A particle moves along the curve r(t) = ⟨cos t, sin t, t⟩. A student computes r′(t) = ⟨−sin t, cos t, 1⟩ and concludes this is the particle's speed. What is wrong with this conclusion?
Ar′(t) is computed incorrectly — the derivative of a vector-valued function is a scalar
Br′(t) is the velocity vector, not the speed; speed is the scalar |r′(t)| = √(sin²t + cos²t + 1) = √2
Cr′(t) gives speed only if the parameter t represents time, which is not stated here
Dr′(t) is correct as the speed, since each component is differentiated independently
r′(t) is a vector — the tangent (velocity) vector — not a scalar speed. Speed is the magnitude |r′(t)|, which here equals √((-sin t)² + (cos t)² + 1²) = √(sin²t + cos²t + 1) = √2, a constant. Confusing a vector with its magnitude is the most common error with vector-valued derivatives. The vector tells you direction AND encodes speed through its magnitude; the magnitude extracts only the scalar speed.
Question 2 Multiple Choice
A particle travels along a space curve r(t) for t ∈ [0, 2π], beginning and ending at the same point. What does ∫₀²π r′(t) dt equal?
AThe total arc length of the curve
BThe magnitude of the average velocity
CThe zero vector ⟨0, 0, 0⟩
DThe position vector r(2π)
The definite integral ∫_a^b r′(t) dt gives net displacement — the vector from starting position to ending position. Since the particle returns to its start, net displacement is the zero vector. This is distinct from arc length, which is ∫₀²π |r′(t)| dt — a positive scalar measuring total distance traveled. A particle can travel a great distance along a winding path while having zero net displacement.
Question 3 True / False
Differentiating a vector-valued function r(t) = ⟨f(t), g(t), h(t)⟩ produces a scalar — the rate at which the magnitude |r(t)| changes.
TTrue
FFalse
Answer: False
r′(t) = ⟨f′(t), g′(t), h′(t)⟩ is a vector, computed component-wise. It represents the velocity (tangent vector) at each point on the curve, not the rate of change of |r(t)|. To find how the magnitude changes, you would need d/dt |r(t)|, a separate calculation using the chain rule. The derivative of a vector-valued function is always a vector of the same dimension.
Question 4 True / False
Integrating a vector-valued function r(t) = ⟨f(t), g(t), h(t)⟩ over an interval [a, b] produces a vector whose components are the definite integrals of each scalar component.
TTrue
FFalse
Answer: True
Integration of vector-valued functions is component-wise: ∫_a^b r(t) dt = ⟨∫_a^b f(t) dt, ∫_a^b g(t) dt, ∫_a^b h(t) dt⟩. The result is a vector, not a scalar. This component-wise principle carries through all of vector calculus — limits, continuity, differentiation, and integration each reduce to the corresponding scalar operation applied independently to each component.
Question 5 Short Answer
Explain the difference between net displacement and arc length for a particle moving along a space curve. When, if ever, are they equal in magnitude?
Think about your answer, then reveal below.
Model answer: Net displacement is the vector ∫_a^b r′(t) dt — how far and in what direction the particle moved from start to finish. Arc length is the scalar ∫_a^b |r′(t)| dt — total distance traveled along the path, regardless of direction. They are equal in magnitude only when the particle moves in a straight line without reversing direction, so every infinitesimal step contributes positively in the same direction.
The distinction mirrors displacement vs. distance in one-dimensional motion. A particle that traces a closed loop has zero net displacement but positive arc length. Net displacement is a vector and can be zero even after extensive travel; arc length is always nonnegative. This difference becomes crucial when studying arc-length parameterization, where the goal is to reparameterize a curve so that |r′(t)| = 1 everywhere, making arc length and parameter advance at the same rate.