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πŸ“ Change of parameter for curves (25 MCQs)

πŸ“– From Calculus β€’ 13. Vector Valued Functions β€’ 25 questions available

What is Change of parameter for curves?

Definition:
Change of parameter replaces tt with a new variable uu via a bijective function t=Ο•(u)t = \phi(u) to alter the curve's description without changing its geometry.

Example:
Substituting t=u3t = u^3 in rβƒ—(t)=⟨t,t⟩\vec{r}(t) = \langle t, t \rangle gives rβƒ—(u)=⟨u3,u3⟩\vec{r}(u) = \langle u^3, u^3 \rangle.

Reason:
Reparameterization can simplify calculations, resolve singularities, or adapt to specific boundary conditions in applied problems.

8
Easy
6
Medium
11
Hard

πŸ“ All Change of parameter for curves MCQs

Q1. A particle moves along a curve defined by r(t)=⟨t,t2⟩\mathbf{r}(t) = \langle t, t^2 \rangle. If the parameter is changed to s=t3s = t^3, how does the orientation and speed of traversal change for t>0t > 0?

A.Orientation reverses; speed decreases uniformly.
B.Orientation remains same; speed increases non-linearly with respect to s. βœ…
C.Orientation remains same; speed becomes constant.
D.Orientation reverses; speed increases linearly with respect to s.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: When substituting t=s1/3t = s^{1/3}, the new vector function becomes R(s)=⟨s1/3,s2/3⟩\mathbf{R}(s) = \langle s^{1/3}, s^{2/3} \rangle. Since s=t3s = t^3 is strictly increasing for positive values, orientation is preserved. However, the derivative magnitude changes because the chain rule introduces a factor of dt/dsdt/ds, causing non-linear speed variation relative to the new parameter.

Q2. Consider two parametrizations of the same geometric path: r1(t)\mathbf{r}_1(t) and r2(u)\mathbf{r}_2(u). If \mathbf{r}_1'(t) \times \mathbf{r}_2'(u) = \mathbf{0} at corresponding points, what can be definitively concluded about the relationship between parameters tt and uu?

A.The parameters are identical.
B.The curve has zero curvature at that point.
C.The tangent vectors are parallel, implying a valid smooth reparametrization exists locally. βœ…
D.The acceleration vectors must also be parallel.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The cross product being zero indicates that the velocity vectors are collinear or parallel. This is a necessary condition for two parametrizations to trace the same geometric curve with consistent orientation. It does not imply identical parameters or equal accelerations, as reparametrization affects higher-order derivatives through the chain rule and product rule.

Q3. A student claims that reparametrizing r(t)=⟨cos⁑t,sin⁑t⟩\mathbf{r}(t) = \langle \cos t, \sin t \rangle using u=t2u = t^2 preserves the unit speed property because the geometric shape is unchanged. Which statement best analyzes this error?

A.The student is correct; geometry dictates speed.
B.The student confuses geometric invariance with kinematic properties; βˆ₯dr/duβˆ₯β‰ 1\| d\mathbf{r}/du \| \neq 1 due to the chain rule factor dt/dudt/du. βœ…
C.The student failed to recognize that u=t2u=t^2 is not one-to-one over all reals.
D.The error lies in assuming sine and cosine remain orthogonal under reparametrization.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Reparametrization preserves the image set (geometry) but alters the velocity vector magnitude unless specifically designed for arc length. Applying the chain rule gives dr/du=(dr/dt)(dt/du)d\mathbf{r}/du = (d\mathbf{r}/dt)(dt/du). Since dt/du=1/(2u)dt/du = 1/(2\sqrt{u}), the speed varies with position. Geometric invariance does not imply kinematic invariance; confusing these is a fundamental misconception in vector calculus.

Q4. Given a smooth curve CC and a bijective differentiable function Ο•(s)=t\phi(s) = t, which condition ensures that the reparametrized curve R(s)=r(Ο•(s))\mathbf{R}(s) = \mathbf{r}(\phi(s)) maintains the exact same orientation as r(t)\mathbf{r}(t)?

