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Arc Length and Curvature

Arc length for a 3D vector-valued function

Let

r(t)=(x(t),y(t),z(t)),a≤t≤b,\mathbf{r}(t) = \bigl(x(t), y(t), z(t)\bigr), \quad a \le t \le b,

with a continuous derivative.

To measure the length of the curve, break the interval [a,b][a,b] into small pieces. On one small piece, from tt to t+Δtt+\Delta t, the position change is

Δr≈r′(t) Δt\Delta \mathbf{r} \approx \mathbf{r}'(t)\,\Delta t

so its length is approximately

∥Δr∥≈∥r′(t)∥ Δt.\|\Delta \mathbf{r}\| \approx \|\mathbf{r}'(t)\|\,\Delta t.

Summing these small lengths and taking a limit gives

L=∫ab∥r′(t)∥ dt.L = \int_a^b \|\mathbf{r}'(t)\|\,dt.

If we write derivatives component-wise,

r′(t)=(x′(t),y′(t),z′(t)),\mathbf{r}'(t) = \bigl(x'(t),y'(t),z'(t)\bigr),

then by the 3D Pythagorean rule,

∥r′(t)∥=(x′(t))2+(y′(t))2+(z′(t))2.\|\mathbf{r}'(t)\| = \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2+\bigl(z'(t)\bigr)^2}.

So the arc-length formula is

L=∫ab(x′(t))2+(y′(t))2+(z′(t))2 dt.\boxed{L = \int_a^b \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2+\bigl(z'(t)\bigr)^2}\,dt.}

For a tiny move on the curve, let

dx=x′(t)dt,dy=y′(t)dt,dz=z′(t)dt.dx = x'(t)dt, \quad dy = y'(t)dt, \quad dz = z'(t)dt.

The tiny displacement vector is (dx,dy,dz)(dx,dy,dz), so its length is

ds=dx2+dy2+dz2ds = \sqrt{dx^2+dy^2+dz^2}

which is exactly the 3D distance formula (Pythagorean theorem in three perpendicular directions). Substitute dx=x′(t)dtdx=x'(t)dt etc.:

ds=(x′(t))2+(y′(t))2+(z′(t))2 dt=∥r′(t)∥dt.ds = \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2+\bigl(z'(t)\bigr)^2}\,dt = \|\mathbf{r}'(t)\|dt.

Integrating dsds from aa to bb gives total length.

Arc-length function

Fix a starting value aa. Define

s(t)=∫at∥r′(u)∥ du,s(t) = \int_a^t \|\mathbf{r}'(u)\|\,du,

where s(t)s(t) is the distance traveled along the curve from u=au=a to u=tu=t.

By the Fundamental Theorem of Calculus,

s′(t)=∥r′(t)∥.\boxed{s'(t)=\|\mathbf{r}'(t)\|.}

So the derivative of arc length is speed (magnitude of velocity / tangent vector).

If ∥r′(t)∥>0\|\mathbf{r}'(t)\|>0, then s(t)s(t) is strictly increasing and can be used as a new parameter (arc-length parameter).

Start with

s(t+h)−s(t)=∫tt+h∥r′(u)∥ du.s(t+h)-s(t)=\int_t^{t+h}\|\mathbf{r}'(u)\|\,du.

Divide by hh:

s(t+h)−s(t)h=1h∫tt+h∥r′(u)∥ du.\frac{s(t+h)-s(t)}{h}=\frac{1}{h}\int_t^{t+h}\|\mathbf{r}'(u)\|\,du.

As h→0h\to 0, the average value of a continuous function over [t,t+h][t,t+h] approaches its value at tt, so

lim⁡h→0s(t+h)−s(t)h=∥r′(t)∥.\lim_{h\to 0}\frac{s(t+h)-s(t)}{h}=\|\mathbf{r}'(t)\|.

Hence s′(t)=∥r′(t)∥s'(t)=\|\mathbf{r}'(t)\|.

