Arc Length of a Space Curve

The arc length of a space curve r(t) = ⟨x(t), y(t), z(t)⟩ from t = a to t = b is L = ∫ ‖r′(t)‖ dt, where ‖r′(t)‖ = √(x′² + y′² + z′²) is the speed. You add up speed times time along the curve. It is the 2D arc length formula with one more term for z.

Before this lesson: Vector Functions and Space Curves, Arc Length of Parametric Curves and Vectors in 3D

What arc length means in space

Picture a point moving along a curve in space, with position r(t) at time t. Its velocity is r′(t), and its speed is the length of that vector, ‖r′(t)‖. Distance traveled is speed times time, so over a tiny time step dt the point moves about ‖r′(t)‖ dt.

Adding up these tiny distances from t = a to t = b gives the length of the curve. It is like walking a winding mountain trail with a step counter: the counter adds every short step, even when the trail doubles back or climbs.

You can also think of the curve as a chain of short straight pieces, called chords. Each chord has a length from the 3D distance formula. As the pieces get shorter, the total of the chord lengths approaches the arc length.

Arc length formula for a space curve

L = \int_a^b \|\mathbf{r}'(t)\|\,dt = \int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 + \left(\dfrac{dz}{dt}\right)^2}\,dt

• The formula assumes r′(t) is continuous and the curve is traced exactly once as t runs from a to b.
• The arc length function measures length from a starting time t = a up to time t: s(t) = \int_a^t \|\mathbf{r}'(u)\|\,du, so ds/dt = \|\mathbf{r}'(t)\|.
• To parametrize by arc length, solve s = s(t) for t and substitute into r(t). The new curve r(s) has speed 1, so the parameter s measures distance along the curve.

Arc length of a space curve examples

Example 1: A line segment

Find the length of r(t) = ⟨1 + 2t, 3 − t, 2t⟩ for 0 ≤ t ≤ 3.

\mathbf{r}'(t) = \langle 2, -1, 2 \rangle

\|\mathbf{r}'(t)\| = \sqrt{2^2 + (-1)^2 + 2^2}

= \sqrt{4 + 1 + 4}

= \sqrt{9}

= 3

L = \int_0^3 3\,dt

= 3(3) - 3(0)

= 9

\mathbf{r}(0) = \langle 1, 3, 0 \rangle, \; \mathbf{r}(3) = \langle 7, 0, 6 \rangle

\sqrt{6^2 + (-3)^2 + 6^2} = \sqrt{81} = 9

Answer: the segment has length 9.

Example 2: One turn of a helix

Find the length of the helix r(t) = ⟨3 cos t, 3 sin t, 4t⟩ for 0 ≤ t ≤ 2π.

\mathbf{r}'(t) = \langle -3\sin t, \; 3\cos t, \; 4 \rangle

\|\mathbf{r}'(t)\|^2 = 9\sin^2 t + 9\cos^2 t + 16

= 9(\sin^2 t + \cos^2 t) + 16

= 9 + 16

= 25

\|\mathbf{r}'(t)\| = 5

L = \int_0^{2\pi} 5\,dt = 5(2\pi) - 5(0)

= 10\pi \approx 31.42

Answer: one turn of the helix has length 10π ≈ 31.42. The picture above shows why: unrolled, one turn is the hypotenuse of a right triangle with legs 6π and 8π.

Example 3: A curve with a perfect-square speed

Find the length of r(t) = ⟨t³, 3t², 6t⟩ for 0 ≤ t ≤ 1.

\mathbf{r}'(t) = \langle 3t^2, \; 6t, \; 6 \rangle

\|\mathbf{r}'(t)\|^2 = 9t^4 + 36t^2 + 36

= 9(t^4 + 4t^2 + 4)

= 9(t^2 + 2)^2

\|\mathbf{r}'(t)\| = 3(t^2 + 2) = 3t^2 + 6

L = \int_0^1 (3t^2 + 6)\,dt

= \left[ t^3 + 6t \right]_0^1

= (1 + 6) - (0 + 0)

= 7

Answer: the length is 7. As a check, the straight-line distance from (0, 0, 0) to (1, 3, 6) is √46 ≈ 6.78, a little less than 7.

