Learning Objectives
Set up and evaluate arc length using the magnitude of a derivative.
Distinguish length along a curve from straight-line endpoint distance.
Construct an arc-length parameter for a constant-speed curve.
Prerequisites: vector derivatives, speed, and definite integration. Assume the curve is continuously differentiable on the interval unless stated otherwise. Angles are in radians.
1. Add Small Distances along the Curve
For r(t) on a ≤ t ≤ b, the length of the traversal is L = ∫ from a to b |r′(t)| dt. In coordinates, the integrand is √[(x′(t))² + (y′(t))² + (z′(t))²]. This is the same speed integral used for total distance in a motion problem.
The formula counts the curve as it is traversed. If the parameterization goes over the same segment twice, that segment contributes twice. For the geometric length of a path counted once, choose an interval that traces it once.
Worked Example: A Helix
Find the length of r(t) = ⟨3 cos t, 3 sin t, 4t⟩ from t = 0 to t = 2π.
The derivative is ⟨−3 sin t, 3 cos t, 4⟩. Its magnitude is √(9 sin²t + 9 cos²t + 16) = 5.
Hence L = ∫ from 0 to 2π 5 dt = 10π. The straight-line distance between the endpoints is only 8π, because the endpoints differ by ⟨0, 0, 8π⟩. The extra length comes from the circular motion.
2. Simplify the Speed before Integrating
Worked Example: A Perfect Square
Let r(t) = ⟨t, (2/3)t^(3/2), 0⟩ for 0 ≤ t ≤ 3. Then r′(t) = ⟨1, √t, 0⟩ and speed is √(1 + t).
The length is ∫ from 0 to 3 √(1 + t) dt = [(2/3)(1 + t)^(3/2)] from 0 to 3.
Therefore L = (2/3)(8 − 1) = 14/3. The parameter restriction t ≥ 0 is needed for the real square-root expression.
Not every arc-length integral has a simple elementary antiderivative. A correct setup is still meaningful. Use numerical integration when a question requests an approximation and an exact elementary evaluation is unavailable.
3. Measure Position by Distance Traveled
Define s(t) = ∫ from a to t |r′(u)| du for t ≥ a. This measures the length traveled from the starting parameter a. If speed is positive, s increases and can locally be used as a new parameter. Solving for t in terms of s gives an arc-length parameterization.
Worked Example: Reparameterize the Helix
The helix above has speed 5 and starts at t = 0, so s = 5t. Thus t = s/5.
Substitute to obtain R(s) = ⟨3 cos(s/5), 3 sin(s/5), 4s/5⟩ for 0 ≤ s ≤ 10π.
Differentiate with respect to s: R′(s) = ⟨−(3/5) sin(s/5), (3/5) cos(s/5), 4/5⟩. Its magnitude is 1, confirming that one unit of s corresponds to one unit of arc length.
Common Mistakes
Integrate the magnitude of the derivative, not the magnitude of position.
The magnitude of a vector integral is generally smaller than the integral of its magnitude when direction changes.
A new parameter requires a new interval as well as a substituted formula.
4. Practice
1. Find the length of r(t) = ⟨6t, 8t, 0⟩ for 0 ≤ t ≤ 3.
2. Find the length of r(t) = ⟨2 cos t, 2 sin t, t⟩ for 0 ≤ t ≤ π.
3. Reparameterize r(t) = ⟨3t, 4t, 0⟩, 0 ≤ t ≤ 2, by arc length measured from t = 0.
Worked Solutions
1. Speed is √(36 + 64) = 10, so the length is 10(3) = 30.
2. Speed is √(4 sin²t + 4 cos²t + 1) = √5. The length is π√5.
3. Speed is 5, so s = 5t and t = s/5. The new parameterization is R(s) = ⟨3s/5, 4s/5, 0⟩ for 0 ≤ s ≤ 10; its derivative has magnitude 1.
5. Summary
- Arc length is the integral of speed over the specified traversal.
- Simplify the magnitude before choosing an integration method.
- An arc-length parameterization has unit speed where it is differentiable.
- Track repeated traversal and update parameter intervals correctly.