Interactive illustrations

1) A 3D helix and its tangent vectors

For

r(t)=(cos⁡t, sin⁡t, t/(2π)),0≤t≤2π,\mathbf{r}(t)=\bigl(\cos t,\,\sin t,\,t/(2\pi)\bigr), \quad 0\le t\le 2\pi,

we have

r′(t)=(−sin⁡t, cos⁡t, 1/(2π)),\mathbf{r}'(t)=\bigl(-\sin t,\,\cos t,\,1/(2\pi)\bigr),

so the speed is

∥r′(t)∥=1+14π2,\|\mathbf{r}'(t)\|=\sqrt{1+\frac{1}{4\pi^2}},

which is constant.

Helix with sample points (constant speed curve)

2) Arc length as accumulated distance

The same helix can be viewed as adding many tiny pieces

ds=∥r′(t)∥dt.ds=\|\mathbf{r}'(t)\|dt.

Since the speed is constant here, arc length grows linearly with tt:

s(t)=∫0t∥r′(u)∥du=t1+14π2.s(t)=\int_0^t \|\mathbf{r}'(u)\|du = t\sqrt{1+\frac{1}{4\pi^2}}.

Use the sample points below to see that equal parameter steps produce equal arc-length increments for this curve.

Equal t-steps on a constant-speed helix

For the helix above,

L[0,2π]=∫02π1+14π2 dt=2π1+14π2.L_{[0,2\pi]}=\int_0^{2\pi}\sqrt{1+\frac{1}{4\pi^2}}\,dt =2\pi\sqrt{1+\frac{1}{4\pi^2}}.

Try confirming numerically by approximating with short straight segments between many nearby parameter values.

Optional: curvature and torsion

Curvature measures how strongly a curve bends.

Let

T(t)=r′(t)∥r′(t)∥\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}

be the unit tangent vector. Curvature is

κ=∥dTds∥.\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|.

Equivalent practical formula (for r′≠0\mathbf{r}'\neq 0):

κ(t)=∥r′(t)×r′′(t)∥∥r′(t)∥3.\boxed{\kappa(t)=\frac{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}.}

For a circle of radius RR, curvature is constant: κ=1/R\kappa=1/R.

Circle of radius 2 (constant curvature k = 1/2)

Curvature describes bending in the osculating plane; torsion describes twisting out of that plane in 3D.

Using

B=T×N,\mathbf{B}=\mathbf{T}\times\mathbf{N},

torsion can be defined by

au=−dBds⋅N. au = -\frac{d\mathbf{B}}{ds}\cdot\mathbf{N}.

A useful computational formula is

τ(t)=det⁡(r′(t),r′′(t),r′′′(t))∥r′(t)×r′′(t)∥2.\boxed{\tau(t)=\frac{\det\bigl(\mathbf{r}'(t),\mathbf{r}''(t),\mathbf{r}'''(t)\bigr)}{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|^2}.}

For the standard helix

r(t)=(acos⁡t, asin⁡t, bt),\mathbf{r}(t)=(a\cos t,\,a\sin t,\,bt),

both curvature and torsion are constants:

κ=aa2+b2,τ=ba2+b2.\kappa=\frac{a}{a^2+b^2}, \qquad \tau=\frac{b}{a^2+b^2}.

So a helix bends and twists at steady rates.

Helix: classic example with nonzero curvature and nonzero torsion

Summary

  1. For r(t)=(x(t),y(t),z(t))\mathbf{r}(t)=(x(t),y(t),z(t)), arc length on [a,b][a,b] is
L=∫ab∥r′(t)∥dt=∫abx′(t)2+y′(t)2+z′(t)2 dt.L=\int_a^b\|\mathbf{r}'(t)\|dt=\int_a^b\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\,dt.
  1. The arc-length function
s(t)=∫at∥r′(u)∥dus(t)=\int_a^t\|\mathbf{r}'(u)\|du

satisfies

s′(t)=∥r′(t)∥.s'(t)=\|\mathbf{r}'(t)\|.
  1. Curvature κ\kappa measures bending and torsion τ\tau measures twisting in 3D.