Arc Length of a Space Curve

Core Definition

Let $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}$ be a smooth space curve defined on the interval $a \le t \le b$ , such that the velocity vector $\vec{r}'(t)$ is continuous and $\vec{r}'(t) \ne \vec{0}$ . The total arc length $s$ traversed by a particle along the curve from $t = a$ to $t = b$ is the definite integral of the scalar speed $\|\vec{r}'(t)\|$ :

s = \int_{a}^{b} \|\vec{r}'(t)\|\,dt = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\,dt

The scalar differential of arc length is

ds = \|\vec{r}'(t)\|\,dt = \sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}\,dt

Concept Explanation

The Geometry of Arc Length

In single-variable calculus, the arc length of a planar curve $y = f(x)$ is computed by summing infinitesimal hypotenuses $ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + (f'(x))^2}\,dx$ . In multivariable calculus, spatial motion extends into three dimensions.

An infinitesimal step along a 3D trajectory $\vec{r}(t)$ has differential displacement:

d\vec{r} = \langle dx, dy, dz \rangle = \langle f'(t), g'(t), h'(t) \rangle\,dt

The scalar distance $ds$ traversed in time interval $dt$ is the Euclidean magnitude of this displacement vector:

ds = \|d\vec{r}\| = \sqrt{dx^2 + dy^2 + dz^2} = \sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}\,dt = \|\vec{r}'(t)\|\,dt

                   z ^
                     |            . P(t + dt)
                     |           /|
                     |     ds   / | dz
                     |         /  |
                     |  P(t) .----+
                     |       | dx/
                     |       |  / dy
                     |       | /
                     +-------+-------------> y
                    /
                   /
                  v x

The Arc Length Function $s(t)$

If we fix an initial reference point at $t = a$ and evaluate the accumulated distance traveled up to a variable time $t$ , we obtain the arc length function $s(t)$ :

s(t) = \int_{a}^{t} \|\vec{r}'(u)\|\,du

By the Fundamental Theorem of Calculus, the derivative of the accumulated arc length with respect to parameter

t

is exactly the scalar speed:

\frac{ds}{dt} = \|\vec{r}'(t)\| = v(t)

Engineering Applications

Structural Routing & Pipeline Design: Determining the exact raw material length of bent conduits, oil pipelines, and subsea cables required to route around subsurface geological obstacles.
Automated CNC Milling: Calculating total toolpath traversal distances to estimate cycle times, spindle wear, and coolant consumption.
Flight Path Navigation: Integrating instantaneous airspeed data to compute cumulative distance traveled during aircraft or drone flight corridors.

Example Problems

Example 1: Arc Length of a 3D Circular Helix

Find the total arc length of the circular helix $\vec{r}(t) = \langle 3\cos(t), 3\sin(t), 4t \rangle$ from $t = 0$ to $t = 2\pi$ .

Solution

State Governing Formula: Arc Length Formula for a Space Curve: $s = \int_{a}^{b} \|\vec{r}'(t)\|\,dt$
Differentiate Position Vector: $\vec{r}'(t) = \frac{d}{dt}\langle 3\cos(t), 3\sin(t), 4t \rangle = \langle -3\sin(t), 3\cos(t), 4 \rangle$
Compute Magnitude (Speed): $\|\vec{r}'(t)\| = \sqrt{(-3\sin(t))^2 + (3\cos(t))^2 + 4^2} = \sqrt{9(\sin^2(t) + \cos^2(t)) + 16} = \sqrt{9(1) + 16} = \sqrt{25} = 5$
Evaluate the Definite Integral: $s = \int_{0}^{2\pi} 5\,dt = [5t]_{0}^{2\pi} = 10\pi$

Example 2: Arc Length Function with Perfect Square Simplification

Find the arc length function $s(t)$ starting at $t = 0$ for $\vec{r}(t) = \langle \frac{1}{3}t^3, t^2, 2t \rangle$ for $t \ge 0$ .

Solution

State Governing Formula: Accumulated Arc Length Function Formula: $s(t) = \int_{a}^{t} \|\vec{r}'(u)\|\,du$
Differentiate Position Vector: $\vec{r}'(u) = \langle u^2, 2u, 2 \rangle$
Compute Magnitude & Factor Perfect Square: $\|\vec{r}'(u)\| = \sqrt{(u^2)^2 + (2u)^2 + 2^2} = \sqrt{u^4 + 4u^2 + 4} = \sqrt{(u^2 + 2)^2} = u^2 + 2$
Integrate from $u = 0$ to $u = t$ : $s(t) = \int_{0}^{t} (u^2 + 2)\,du = \left[ \frac{1}{3}u^3 + 2u \right]_{0}^{t} = \frac{1}{3}t^3 + 2t$

Connections & References

Parent Concepts: Derivatives of Vector-Valued Functions | Definite Integrals
Related Notes: Arc Length Parametrization | Curvature of a Space Curve | Unit Tangent Vector
Course Sessions: Calc-3 Session 06

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