A.Ο•(s)\phi(s) must be continuous.
B.\phi'(s) > 0 for all s in the domain. βœ…
C.\phi''(s) = 0 for all s.
D.\|\mathbf{r}'(t)\| = \|\mathbf{R}'(s)\|.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Orientation preservation in reparametrization depends solely on the sign of the derivative of the parameter transformation function. If \phi'(s) > 0, the mapping is strictly increasing, preserving direction. A negative derivative would reverse orientation. Continuity alone allows decreasing functions, and equal speed is unrelated to directional consistency along the curve.

Q5. An engineer models a roller coaster track using r(t)\mathbf{r}(t). To simulate realistic g-forces, they must reparametrize by arc length ss. Why is solving s(t) = \int_0^t \|\mathbf{r}'(u)\| du analytically often impossible for cubic splines?

A.Cubic splines always produce discontinuous derivatives.
B.The integrand involves the square root of a quartic polynomial, leading to elliptic integrals without elementary antiderivatives. βœ…
C.Arc length parametrization only exists for planar curves.
D.Numerical methods cannot handle vector-valued functions.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: For a cubic spline component, \|\mathbf{r}'(t)\| typically involves (at2+bt+c)2+...\sqrt{(at^2+bt+c)^2 + ...}, resulting in the square root of a degree-four polynomial. Such integrals generally yield elliptic integrals, which lack closed-form expressions in elementary functions. This necessitates numerical inversion or approximation techniques for practical arc-length parametrization in engineering simulations involving complex trajectories.

Q6. If r(t)\mathbf{r}(t) describes a helix and we apply the transformation u=eβˆ’tu = e^{-t}, how does the behavior of the reparametrized curve R(u)\mathbf{R}(u) differ from r(t)\mathbf{r}(t) as tβ†’βˆžt \to \infty?

A.R(u)\mathbf{R}(u) approaches the origin asymptotically while r(t)\mathbf{r}(t) extends infinitely. βœ…
B.Both extend infinitely but at different rates.
C.R(u)\mathbf{R}(u) becomes undefined as tβ†’βˆžt \to \infty.
D.The pitch of the helix changes in R(u)\mathbf{R}(u).
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: As tβ†’βˆžt \to \infty, u=eβˆ’tβ†’0+u = e^{-t} \to 0^+. The original helix extends indefinitely along the z-axis. The reparametrized version maps this infinite extension to a finite interval near u=0u=0. While the geometric set traced is identical for corresponding domains, the limiting behavior in parameter space differs fundamentally, compressing infinite traversal into a boundary point.

Q7. Examine a graph showing speed v(t)v(t) versus time for a particle. After reparametrization by arc length ss, what would the graph of speed versus ss look like?

A.Identical shape to v(t)v(t).
B.A horizontal line at height 1. βœ…
C.A horizontal line at height equal to total arc length.
D.A curve with area equal to displacement.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: By definition, arc length parametrization means the parameter itself measures distance traveled. Therefore, the rate of change of position with respect to arc length is identically unity: βˆ₯dr/dsβˆ₯=1\| d\mathbf{r}/ds \| = 1. Regardless of the original speed profile v(t)v(t), plotting speed against arc length always yields a constant value of one, distinguishing it from time-based representations.

Q8. Two students reparametrize r(t)=⟨t,t⟩\mathbf{r}(t)=\langle t,t\rangle. Student A uses u=2tu=2t; Student B uses u=t+5u=t+5. They argue about whether the unit tangent vector T\mathbf{T} changes. Who is correct and why?

A.Student A, because scaling changes direction.
B.Student B, because translation affects tangency.
C.Neither; T\mathbf{T} is invariant under any smooth, orientation-preserving reparametrization. βœ…
D.Both are wrong; T\mathbf{T} depends on the specific formula used.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The unit tangent vector \mathbf{T} = \mathbf{r}'(t)/\|\mathbf{r}'(t)\| is a geometric property independent of parametrization. Under any smooth bijection with positive derivative, the chain rule introduces a scalar factor that cancels during normalization. Both linear transformations preserve orientation and smoothness, so T\mathbf{T} remains identical at corresponding geometric points regardless of scaling or translation.