Example 4: Arc length function and reparametrizing

For the helix r(t) = ⟨3 cos t, 3 sin t, 4t⟩, find the arc length function s(t) starting at t = 0. Then parametrize the helix by arc length, and find the point 5π units along the curve from (3, 0, 0).

s(t) = \int_0^t \|\mathbf{r}'(u)\|\,du = \int_0^t 5\,du

= 5t

t = \dfrac{s}{5}

\mathbf{r}(s) = \left\langle 3\cos\dfrac{s}{5}, \; 3\sin\dfrac{s}{5}, \; \dfrac{4s}{5} \right\rangle

\mathbf{r}'(s) = \left\langle -\dfrac{3}{5}\sin\dfrac{s}{5}, \; \dfrac{3}{5}\cos\dfrac{s}{5}, \; \dfrac{4}{5} \right\rangle

\|\mathbf{r}'(s)\|^2 = \dfrac{9}{25} + \dfrac{16}{25} = 1

\mathbf{r}(5\pi) = \langle 3\cos\pi, \; 3\sin\pi, \; 4\pi \rangle

= \langle -3, \; 0, \; 4\pi \rangle

Answer: s(t) = 5t, r(s) = ⟨3 cos(s/5), 3 sin(s/5), 4s/5⟩, and the point 5π units along is (−3, 0, 4π), half a turn up the helix.

Example 5: Length of a stair handrail (word problem)

A spiral staircase has a handrail shaped like the helix r(t) = ⟨1.2 cos t, 1.2 sin t, 3t/(2π)⟩ in meters. It makes 2 full turns (0 ≤ t ≤ 4π) and rises 3 m per turn. How long is the handrail?

\mathbf{r}'(t) = \left\langle -1.2\sin t, \; 1.2\cos t, \; \dfrac{3}{2\pi} \right\rangle

\|\mathbf{r}'(t)\|^2 = 1.44(\sin^2 t + \cos^2 t) + \dfrac{9}{4\pi^2}

= 1.44 + \dfrac{9}{4\pi^2}

\approx 1.44 + 0.2280 = 1.6680

\|\mathbf{r}'(t)\| \approx 1.2915

L = \int_0^{4\pi} 1.2915\,dt \approx 1.2915(4\pi)

\approx 16.23

Answer: the handrail is about 16.23 m long. Check by unrolling one turn: the legs are 2π(1.2) ≈ 7.54 m and 3 m, so one turn is √(7.54² + 3²) ≈ 8.11 m, and two turns are about 16.23 m.

Common mistakes

Related methods

Arc length in the plane

For a plane curve, drop the z-term: L = ∫ √(x′² + y′²) dt, as in arc length of parametric curves. For a graph y = f(x), it becomes ∫ √(1 + f′(x)²) dx, from arc length of a curve.

When the integral has no nice antiderivative

Most speeds do not simplify, so most arc length integrals must be estimated. For example, ⟨t, t², t³⟩ on 0 ≤ t ≤ 1 has speed √(1 + 4t² + 9t⁴), and its length is about 1.863. Use a calculator’s numerical integration or Simpson’s rule.

Why the arc length parameter matters

With a curve written in terms of s, the speed is 1, so the derivative is already a unit vector. The next lessons use this idea to find the unit tangent vector and to measure curvature.

Practice problems

1. Find the length of r(t) = ⟨2t, 3t, 6t⟩ for 0 ≤ t ≤ 2.
2. Find the length of r(t) = ⟨cos t, sin t, t⟩ for 0 ≤ t ≤ 2π.
3. Find the length of r(t) = ⟨5 sin t, 5 cos t, 12t⟩ for 0 ≤ t ≤ π.
4. Find the length of r(t) = ⟨t, t², (2/3)t³⟩ for 0 ≤ t ≤ 1.
5. Find the length of r(t) = ⟨t², 2t, ln t⟩ for 1 ≤ t ≤ e.
6. Find the length of r(t) = ⟨eᵗ cos t, eᵗ sin t, eᵗ⟩ for 0 ≤ t ≤ ln 2.
7. Find the arc length function s(t), starting at t = 0, for r(t) = ⟨4 cos t, 4 sin t, 3t⟩.
8. Parametrize r(t) = ⟨4 cos t, 4 sin t, 3t⟩ by arc length, measured from t = 0.
9. On the helix in Problem 7, find the point 10π units along the curve from (4, 0, 0), moving in the direction of increasing t.
10. A drone’s position after t seconds is r(t) = ⟨6t, 3t², t³⟩ meters. How far does it fly in the first 2 seconds?

