Arc Length of Standard Space Curves

Length from a Space-Curve Parametrization

A vector parametrization describes a curve by giving the position of a point as the parameter varies. In three dimensions, write the position as \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\). The derivative \(\mathbf{r}'(t)\) is the velocity vector, and its magnitude is the speed along the curve.

Arc length accumulates this speed over the parameter interval. This is especially useful for standard space curves: differentiating their coordinates often produces terms that simplify before the integral is evaluated.

The formula uses the whole velocity vector: motion in any coordinate direction contributes to the distance traveled. A zero velocity at an isolated parameter value does not invalidate the integral. For the examples below, the coordinates are measured in units and the parameter is dimensionless, so the resulting lengths are in units.

Why the Speed Integral Gives Length

Over a small parameter change \(\Delta t\), the change in position is approximately \(\mathbf{r}'(t)\Delta t\). Its magnitude is approximately \(\|\mathbf{r}'(t)\|\Delta t\), the speed multiplied by the parameter change. Adding these small distances along the curve leads to the integral of speed. Equivalently, in the limit of finer partitions, the sum of the lengths of the short chords approaches the curve’s length, and those chord lengths are approximated by speed times parameter increment.

For calculating standard curves, the practical steps are direct: differentiate each coordinate, compute the magnitude of the resulting vector, simplify that magnitude, and integrate on the stated interval. The interval matters: a parametrization that traces the same geometric curve several times accumulates length on each traversal.

Line Segments in Space

A line parametrization has constant velocity. If \(\mathbf{r}(t)=\mathbf{p}+t\mathbf{v}\) for \(a\le t\le b\), where \(\mathbf{p}\) and \(\mathbf{v}\) are fixed vectors, then \(\mathbf{r}'(t)=\mathbf{v}\). Its speed is constant, so the length is the speed multiplied by the parameter interval.

Circular Arcs in Space

A circle of radius \(R>0\) in a horizontal plane can be parametrized as \(\mathbf{r}(t)=\langle h+R\cos t,k+R\sin t,z_0\rangle\), where \((h,k,z_0)\) is the center. Differentiating gives \(\langle-R\sin t,R\cos t,0\rangle\). Its magnitude is constant because \(\sin^2 t+\cos^2 t=1\). Thus an interval of angular width \(|b-a|\) traces an arc of length \(R|b-a|\). The same reasoning works for a circle in any plane when its two varying coordinate directions are perpendicular unit vectors.

Circular Helices

A circular helix winds around a cylinder while changing height. A standard parametrization is \(\mathbf{r}(t)=\langle R\cos t,R\sin t,ct\rangle\), where \(R>0\) is the cylinder’s radius and \(c\) is the vertical change per unit of parameter. Its derivative has both circular and vertical components. These components are perpendicular, so their squared magnitudes add when finding speed.

To see why the simplification works, differentiate and compute the squared magnitude. The first two coordinates contribute the circular motion, and the third contributes the vertical motion:

Since \(R^2+c^2\) is constant, the speed is \(\sqrt{R^2+c^2}\) throughout the interval. Integrating that constant speed from \(a\) to \(b\) gives \(\sqrt{R^2+c^2}(b-a)\) when \(a\le b\), which is the stated formula.

Interpreting the Parameter Interval

For a helix, a parameter interval of \(2\pi\) corresponds to one complete turn because the sine and cosine coordinates complete one period. An interval of \(4\pi\) corresponds to two turns, even though the curve lies on the same cylinder. The length formula includes both turns. The vertical change over an interval is \(c(b-a)\), while the angle traversed is \(b-a\) radians.

For example, changing the previous interval to \(0\le t\le4\pi\) doubles the length to \(4\pi\sqrt{13}\) units. This is not a different formula: it follows from integrating the same constant speed over twice the parameter interval. Always use the interval that is actually specified rather than assuming the parametrization traces only one copy of the curve.

A Common Speed Error

A frequent mistake is to calculate only the speed of the circular projection of a helix and omit the vertical coordinate. For the example \(\mathbf{r}(t)=\langle2\cos t,2\sin t,3t\rangle\), that incorrect calculation would be

This is the circumference of the circle traced in the horizontal plane, not the length of the rising helix. The third coordinate has derivative \(3\), so its squared contribution must be included:

The corrected result is greater than \(4\pi\), as it must be: the helix travels around the cylinder and also rises. In general, when computing speed in three dimensions, square and add all three coordinate derivatives before taking the square root.

Choosing an Efficient Calculation

For the standard curves in this tutorial, the key simplification occurs before integration. A line has constant derivative. A circle has a speed that simplifies using \(\sin^2 t+\cos^2 t=1\). A circular helix has the same circular contribution plus the square of its vertical derivative. Once the speed is constant, the integral reduces to speed times the parameter-interval length.

These calculations also illustrate why a parametrization and its interval must be read together. The parametrization identifies the path and its direction of motion; the interval specifies how much of that path is traced. The length integral accounts for the traveled curve, including repeated turns, rather than merely measuring the distance between its endpoints.