Q9. In modeling fluid flow along a streamline r(t)\mathbf{r}(t), a researcher needs the curvature ΞΊ\kappa. If they accidentally use a non-arc-length parameter without adjusting the formula, what is the likely consequence?

A.Curvature will be underestimated everywhere.
B.Curvature calculation remains valid since it is geometric.
C.The standard formula \|\mathbf{r}''\| yields incorrect results; the full formula involving cross products is required. βœ…
D.Only torsion is affected, not curvature.
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Curvature is geometrically intrinsic, but its computational formula depends on parametrization. The simple form \kappa = \|\mathbf{r}''(s)\| applies only to arc length. For arbitrary tt, one must use \kappa = \|\mathbf{r}' \times \mathbf{r}''\| / \|\mathbf{r}'\|^3. Using the unadjusted formula with a general parameter produces erroneous values that vary with speed, misleading physical interpretations of flow dynamics.

Q10. Suppose r(t)\mathbf{r}(t) has a cusp at t=0t=0 where \mathbf{r}'(0)=\mathbf{0}. Can a smooth reparametrization R(s)\mathbf{R}(s) eliminate this singularity to make \mathbf{R}'(s) \neq \mathbf{0} everywhere?

A.Yes, by choosing s=t3s = t^3.
B.Yes, arc length parametrization always regularizes cusps.
C.No, a cusp is a geometric singularity invariant under diffeomorphisms. βœ…
D.Only if the cusp is removable via coordinate rotation.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: A cusp represents a genuine geometric singularity where the curve fails to have a well-defined tangent line. Reparametrization by a smooth bijection cannot create a non-zero velocity where none existed geometrically. While some algebraic singularities are artifacts of parametrization, true cusps persist because no smooth invertible transformation can map a zero-velocity point to a regular point while preserving the curve's local structure.

Q11. A satellite orbit is modeled by r(t)\mathbf{r}(t). Mission control wants to reparametrize so that equal parameter increments correspond to equal areas swept (Kepler’s second law). What mathematical object defines this new parameter?

A.Arc length integral.
B.Time itself, since orbits are naturally area-parametrized.
C.The integral of \|\mathbf{r}(t) \times \mathbf{r}'(t)\|. βœ…
D.The reciprocal of angular momentum magnitude.
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Kepler’s second law states that radius vectors sweep equal areas in equal times for gravitational orbits. However, for arbitrary parametrization, the area sweep rate is \frac{1}{2}\|\mathbf{r} \times \mathbf{r}'\|. To achieve uniform area parametrization, one must integrate this quantity. This differs from arc length and requires computing the magnitude of the cross product, linking geometry directly to the desired physical constraint.

Q12. When analyzing the graph of \|\mathbf{r}'(t)\| versus tt, a student observes symmetry about t=at=a. They conclude the reparametrized curve R(s)\mathbf{R}(s) must exhibit spatial symmetry. Why might this reasoning fail?

A.Speed symmetry never implies geometric symmetry.
B.Symmetry in speed does not guarantee symmetry in position; integration constants and initial conditions matter. βœ…
C.The graph should be of acceleration, not velocity.
D.Spatial symmetry requires \mathbf{r}''(t) symmetry instead.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: While symmetric speed profiles suggest potential geometric symmetry, the actual position depends on integrating velocity from a specific starting point. Even if \|\mathbf{r}'(a+h)\| = \|\mathbf{r}'(a-h)\|, the directions of \mathbf{r}' may differ asymmetrically. Additionally, the base point r(a)\mathbf{r}(a) might not lie on an axis of symmetry. Thus, speed symmetry is necessary but insufficient for concluding spatial symmetry without verifying directional and positional constraints.

Q13. Which reparametrization strategy is most appropriate when numerically integrating a line integral ∫Cfds\int_C f ds where ff varies rapidly near one endpoint?

A.Uniform arc length sampling.
B.Adaptive parameter clustering near the region of high variation. βœ…
C.Linear time rescaling.
D.Inverse tangent transformation globally.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Line integrals depend on geometric measure dsds, but numerical efficiency benefits from adaptive sampling. When the integrand ff changes rapidly in a localized region, concentrating parameter samples there reduces discretization error. Uniform arc length ignores integrand behavior, potentially missing critical features. Adaptive reparametrization aligns computational resources with functional complexity, optimizing accuracy without increasing total sample count unnecessarily.