\mathbf{r}'(t) = \langle 2, 3, 6 \rangle

\|\mathbf{r}'(t)\| = \sqrt{4 + 9 + 36}

= \sqrt{49} = 7

L = \int_0^2 7\,dt = 7(2) - 7(0) = 14

\mathbf{r}'(t) = \langle -\sin t, \; \cos t, \; 1 \rangle

\|\mathbf{r}'(t)\|^2 = \sin^2 t + \cos^2 t + 1

= 1 + 1 = 2

\|\mathbf{r}'(t)\| = \sqrt{2}

L = \int_0^{2\pi} \sqrt{2}\,dt = 2\pi\sqrt{2} \approx 8.89

\mathbf{r}'(t) = \langle 5\cos t, \; -5\sin t, \; 12 \rangle

\|\mathbf{r}'(t)\|^2 = 25\cos^2 t + 25\sin^2 t + 144

= 25 + 144 = 169

\|\mathbf{r}'(t)\| = 13

L = \int_0^{\pi} 13\,dt = 13\pi \approx 40.84

\mathbf{r}'(t) = \langle 1, \; 2t, \; 2t^2 \rangle

\|\mathbf{r}'(t)\|^2 = 1 + 4t^2 + 4t^4

= (1 + 2t^2)^2

\|\mathbf{r}'(t)\| = 1 + 2t^2

L = \left[ t + \dfrac{2t^3}{3} \right]_0^1

= 1 + \dfrac{2}{3} - 0 = \dfrac{5}{3}

\mathbf{r}'(t) = \left\langle 2t, \; 2, \; \dfrac{1}{t} \right\rangle

\|\mathbf{r}'(t)\|^2 = 4t^2 + 4 + \dfrac{1}{t^2}

= \left(2t + \dfrac{1}{t}\right)^2

\|\mathbf{r}'(t)\| = 2t + \dfrac{1}{t}

L = \left[ t^2 + \ln t \right]_1^e

= (e^2 + 1) - (1 + 0)

= e^2 \approx 7.39

x' = e^t(\cos t - \sin t)

y' = e^t(\sin t + \cos t)

z' = e^t

(\cos t - \sin t)^2 + (\sin t + \cos t)^2 = 2

\|\mathbf{r}'(t)\|^2 = e^{2t}(2 + 1) = 3e^{2t}

\|\mathbf{r}'(t)\| = \sqrt{3}\,e^t

L = \left[ \sqrt{3}\,e^t \right]_0^{\ln 2} = \sqrt{3}(2 - 1)

= \sqrt{3} \approx 1.73

\mathbf{r}'(t) = \langle -4\sin t, \; 4\cos t, \; 3 \rangle

\|\mathbf{r}'(t)\| = \sqrt{16 + 9} = 5

s(t) = \int_0^t 5\,du = 5t

t = \dfrac{s}{5}

\mathbf{r}(s) = \left\langle 4\cos\dfrac{s}{5}, \; 4\sin\dfrac{s}{5}, \; \dfrac{3s}{5} \right\rangle

\|\mathbf{r}'(s)\|^2 = \dfrac{16}{25} + \dfrac{9}{25} = 1

5t = 10\pi

t = 2\pi

\mathbf{r}(2\pi) = \langle 4\cos 2\pi, \; 4\sin 2\pi, \; 6\pi \rangle = \langle 4, 0, 6\pi \rangle

\mathbf{r}'(t) = \langle 6, \; 6t, \; 3t^2 \rangle

\|\mathbf{r}'(t)\|^2 = 36 + 36t^2 + 9t^4

= 9(t^2 + 2)^2

\|\mathbf{r}'(t)\| = 3t^2 + 6

L = \left[ t^3 + 6t \right]_0^2 = 8 + 12

= 20

Arc length of a space curve FAQ

What is the formula for the arc length of a space curve?

For r(t) = ⟨x(t), y(t), z(t)⟩ with a ≤ t ≤ b, the length is L = ∫ from a to b of √(x′(t)² + y′(t)² + z′(t)²) dt, the integral of the speed ‖r′(t)‖.

How is arc length in 3D different from arc length in 2D?

Only by one term. In 3D you add (dz/dt)² under the square root. Everything else, including the idea of integrating speed, stays the same.

What does it mean to parametrize a curve by arc length?

It means rewriting r in terms of s, the distance traveled along the curve. Then ‖r′(s)‖ = 1, and the point r(s) is exactly s units from the start.

Why do so many arc length integrals have no exact answer?

The square root of a sum of squares rarely simplifies. Textbook problems are chosen so that it does, often as a perfect square or a constant; in real work you estimate the integral numerically.

Summary

• Arc length of a space curve: L = ∫ from a to b of ‖r′(t)‖ dt, the integral of speed.
• ‖r′(t)‖ = √(x′² + y′² + z′²); in 2D the z-term drops out.
• Look for a constant speed (helices) or a perfect square under the root.
• The arc length function is s(t) = ∫ from a to t of ‖r′(u)‖ du, and ds/dt = ‖r′(t)‖.
• Parametrizing by arc length gives a curve with speed 1.

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