1.4 Arc length and curvature

Arc Length and Curvature

Arc length and curvature give you the tools to describe curves in space geometrically rather than just algebraically. Arc length tells you how far you've traveled along a curve, while curvature tells you how sharply the curve bends at each point. Together with torsion (which measures twisting), these quantities are fundamental for analyzing motion along paths in physics, engineering, and computer graphics.

Arc Length and Parameterization

Calculating Arc Length

Arc length measures the actual distance traveled along a curve $C$ between two points $P(a)$ and $P(b)$. The idea is to sum up infinitesimal straight-line segments along the curve, which leads to the integral formula:

$L = \int_{a}^{b} \left\|\vec{r}\,'(t)\right\| \, dt = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt$

The curve must be smooth (continuously differentiable) on $[a, b]$ for this to work. Each term under the radical is the derivative of a coordinate function with respect to the parameter $t$.

Example: Find the arc length of $\vec{r}(t) = (t,\, t^2,\, t^3)$ for $0 \leq t \leq 1$.

1. Compute the derivatives: $\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 2t, \quad \frac{dz}{dt} = 3t^2$
2. Form the integrand: $\left\|\vec{r}\,'(t)\right\| = \sqrt{1 + 4t^2 + 9t^4}$
3. Set up the integral: $L = \int_{0}^{1} \sqrt{1 + 4t^2 + 9t^4} \, dt$

This integral doesn't have a nice closed form, so in practice you'd evaluate it numerically ($L \approx 1.8519$).

Arc Length Parameterization

Most parameterizations traverse a curve at varying speed. Arc length parameterization reparameterizes the curve using the arc length $s$ itself as the parameter, so the curve is traversed at a constant speed of exactly 1 unit per unit of $s$.

How to find it:

1. Compute the arc length function: $s(t) = \int_{t_0}^{t} \left\|\vec{r}\,'(u)\right\| \, du$
2. Solve for $t$ in terms of $s$ (i.e., invert the function to get $t = t(s)$).
3. Substitute back into $\vec{r}(t)$ to get $\vec{r}(s)$.

Example: Find the arc length parameterization of $\vec{r}(t) = (\cos t,\, \sin t)$ for $0 \leq t \leq 2\pi$.

1. Derivatives: $\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t$
2. Speed: $\left\|\vec{r}\,'(t)\right\| = \sqrt{\sin^2 t + \cos^2 t} = 1$
3. Arc length function: $s = \int_{0}^{t} 1 \, du = t$, so $t = s$.
4. The arc length parameterization is $\vec{r}(s) = (\cos s,\, \sin s)$ for $0 \leq s \leq 2\pi$.

This worked out so cleanly because the unit circle already has unit speed. For most curves, the inversion step is much harder.

Natural Parameterization

Natural parameterization (also called unit speed parameterization) is just another name for arc length parameterization. The defining property is that $\left\|\vec{r}\,'(s)\right\| = 1$ everywhere along the curve.

Why care? When a curve is naturally parameterized, many formulas simplify dramatically. For instance, the curvature reduces to $\kappa = \left\|\vec{r}\,''(s)\right\|$, and the unit tangent vector is simply $\vec{T}(s) = \vec{r}\,'(s)$.

Verification example: For $\vec{r}(s) = (\cos s,\, \sin s)$, compute $\vec{r}\,'(s) = (-\sin s,\, \cos s)$, and $\left\|\vec{r}\,'(s)\right\| = \sqrt{\sin^2 s + \cos^2 s} = 1$. Confirmed: this is a natural parameterization.

Curvature

Measuring Curvature

Curvature, denoted $\kappa$, quantifies how rapidly the direction of the curve is changing at a given point. A straight line has $\kappa = 0$ everywhere; a tight turn has large $\kappa$.

The formal definition is the magnitude of the rate of change of the unit tangent vector $\vec{T}$ with respect to arc length:

$\kappa = \left\|\frac{d\vec{T}}{ds}\right\|$

Since most curves aren't parameterized by arc length, you'll typically use this equivalent formula for an arbitrary parameter $t$:

$\kappa = \frac{\left\|\vec{r}\,'(t) \times \vec{r}\,''(t)\right\|}{\left\|\vec{r}\,'(t)\right\|^3}$

Example: Find the curvature of $\vec{r}(t) = (t,\, t^2,\, t^3)$ at $t = 1$.

1. First derivative: $\vec{r}\,'(t) = (1,\, 2t,\, 3t^2)$, so $\vec{r}\,'(1) = (1, 2, 3)$
2. Second derivative: $\vec{r}\,''(t) = (0,\, 2,\, 6t)$, so $\vec{r}\,''(1) = (0, 2, 6)$
3. Cross product: $\vec{r}\,'(1) \times \vec{r}\,''(1) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 3 \\ 0 & 2 & 6 \end{vmatrix} = (12-6,\, -(6-0),\, 2-0) = (6, -6, 2)$
4. Magnitudes: $\left\|(6, -6, 2)\right\| = \sqrt{36+36+4} = \sqrt{76} = 2\sqrt{19}$ and $\left\|(1,2,3)\right\| = \sqrt{14}$
5. Curvature: $\kappa = \frac{2\sqrt{19}}{(\sqrt{14})^3} = \frac{2\sqrt{19}}{14\sqrt{14}}$

Radius of Curvature

The radius of curvature $\rho$ is the reciprocal of curvature:

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

Geometrically, $\rho$ is the radius of the osculating circle (the best-fit circle) at that point. Large $\rho$ means the curve is nearly straight; small $\rho$ means a tight bend.