Q14. If r(t)\mathbf{r}(t) is periodic with period TT, and we define u=t+sin⁑(t)u = t + \sin(t), under what condition does R(u)\mathbf{R}(u) remain a valid global parametrization of the same closed curve?

A.Never, because u(t)u(t) is not monotonic.
B.Always, since addition of sine preserves periodicity.
C.Only if 1+cos⁑(t)>01 + \cos(t) > 0 for all t, ensuring bijectivity within each period. βœ…
D.Only if the amplitude of sine is less than T/2Ο€T/2\pi.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: For u(t)=t+sin⁑(t)u(t) = t + \sin(t) to serve as a valid reparametrization, it must be strictly monotonic (bijective) on the domain. The derivative u'(t) = 1 + \cos(t) must be positive everywhere. Since cos⁑(t)β‰₯βˆ’1\cos(t) \geq -1, u'(t) \geq 0, but equals zero at odd multiples of Ο€\pi. Strict positivity requires avoiding these points or restricting domain. Without monotonicity, the mapping folds, violating the definition of a proper parametrization.

Q15. A curve is given by r(t)=⟨t3,t2⟩\mathbf{r}(t) = \langle t^3, t^2 \rangle. At t=0t=0, \mathbf{r}'(0)=\mathbf{0}. A student proposes s=t2s = t^2 to regularize it. Does R(s)\mathbf{R}(s) have a well-defined tangent at s=0s=0?

A.Yes, \mathbf{R}'(0) = \langle 0, 1 \rangle.
B.No, the cusp persists because s=t2s=t^2 is not invertible near zero.
C.Yes, but only from the right side since sβ‰₯0s \geq 0.
D.No, because t3t^3 becomes s3/2s^{3/2} which is non-differentiable at 0. βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: Substituting t=Β±st = \pm\sqrt{s} leads to ambiguity, but even considering tβ‰₯0t \geq 0, x=s3/2x = s^{3/2} and y=sy = s. Then dx/ds=(3/2)s1/2β†’0dx/ds = (3/2)s^{1/2} \to 0 and dy/ds=1dy/ds = 1. So \mathbf{R}'(0) = \langle 0,1 \rangle seems defined. However, the original curve has a cusp: left and right derivatives differ in direction. The reparametrization s=t2s=t^2 collapses both branches onto sβ‰₯0s \geq 0, losing directional information. Thus, while R(s)\mathbf{R}(s) appears differentiable, it misrepresents the original geometry’s singularity.

Q16. In computer graphics, BΓ©zier curves are often reparametrized for animation timing. If an animator desires ease-in-ease-out motion, which characteristic should the reparametrization function Ο•(t)\phi(t) possess?

A.Constant first derivative.
B.Zero second derivative at endpoints and maximum first derivative at midpoint. βœ…
C.Linear increase throughout.
D.Discontinuous jump at midpoint.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Ease-in-ease-out requires smooth acceleration and deceleration. Mathematically, this corresponds to \phi'(0) = \phi'(1) = 0 (zero initial/final velocity) and \phi''(t) changing sign appropriately. A sigmoidal or polynomial function satisfying boundary derivative conditions achieves this. Constant derivative implies uniform motion, linear implies constant acceleration, and discontinuities cause jerky motion. Matching human perceptual expectations demands careful design of the parameter transformation’s differential properties.

Q17. Two curves r1(t)\mathbf{r}_1(t) and r2(u)\mathbf{r}_2(u) trace the same path. If \mathbf{r}_1''(t) \cdot \mathbf{T} \neq \mathbf{r}_2''(u) \cdot \mathbf{T} at corresponding points, what does this indicate?

