2.3 Arc Length and Curvature

Arc Length and Parameterization

Arc length of space curves

Arc length lets you measure the actual distance traveled along a curve, rather than just the straight-line distance between endpoints. For a curve described parametrically as $(x(t), y(t), z(t))$, the arc length formula is:

$s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt$

The expression under the square root is just $|\mathbf{r}'(t)|$, the magnitude of the velocity vector. So you're integrating speed over time, which gives total distance.

To calculate arc length:

1. Write the curve in parametric form $(x(t), y(t), z(t))$
2. Compute the derivatives $\frac{dx}{dt}$, $\frac{dy}{dt}$, $\frac{dz}{dt}$
3. Plug them into the arc length formula (square each, sum, take the square root)
4. Evaluate the definite integral

Special cases:

• For planar curves (no z component): $s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
• For curves given as $y = f(x)$: $s = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2} \, dx$

Quick example: For the helix $\mathbf{r}(t) = (\cos t, \sin t, t)$ from $t = 0$ to $t = 2\pi$, you get $\mathbf{r}'(t) = (-\sin t, \cos t, 1)$, so $|\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$. The arc length is $\int_0^{2\pi} \sqrt{2} \, dt = 2\pi\sqrt{2}$.

Arc length parameter for reparameterization

Arc length parameterization means re-describing a curve so that the parameter itself represents distance traveled along the curve. When a curve is parameterized by arc length $s$, the speed is always 1: $|\mathbf{r}'(s)| = 1$. This is sometimes called unit-speed parameterization.

How to reparameterize by arc length:

1. Start with the original parameterization $\mathbf{r}(t)$
2. Define the arc length function: $s(t) = \int_{t_0}^{t} |\mathbf{r}'(u)| \, du$
3. Solve for $t$ as a function of $s$ (invert $s(t)$ to get $t(s)$)
4. Substitute $t(s)$ back into $\mathbf{r}(t)$ to get $\mathbf{r}(s)$

Step 3 is often the hardest part. Inverting $s(t)$ analytically is only possible for certain curves (like helices or lines). In practice, arc length parameterization is more of a theoretical tool than a computational one.

Why bother? Arc length parameterization simplifies many formulas. For instance, curvature becomes just $\kappa = |\mathbf{r}''(s)|$ when the curve is parameterized by arc length. It also describes motion at constant speed, which is useful for applications like animating objects along a path or designing roller coaster tracks.

Concept of curvature

Curvature $\kappa$ measures how sharply a curve is turning at a given point. A straight line has zero curvature everywhere. A tight turn has high curvature; a gentle bend has low curvature.

For a general parameterization, curvature is computed as:

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

If the curve is already parameterized by arc length, this simplifies to $\kappa = |\mathbf{r}''(s)|$.

Geometric interpretation: At any point on a curve, you can fit a circle that best matches the curve's behavior at that point. This is the osculating circle ("osculating" means "kissing"). The curvature is the reciprocal of the osculating circle's radius:

$\kappa = \frac{1}{R}$

So a circle of radius 5 has curvature $\frac{1}{5}$ everywhere. A smaller osculating circle means a sharper turn and higher curvature.

For a plane curve $y = f(x)$, there's a handy formula:

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

Unit vectors of curves

Three unit vectors form a natural coordinate system that moves along with a curve. Together they're called the Frenet-Serret frame (or TNB frame).

Unit tangent vector $\mathbf{T}(t)$ points in the direction of motion:

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

Unit normal vector $\mathbf{N}(t)$ points toward the center of the osculating circle, perpendicular to $\mathbf{T}$:

$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}$

Note that you differentiate $\mathbf{T}$, not $\mathbf{r}$. Since $\mathbf{T}$ has constant magnitude 1, its derivative is always perpendicular to it.

Binormal vector $\mathbf{B}(t)$ is perpendicular to both $\mathbf{T}$ and $\mathbf{N}$:

$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$

Together, $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ form a right-handed orthonormal basis at each point of the curve. Think of it as a coordinate system that rides along the curve, rotating as the curve bends and twists.

The Frenet-Serret formulas describe how these vectors change:

• $\mathbf{T}' = \kappa |\mathbf{r}'| \, \mathbf{N}$
• $\mathbf{N}' = |\mathbf{r}'|(-\kappa \mathbf{T} + \tau \mathbf{B})$
• $\mathbf{B}' = -\tau |\mathbf{r}'| \, \mathbf{N}$

(When parameterized by arc length, the $|\mathbf{r}'|$ factors become 1, giving the cleaner forms $\mathbf{T}' = \kappa \mathbf{N}$, etc.)

Here $\tau$ is the torsion, which measures how much the curve twists out of its osculating plane. A plane curve has $\tau = 0$ everywhere. A helix has constant nonzero torsion. While curvature tells you how much the curve bends, torsion tells you how much it twists.