Arc Length in Space

In Calculus 2 you found the length of a curve in the plane. In space the idea is the same, and the velocity vector makes it look natural: distance traveled is speed times time, and the speed of a moving point is the length of its velocity, $|\mathbf{r}'(t)|$.

Key idea

The length of the curve $\mathbf{r}(t)$, $a \le t \le b$, traced once, is $$L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2 + h'(t)^2}\,dt.$$ The arc length function $s(t) = \int_a^t |\mathbf{r}'(u)|\,du$ measures distance along the curve, and $\frac{ds}{dt} = |\mathbf{r}'(t)|$.

Why it works

Chop the time interval into small pieces of length $\Delta t$. Over one piece the point moves about $\mathbf{r}'(t)\,\Delta t$ (velocity times time), a straight step of length $|\mathbf{r}'(t)|\,\Delta t$. Adding the steps gives a Riemann sum $\sum |\mathbf{r}'(t_i)|\,\Delta t$, which becomes the integral as $\Delta t \to 0$. Speed, not velocity: a length is never negative and does not care about direction.

A helix with constant speed. The readout shows the distance traveled since $t = 0$: it grows at the same rate at every moment, so arc length is simply speed × time here.

Two facts are worth knowing. First, the length does not depend on how the curve is parametrized, as long as it is traced once: going twice as fast doubles $|\mathbf{r}'(t)|$ but halves the time, and the substitution rule shows that the integral is unchanged. Second, the length is at least the straight-line distance between the endpoints, since $$|\mathbf{r}(b) - \mathbf{r}(a)| = \left|\int_a^b \mathbf{r}'(t)\,dt\right| \le \int_a^b |\mathbf{r}'(t)|\,dt.$$ Equality holds only for a straight segment traced without turning back.

Worked examples

Worked example: A helix

Find the length of $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangle$ for ${}0 \le t \le 2\pi$.

$\mathbf{r}'(t) = \langle -3\sin t, 3\cos t, 4 \rangle$.
$$|\mathbf{r}'(t)| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5.$$
$L = \int_0^{2\pi} 5\,dt = 10\pi$.

Answer: ${}10\pi$

Worked example: A perfect square under the root

Find the length of $\mathbf{r}(t) = \left\langle t, t^2, \frac{2}{3}t^3 \right\rangle$ for ${}0 \le t \le 1$.

$\mathbf{r}'(t) = \langle 1, 2t, 2t^2 \rangle$.
$|\mathbf{r}'(t)|^2 = 1 + 4t^2 + 4t^4 = (1 + 2t^2)^2$, so $|\mathbf{r}'(t)| = 1 + 2t^2$.
$L = \int_0^1 (1 + 2t^2)\,dt = 1 + \frac{2}{3} = \frac{5}{3}$.

Answer: $\frac{5}{3}$

Worked example: Measuring by arc length

For the helix of the first example, where is the point after traveling a distance $\frac{5\pi}{2}$ from $t = 0$?

Speed ${}5$, so $s = 5t$ and $t = \frac{s}{5}$.
$s = \frac{5\pi}{2}$ gives $t = \frac{\pi}{2}$.
$\mathbf{r}\left(\frac{\pi}{2}\right) = (0, 3, 2\pi)$.

Answer: $(0, 3, 2\pi)$

Rewriting a curve with $s$ in place of $t$, here $\mathbf{r}(s) = \left\langle 3\cos\frac{s}{5}, 3\sin\frac{s}{5}, \frac{4s}{5} \right\rangle$, gives a parametrization with speed exactly ${}1$. Arc length is the parameter the curve itself prefers, which is why curvature is defined with it.

Common mistake

Integrating the velocity instead of the speed. $\int_a^b \mathbf{r}'(t)\,dt = \mathbf{r}(b) - \mathbf{r}(a)$ is a displacement vector, not a length. You must take the length $|\mathbf{r}'(t)|$ first, then integrate. Check: the arc length is never shorter than the straight-line distance $|\mathbf{r}(b) - \mathbf{r}(a)|$.

Arc Length in Space

Worked examples

Practice

Builds on

Leads to

Sources and further reading