2.4 Arc Length

How long is a curve? The honest answer starts with something you can actually measure: straight segments. Pick sample points along the curve, connect neighbors by chords, and add up the chord lengths. The result — the length of an inscribed polygon — is a slight underestimate, since each chord shortcuts its arc. Refine the sampling and the polygon hugs the curve ever more closely; arc length is the limit of the chord sums.

The figure below stages the definition on the helix from earlier in this chapter, drawn twice: once finely sampled, so it reads as the smooth curve, and once from only eight sample points, so it is the inscribed polygon.

A smooth red helix winding two full turns up around the z-axis, threaded by a blue chain of seven straight segments whose corners all lie exactly on the helix: the same curve sampled at only eight points, so the blue path is the inscribed chord polygon that arc length is defined from. The blue shortcut visibly hugs the inside of every red turn, making the polygon slightly shorter than the curve it approximates. — The helix $open parenthesis cos t, sin t, t / 4 close parenthesis$ over two full turns, drawn smooth (red, 500 samples) and as its own inscribed chord polygon (blue, 8 samples — seven chords). The blue path cuts inside every turn of the red one.

Explore

Orbit until you can sight along one turn of the helix. Where does each blue chord sit relative to the red arc it replaces — inside or outside? What does that say about the sign of the error in a chord-sum estimate?
The seven chords look congruent. Use the symmetry of the helix to explain why they must be.
Roughly estimate one chord's length against the grid, multiply by seven, and hold onto the number — the example below computes what the fine curve converges to.

From chord sum to integral

Now let calculus take the limit. On a fine partition $t sub 0, t sub 1, …, t sub n$ , the chord from $𝐫 open parenthesis t sub k minus 1 close parenthesis$ to $𝐫 (t_{k})$ has length approximately $‖ 𝐫^{'} (t_{k}) ‖ Δ t$ — displacement is velocity times elapsed time, when the elapsed time is short. The chord sum is then a Riemann sum, and its limit is

$L = \int_{a}^{b} ‖ 𝐫^{'} (t) ‖ d t = \int_{a}^{b} \sqrt{f^{'} (t)^{2} + g^{'} (t)^{2} + h^{'} (t)^{2}} d t .$

The integrand $‖ 𝐫^{'} (t) ‖$ is the speed of the moving point, so the formula says something you already believed: distance traveled is speed integrated over time. For plane curves $h equals 0$ and the same formula covers the cycloid and rose of the playground section.

Example 1 Exact length of the helix, and the polygon's verdict

Find the length of $𝐫 open parenthesis t close parenthesis equals open parenthesis cos t, sin t, t / 4 close parenthesis$ for $t ∈ [ 0, 4 π ]$ , and compare with the seven-chord polygon in the figure above.

Differentiate componentwise: $𝐫^{'} (t) = (- \sin t, \cos t, 1 / 4)$ , so the speed is

$‖ 𝐫^{'} (t) ‖ = \sqrt{\sin^{2} t + \cos^{2} t + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4},$

a constant — the helix is traced at uniform speed, which is also why the polygon's chords are congruent. Therefore

$L = \int_{0}^{4 π} \frac{\sqrt{17}}{4} d t = π \sqrt{17} \approx 12.95 .$

For the polygon, the eight samples are equally spaced in $t$ with step $Δ t equals 4 π / 7$ , and the chord between consecutive samples has length $\sqrt{2 - 2 \cos Δ t + (π / 7)^{2}} \approx 1.63$ , so the polygon measures about $7 \times 1.63 \approx 11.4$ — noticeably under the true $12.95$ , exactly the shortfall your eye predicted from the chords cutting inside the turns. At three and a half chords per turn the polygon is still crude; refining to a few dozen samples closes the gap to within a pixel's worth of length.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.