Math Core

How long is a coiled spring if you stretch it out straight? How sharply does a road bend at a given point? Both questions are about the shape of a curve rather than how it happens to be parametrized, and the tangent vector from the last lesson is the tool that answers them.

Arc length

Chop the parameter interval $[a, b]$ into tiny pieces of width $\Delta t$ . Over one piece, the point moves by approximately $\mathbf{r}'(t)\,\Delta t$ , a displacement of length $|\mathbf{r}'(t)|\,\Delta t$ . Adding up these lengths and letting $\Delta t \to 0$ gives an integral.

Arc length of a space curve

If $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ is smooth on $[a, b]$ and traces its curve once, the length of the curve is

L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2 + h'(t)^2}\,dt.

This is the same formula you used for plane parametric curves, with a third term under the root. The integrand $|\mathbf{r}'(t)|$ is the rate at which length accumulates, which you'll soon call speed.

Worked example: Unrolling a helix

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

$\mathbf{r}'(t) = \langle -3\sin t,\; 3\cos t,\; 4 \rangle$ , so

|\mathbf{r}'(t)| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5.

Then $L = \displaystyle\int_0^{2\pi} 5\,dt = 10\pi$ . That makes sense: one turn goes around a circle of circumference $6\pi$ while rising $8\pi$ , and unrolling the cylinder turns the helix into the hypotenuse of a right triangle with legs $6\pi$ and $8\pi$ .

Most arc length integrals can't be done by hand. The ones that can usually have a perfect square hiding under the root, so always expand and look for one.

The arc length function and reparametrization

Measuring length from a fixed starting time $a$ gives the arc length function

s(t) = \int_a^t |\mathbf{r}'(u)|\,du, \qquad \frac{ds}{dt} = |\mathbf{r}'(t)|.

If you solve for $t$ in terms of $s$ and substitute, you get a parametrization by arc length, in which the parameter value equals the distance traveled along the curve. For the helix above, $s = 5t$ , so $t = s/5$ and

\mathbf{r}(s) = \left\langle 3\cos\tfrac{s}{5},\; 3\sin\tfrac{s}{5},\; \tfrac{4s}{5} \right\rangle.

In an arc length parametrization the tangent vector always has length 1. That's what makes it the natural parametrization for describing shape: it strips away how fast you happen to travel.

Curvature

A straight line never changes direction. A small circle changes direction quickly. Curvature measures how fast the unit tangent vector turns per unit of distance traveled.

Definition

Curvature

The curvature of a smooth curve is

\kappa = \left| \frac{d\mathbf{T}}{ds} \right|,

the magnitude of the rate of change of the unit tangent vector with respect to arc length.

Since $\dfrac{d\mathbf{T}}{dt} = \dfrac{d\mathbf{T}}{ds}\dfrac{ds}{dt}$ , you can compute curvature in any parametrization with $\kappa(t) = \dfrac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}$ . Computing $\mathbf{T}'$ is often messy, so the following formula is usually faster:

\kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.

For a circle of radius $a$ , $\kappa = \dfrac{1}{a}$ everywhere: bigger circles bend more gently. This motivates the radius of curvature $\rho = \dfrac{1}{\kappa}$ . At each point, the osculating circle is the circle of radius $\rho$ that sits on the concave side of the curve, shares its tangent line and bends by the same amount. It's the circle that fits the curve best at that point.

Worked example: Curvature of the twisted cubic

Find the curvature of $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ at $t = 0$ .

$\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle$ and $\mathbf{r}''(t) = \langle 0, 2, 6t \rangle$ . At $t = 0$ these are $\langle 1, 0, 0 \rangle$ and $\langle 0, 2, 0 \rangle$ , with cross product $\langle 0, 0, 2 \rangle$ . So

\kappa(0) = \frac{|\langle 0, 0, 2 \rangle|}{|\langle 1, 0, 0 \rangle|^3} = \frac{2}{1} = 2.

Worked example: Curvature of a helix

For $\mathbf{r}(t) = \langle a\cos t,\; a\sin t,\; bt \rangle$ with $a > 0$ , you get $\mathbf{r}' = \langle -a\sin t, a\cos t, b \rangle$ and $\mathbf{r}'' = \langle -a\cos t, -a\sin t, 0 \rangle$ . The cross product is $\langle ab\sin t,\; -ab\cos t,\; a^2 \rangle$ , with length $a\sqrt{a^2 + b^2}$ , and $|\mathbf{r}'| = \sqrt{a^2 + b^2}$ . Therefore

\kappa = \frac{a\sqrt{a^2 + b^2}}{(a^2 + b^2)^{3/2}} = \frac{a}{a^2 + b^2}.

The curvature is constant. For $\langle 3\cos t, 3\sin t, 4t \rangle$ it's $\dfrac{3}{25}$ , smaller than the $\dfrac{1}{3}$ of the circle it wraps around, because stretching the spring upward makes it bend less.

Curvature of a plane graph

A graph $y = f(x)$ can be written as $\mathbf{r}(x) = \langle x, f(x), 0 \rangle$ . Plugging into the cross product formula gives

\kappa(x) = \frac{|f''(x)|}{\left(1 + f'(x)^2\right)^{3/2}}.

For $y = x^2$ at the origin, $f'(0) = 0$ and $f''(0) = 2$ , so $\kappa = 2$ and the osculating circle has radius $\tfrac{1}{2}$ and center $\left(0, \tfrac{1}{2}\right)$ .

The parabola y = x² and its osculating circle at the origin, radius 1/2.Open in grapher →

Common mistake

Curvature is not the same as $|f''(x)|$ . The second derivative ignores how steep the graph is. Where the graph is steep, the denominator $(1 + f'^2)^{3/2}$ is large and the true curvature is much smaller than $|f''|$ .

Tip

The vector $\mathbf{T}'(t)$ points toward the concave side of the curve. Normalizing it gives the principal unit normal $\mathbf{N} = \dfrac{\mathbf{T}'}{|\mathbf{T}'|}$ , and $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ is the binormal. These three perpendicular unit vectors form a moving frame that rides along the curve.

Practice

Practice 1

Find the length of the line segment $\mathbf{r}(t) = \langle 2t,\; 3t,\; 6t \rangle$ for $0 \le t \le 2$ .