A.The curves are geometrically distinct.
B.Tangential acceleration is parametrization-dependent, unlike normal acceleration. βœ…
C.There is an error in computation since tangential components must match.
D.Normal acceleration differs due to reparametrization.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Acceleration decomposes into tangential and normal components: a=aTT+aNN\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}. The normal component aN=ΞΊv2a_N = \kappa v^2 relates to geometry and speed squared, while tangential component aT=dv/dta_T = dv/dt depends explicitly on how speed changes with the chosen parameter. Different parametrizations yield different dv/dtdv/dt even at the same geometric point, making tangential acceleration non-invariant, whereas curvature and normal direction remain geometric.

Q18. A physicist argues that since arc length ss is intrinsic, any observable derived from r(s)\mathbf{r}(s) must be frame-independent. Identify the flaw in linking reparametrization to reference frames.

A.Reparametrization is a mathematical relabeling, not a physical Lorentz or Galilean transformation. βœ…
B.Arc length is only invariant in Euclidean space.
C.Observables always depend on parametrization choice.
D.Frame independence requires tensor formulation, not just arc length.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Reparametrization changes only the label assigned to points on a curve, leaving geometric quantities invariant. Reference frame transformations involve changing coordinates in spacetime, affecting measurements like time intervals and lengths relativistically. Confusing internal parameter freedom with external observer dependence is a category error. Arc length parametrization simplifies equations but does not confer physical frame invariance; that requires proper tensorial representation under coordinate transformations.

Q19. Given r(t)=⟨etcos⁑t,etsin⁑t⟩\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t \rangle, find the reparametrization R(s)\mathbf{R}(s) in terms of arc length measured from t=0t=0. Which step is most computationally demanding?

A.Computing \|\mathbf{r}'(t)\|.
B.Integrating 2et\sqrt{2}e^t to get s(t)s(t).
C.Inverting s(t)s(t) to obtain t(s)t(s). βœ…
D.Substituting t(s)t(s) back into r\mathbf{r}.
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Here, \|\mathbf{r}'(t)\| = \sqrt{2}e^t, so s=2(etβˆ’1)s = \sqrt{2}(e^t - 1). Inversion gives t=ln⁑(1+s/2)t = \ln(1 + s/\sqrt{2}), which is straightforward. However, for more complex curves like elliptical helices, s(t)s(t) involves non-elementary integrals, making analytical inversion impossible. Even when integrable, transcendental equations often resist closed-form solutions. Thus, inversion is typically the bottleneck in explicit arc-length parametrization, requiring numerical methods in practice.

Q20. A student computes curvature using \kappa = \|\mathbf{r}''(t)\| / \|\mathbf{r}'(t)\|^2 instead of the correct formula. For r(t)=⟨t,t2⟩\mathbf{r}(t) = \langle t, t^2 \rangle at t=1t=1, what type of error occurs?

A.Overestimation by factor of \|\mathbf{r}'\|. βœ…
B.Underestimation by factor of \|\mathbf{r}'\|.
C.Correct result coincidentally due to planarity.
D.Error depends on concavity direction.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The correct formula is \kappa = \|\mathbf{r}' \times \mathbf{r}''\| / \|\mathbf{r}'\|^3. In 2D, \|\mathbf{r}' \times \mathbf{r}''\| = |x'y'' - y'x''|. The student’s denominator lacks one power of speed. At t=1t=1, \|\mathbf{r}'\| = \sqrt{5}, so their result exceeds true curvature by exactly 5\sqrt{5}. This systematic overestimation arises from dimensional inconsistency: curvature has units of inverse length, but their expression has inverse length divided by speed, revealing fundamental misunderstanding of derivative scaling.

Q21. When reparametrizing a piecewise-defined curve for CNC machining, continuity of which derivative is essential to prevent tool chatter?

A.Position only.
B.Velocity (C1C^1).
C.Acceleration (C2C^2). βœ…
D.Jerk (C3C^3).
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Tool chatter arises from abrupt force changes, which relate to acceleration discontinuities. While C1C^1 continuity prevents instantaneous velocity jumps, C2C^2 ensures smooth acceleration profiles, minimizing inertial shocks. In precision machining, jerk continuity (C3C^3) further improves surface finish, but C2C^2 is the minimum threshold to avoid destructive vibrations. Reparametrization must therefore preserve or enhance smoothness class beyond mere geometric continuity to satisfy dynamic mechanical constraints.

