Questions: Vector-Valued Functions and Parametric Curves

Question 1

Two vector-valued functions r₁(t) = ⟨cos t, sin t, 0⟩ and r₂(t) = ⟨cos 2t, sin 2t, 0⟩ both trace curves. Which statement is correct?

• A. They trace different curves because their derivatives are different
• B. They trace the same curve at the same speed
• C. They trace the same curve, but r₂ traverses it twice as fast
• D. They have the same tangent vectors at corresponding parameter values

Answer: (C)

Both functions trace the unit circle in the xy-plane — the same set of points. But r₂(t) has a parameter that runs at double the angular rate, so it completes a full revolution in half the time. The geometric curve (the set of points) is unchanged by reparametrization; the velocity vectors and speed change. This is the core distinction: the parametrization controls how fast you travel, not where you go.

Question 2

A particle moves along r(t) from t = 0 to t = T. What does |∫₀ᵀ r'(t) dt| represent?

• A. The total arc length of the path traveled
• B. The magnitude of the net displacement
• C. The average speed of the particle
• D. The total distance traveled, accounting for backtracking

Answer: (B)

∫₀ᵀ r'(t) dt = r(T) − r(0) by the Fundamental Theorem of Calculus, which is the net displacement vector — a straight-line arrow from start to end. Its magnitude is the straight-line distance between endpoints, not the path length. To get arc length (total distance traveled, accounting for the curve's shape), you must integrate the speed: L = ∫₀ᵀ |r'(t)| dt. Confusing these two is the most common error when students first encounter vector integration.

Question 3

The derivative r'(t) of a vector-valued function is a vector that is tangent to the curve and points in the direction of increasing t.

• T. True
• F. False

Answer: True

r'(t) = ⟨f'(t), g'(t), h'(t)⟩ is the velocity vector of the particle at time t. Geometrically, it is tangent to the curve at the point r(t) — it points in the direction the curve is heading at that instant — and it specifically points in the direction of increasing t (the orientation given by the parametrization). Dividing by its magnitude gives the unit tangent vector T(t) = r'(t)/|r'(t)|, which strips away speed and keeps only direction.

Question 4

Integrating a vector-valued function r(t) component-by-component from a to b gives the total arc length of the curve traced on that interval.

• T. True
• F. False

Answer: False

∫ₐᵇ r(t) dt computes an antiderivative (or, with limits, a displacement-like quantity) — it is a vector, not a scalar arc length. Arc length requires integrating the scalar speed: L = ∫ₐᵇ |r'(t)| dt. This is a crucial distinction: integrating r itself evaluates its antiderivative; integrating |r'| accumulates the total distance traveled. These are different operations with different outputs — one a vector, one a positive real number.

Question 5

Why must you integrate |r'(t)| to compute arc length rather than simply computing |∫r'(t) dt|?

Model answer: Because ∫r'(t) dt gives the net displacement vector (start to end), whose magnitude is the straight-line distance between endpoints — and that equals arc length only if the curve is a straight line with no backtracking. For any curved or winding path, the straight-line distance is shorter than the total distance traveled. Arc length accumulates tiny increments of actual path distance at each moment, which is |r'(t)| dt — the speed times the time step. Integrating speed gives total distance; integrating velocity gives net displacement.

The distinction mirrors the one you know from single-variable calculus: ∫₀ᵀ v(t) dt gives net displacement (can be zero if you return to start), while ∫₀ᵀ |v(t)| dt gives total odometer distance. In 3D, the speed |r'(t)| is the instantaneous rate at which arc length is accumulating, so integrating it gives the total path length regardless of how the curve winds through space.