Smoothness allows self-intersections as long as each passage has well-defined non-zero tangent. At intersection point $P = \mathbf{r}(t_1) = \mathbf{r}(t_2)$ , we need $\mathbf{r}'(t_1) \neq \mathbf{0}$ and $\mathbf{r}'(t_2) \neq \mathbf{0}$ , typically with different directions. This graph interpretation question clarifies that smoothness is local in parameter, not global in image space.

Q23. Why is the parametrization $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ considered 'better' than $\mathbf{R}(u) = \langle \cos(u^3), \sin(u^3), u^3 \rangle$ for studying helix properties, even though both trace the same set?

A.

\mathbf{R}(u)

has vanishing velocity at

u=0

, losing regularity despite geometric equivalence. ✅

B.

\mathbf{R}(u)

reverses orientation for negative

u

, complicating signed curvature.

C.Helices require linear z-dependence for constant pitch; cubic distortion alters intrinsic geometry.

D.

\mathbf{R}(u)

is not differentiable at

u=0

due to chain rule failure.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Both trace identical point sets, but $\mathbf{R}'(u) = 3u^2 \langle -\sin(u^3), \cos(u^3), 1 \rangle$ vanishes at origin. Regular parametrizations enable consistent tangent/frame definitions everywhere. This emphasizes that parametrization quality affects analytical tractability independent of geometry, a key insight for choosing appropriate representations in research.

Q24. A student argues that since $\mathbf{r}(t) = \langle t^4, t^6 \rangle$ has $\mathbf{r}'(0)=\mathbf{0}$ , the curve $y^{2}=x^{3}$ has no tangent at origin. Refute this using reparametrization concepts.

A.The curve has tangent

y=0

; reparametrizing as

\langle u^2, u^3 \rangle

shows

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

still, but limiting direction exists.

B.Actually the student is correct; semicubical parabolas have cusps with no tangent.

C.Reparametrizing as

\langle u^3, u^2 \rangle

gives non-zero derivative

\langle 0,0 \rangle

at origin, proving tangent exists.

D.Tangents are defined geometrically via secant limits, independent of parametrization; here secants approach horizontal axis. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Geometric tangent exists as limit of secant lines regardless of parametrization. For $y^2=x^3$ , solving $y=\pm x^{3/2}$ shows both branches have horizontal tangent at origin. Parametric singularity reflects poor choice, not geometric defect. This challenges overreliance on parametric criteria and reinforces intrinsic geometric definitions, crucial for advanced curve theory.

📝 Smooth parametrizations of curves (24 MCQs)

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