Learning Objectives
Connect vector-function domains, derivatives, motion, and arc length.
Select scalar or vector operations that match the requested quantity.
Check tangent directions, initial conditions, and unit-speed parameterizations.
Prerequisites: the four vector-function lessons in this section and single-variable calculus. Angles are in radians. This review focuses on regular parameterized curves; curvature and moving normal frames are later extensions.
1. Follow the Chain of Quantities
Quantity | Calculation | Type |
|---|---|---|
Position | r(t) | Vector |
Velocity or tangent direction | r′(t) | Vector |
Acceleration | r″(t) | Vector |
Speed | |r′(t)| | Nonnegative scalar |
Displacement on [a,b] | r(b) − r(a) | Vector |
Traversal length | ∫ from a to b |r′(t)| dt | Nonnegative scalar |
The distinction between a vector and a scalar is part of the answer. For example, the distance traveled cannot be ⟨3, 4, 0⟩. That vector could be a displacement; its magnitude is 5, but the total distance may be larger if the path is curved or retraced.
2. One Curve, Several Questions
Worked Example: A Circular Path
Let r(t) = ⟨2 cos t, 2 sin t, 1⟩ for 0 ≤ t ≤ π. At t = π/2, position is ⟨0, 2, 1⟩.
The derivative is ⟨−2 sin t, 2 cos t, 0⟩, giving tangent direction ⟨−2, 0, 0⟩ at that instant. A tangent line is L(s) = ⟨0, 2, 1⟩ + s⟨−2, 0, 0⟩.
Speed is 2 throughout. The distance over the interval is 2π, but displacement is ⟨−2, 0, 1⟩ − ⟨2, 0, 1⟩ = ⟨−4, 0, 0⟩.
The displacement magnitude is 4, the diameter of the circle. The distance is the longer semicircular arc.
3. Check Before Applying a Formula
Common Mistakes
A tangent-direction formula using r′/|r′| requires a nonzero derivative.
A vector antiderivative includes one constant per coordinate.
For speed change, use v · a rather than inspecting the sign of one acceleration component.
Do not confuse the parameter interval with the range of a coordinate.
4. Mixed Practice
1. Find the domain of ⟨ln(t − 1), √(5 − t), 1/(t − 3)⟩.
2. Find r′(t) and r″(t) for r(t) = ⟨t², t³, e^(2t)⟩.
3. Find a tangent line to r(t) = ⟨t², t, 0⟩ at t = 2.
4. Given v(t) = ⟨2, 6t, 0⟩ and r(0) = ⟨1, −1, 4⟩, find r(t).
5. Find the length of r(t) = ⟨cos t, sin t, 2t⟩ on 0 ≤ t ≤ 2π.
6. At an instant, v = ⟨1, 2, 2⟩ and a = ⟨2, −1, 0⟩. Find speed and its instantaneous rate of change.
7. Write an arc-length parameterization of r(t) = ⟨0, 5t, 12t⟩ for 0 ≤ t ≤ 1, starting at t = 0.
Worked Solutions
1. The restrictions are t > 1, t ≤ 5, and t ≠ 3. The domain is (1, 3) ∪ (3, 5].
2. r′(t) = ⟨2t, 3t², 2e^(2t)⟩ and r″(t) = ⟨2, 6t, 4e^(2t)⟩. Both exponential derivatives require the chain-rule factor 2.
3. The point is ⟨4, 2, 0⟩ and direction is ⟨4, 1, 0⟩. Thus L(s) = ⟨4, 2, 0⟩ + s⟨4, 1, 0⟩.
4. Integrate and apply the initial position: r(t) = ⟨1 + 2t, −1 + 3t², 4⟩.
5. Speed is √(sin²t + cos²t + 4) = √5. The length is 2π√5.
6. Speed is √(1 + 4 + 4) = 3. Since v · a = 2 − 2 = 0, the instantaneous rate of change of speed is 0/3 = 0. Acceleration itself is not zero.
7. Speed is 13, so s = 13t. The new curve is R(s) = ⟨0, 5s/13, 12s/13⟩ for 0 ≤ s ≤ 13, and |R′(s)| = 1.
5. Summary
- Component calculus produces vectors; magnitudes produce scalars.
- Use derivatives for tangent directions and motion, and integrate speed for length.
- Check domain, nonzero-speed conditions, constants, and parameter intervals.
- Interpret the result geometrically rather than stopping at the calculation.