Q22. Consider r(t)=⟨cos⁑(t3),sin⁑(t3)⟩\mathbf{r}(t) = \langle \cos(t^3), \sin(t^3) \rangle. As tβ†’0t \to 0, how does the instantaneous angular speed behave compared to uniform circular motion?

A.It approaches zero, unlike constant angular speed. βœ…
B.It diverges to infinity.
C.It matches uniform motion since t3β‰ˆtt^3 \approx t near zero.
D.It oscillates indefinitely.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Angular position is ΞΈ=t3\theta = t^3, so angular speed is dΞΈ/dt=3t2d\theta/dt = 3t^2. As tβ†’0t \to 0, this tends to zero quadratically. Uniform circular motion has constant dΞΈ/dtd\theta/dt. Graphically, plotting angular speed versus time shows a parabola touching zero at origin, contrasting sharply with a horizontal line for uniform motion. This illustrates how nonlinear reparametrization distorts temporal perception of rotational dynamics despite identical geometric trajectory.

Q23. In proving that curvature is independent of parametrization, which mathematical tool is indispensable for handling the chain rule effects on second derivatives?

A.Integration by parts.
B.FaΓ  di Bruno’s formula or careful application of product and chain rules. βœ…
C.L’HΓ΄pital’s rule.
D.Green’s theorem.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Under reparametrization t=Ο•(s)t = \phi(s), the first derivative transforms via chain rule, but the second derivative involves both \phi''(s) and (\phi'(s))^2 terms due to product rule. Showing that extraneous terms cancel in the curvature formula requires meticulous expansion. FaΓ  di Bruno generalizes this, but even basic multivariable chain rule suffices. Ignoring these higher-order transformation terms is a common pitfall; recognizing their structured cancellation confirms geometric invariance rigorously.

Q24. A robotics path planner generates a trajectory r(t)\mathbf{r}(t) with variable speed. To execute it with constant motor RPM, they reparametrize by arc length. What hidden assumption could cause failure?

A.Motors can respond instantaneously to commanded speed.
B.Arc length parametrization guarantees bounded jerk.
C.The original path is everywhere regular (\mathbf{r}' \neq \mathbf{0}). βœ…
D.Constant RPM implies constant linear speed regardless of gear ratio.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Arc length parametrization requires \|\mathbf{r}'(t)\| > 0 everywhere to define s(t)s(t) as a strictly increasing bijection. If the original path contains stationary points (e.g., stops or cusps), s(t)s(t) plateaus, making inversion impossible. Assuming regularity without verification leads to division by zero or non-invertible mappings. Real-world paths often include dwell periods, necessitating preprocessing to remove or handle singularities before attempting arc-length conversion for constant-speed execution.

Q25. For the cycloid r(t)=⟨tβˆ’sin⁑t,1βˆ’cos⁑t⟩\mathbf{r}(t) = \langle t - \sin t, 1 - \cos t \rangle, the speed vanishes at t=2Ο€nt = 2\pi n. Despite this, the curve has well-defined tangents at cusps. How is this reconciled with reparametrization theory?

A.Cusps have no tangents; the premise is false.
B.The limit of T(t)\mathbf{T}(t) as tβ†’2Ο€nt \to 2\pi n exists from both sides and agrees. βœ…
C.Arc length parametrization removes the cusp entirely.
D.Tangent existence depends on \mathbf{r}'', not \mathbf{r}'.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Although \mathbf{r}'(2\pi n) = \mathbf{0}, the unit tangent \mathbf{T}(t) = \mathbf{r}'(t)/\|\mathbf{r}'(t)\| approaches ⟨1,0⟩\langle 1, 0 \rangle as tβ†’2Ο€nΒ±t \to 2\pi n^\pm. The geometric tangent exists as a limiting direction despite vanishing velocity. Reparametrization by arc length yields R(s)\mathbf{R}(s) with \mathbf{R}'(s) continuous and non-zero at corresponding points, effectively regularizing the parametrization while preserving the geometric tangent. This distinguishes parametric singularity from geometric singularity.

πŸ”— Related Topics (MCQs)