Example: For the curve $\vec{r}(t) = (t,\, t^2,\, t^3)$ at $t = 1$, using the corrected curvature above:

$\rho = \frac{1}{\kappa} = \frac{14\sqrt{14}}{2\sqrt{19}} = \frac{7\sqrt{14}}{\sqrt{19}}$

Osculating Circle

The osculating circle at a point on a curve is the circle that best approximates the curve there. It satisfies three conditions simultaneously:

• It passes through the point on the curve.
• It shares the same tangent direction as the curve at that point.
• It has the same curvature $\kappa$ as the curve at that point.

The center of the osculating circle lies along the principal unit normal vector $\vec{N}$, at a distance $\rho = 1/\kappa$ from the point on the curve. In other words:

$\text{center} = \vec{r}(t_0) + \rho\,\vec{N}(t_0)$

Example: Find the osculating circle for $\vec{r}(t) = (t,\, t^2)$ at $t = 1$.

1. Compute: $\vec{r}(1) = (1, 1)$, $\vec{r}\,'(1) = (1, 2)$, $\vec{r}\,''(1) = (0, 2)$

2. For a 2D curve, the cross product gives a scalar: $\left\|\vec{r}\,' \times \vec{r}\,''\right\| = |1 \cdot 2 - 2 \cdot 0| = 2$

3. Curvature: $\kappa = \frac{2}{(\sqrt{5})^3} = \frac{2}{5\sqrt{5}}$, so $\rho = \frac{5\sqrt{5}}{2}$

4. Unit tangent: $\vec{T} = \frac{(1,2)}{\sqrt{5}} = \left(\frac{1}{\sqrt{5}},\, \frac{2}{\sqrt{5}}\right)$

5. Unit normal (rotate $\vec{T}$ by 90° toward the concave side): $\vec{N} = \left(-\frac{2}{\sqrt{5}},\, \frac{1}{\sqrt{5}}\right)$

6. Center: $(1,1) + \frac{5\sqrt{5}}{2}\left(-\frac{2}{\sqrt{5}},\, \frac{1}{\sqrt{5}}\right) = (1 - 5,\; 1 + \tfrac{5}{2}) = (-4,\; \tfrac{7}{2})$

7. Equation: $(x + 4)^2 + \left(y - \tfrac{7}{2}\right)^2 = \left(\frac{5\sqrt{5}}{2}\right)^2 = \frac{125}{4}$

Torsion

Understanding Torsion

While curvature measures bending within a plane, torsion measures how much a space curve twists out of its osculating plane. It's denoted $\tau$.

• A curve with $\tau = 0$ at every point is a planar curve (it lies entirely in one plane).
• Positive $\tau$ means the curve twists in one direction; negative $\tau$ means it twists the other way. The sign convention depends on orientation: $\tau > 0$ corresponds to the binormal vector rotating toward $-\vec{N}$.

The formal definition involves the rate of change of the binormal vector $\vec{B}$ with respect to arc length:

$\tau = -\frac{d\vec{B}}{ds} \cdot \vec{N}$

For computation with an arbitrary parameter $t$, use:

$\tau = \frac{(\vec{r}\,'(t) \times \vec{r}\,''(t)) \cdot \vec{r}\,'''(t)}{\left\|\vec{r}\,'(t) \times \vec{r}\,''(t)\right\|^2}$

Example: Find the torsion of the helix $\vec{r}(t) = (\cos t,\, \sin t,\, t)$ at $t = \frac{\pi}{2}$.

1. Derivatives:

   • $\vec{r}\,'(t) = (-\sin t,\, \cos t,\, 1)$
   • $\vec{r}\,''(t) = (-\cos t,\, -\sin t,\, 0)$
   • $\vec{r}\,'''(t) = (\sin t,\, -\cos t,\, 0)$

2. Evaluate at $t = \frac{\pi}{2}$:

   • $\vec{r}\,' = (-1, 0, 1)$
   • $\vec{r}\,'' = (0, -1, 0)$
   • $\vec{r}\,''' = (1, 0, 0)$

3. Cross product $\vec{r}\,' \times \vec{r}\,'' = (-1, 0, 1) \times (0, -1, 0) = (0 \cdot 0 - 1 \cdot (-1),\; 1 \cdot 0 - (-1) \cdot 0,\; (-1)(-1) - 0 \cdot 0) = (1, 0, 1)$

4. Scalar triple product: $(1, 0, 1) \cdot (1, 0, 0) = 1$

5. Magnitude squared: $\left\|(1, 0, 1)\right\|^2 = 2$

6. Torsion: $\tau = \frac{1}{2}$

For a circular helix, the torsion is actually constant everywhere (not just at this point). The value $\tau = \frac{1}{2}$ reflects how tightly the helix winds upward relative to its radius.