Curvature
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.3: “Arc Length and Curvature” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature |
| Textbook used in class | Stewart, Calculus, Section 13.3: “Arc Length and Curvature” (Examples 3, 5) |
Quick Reference
Definition: $\kappa = \left|\dfrac{d\mathbf{T}}{ds}\right|$, where $\mathbf{T}$ is the unit tangent vector and $s$ is arc length.
Vector formula (most useful): $$\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}$$
Scalar formula for a plane curve $y = f(x)$: $$\kappa = \frac{|y''|}{\bigl(1 + (y')^2\bigr)^{3/2}}$$
Radius of curvature: $\rho = \dfrac{1}{\kappa}$
Motivation
Imagine driving along a winding road. On a straight stretch, the road does not bend at all -- the curvature is zero. On a sharp hairpin turn, the road bends severely -- the curvature is large. Curvature captures exactly this notion: how rapidly is the curve turning per unit distance traveled?
The key insight is to measure bending using the unit tangent vector $\mathbf{T}$. Because $\mathbf{T}$ has constant length 1, it can only change by rotating. The faster $\mathbf{T}$ rotates with respect to arc length, the more the curve bends.
Key Concepts
1. The Definition
The unit tangent vector is $\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$.
Curvature is the magnitude of the rate of change of $\mathbf{T}$ with respect to arc length $s$: $$\kappa = \left|\frac{d\mathbf{T}}{ds}\right|.$$
Because $s$ and $t$ are related by $ds/dt = |\mathbf{r}'(t)|$, the chain rule gives $$\frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}/dt}{ds/dt} = \frac{\mathbf{T}'(t)}{|\mathbf{r}'(t)|}.$$
So $\kappa = \dfrac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}$.
2. The Vector Formula
Differentiating $\mathbf{T}$ directly is messy. A cleaner route uses the cross product formula: $$\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.$$
This formula works for any smooth space curve and is typically the fastest route in computations.
3. The Osculating Circle
At each point on a curve, the osculating circle (from the Latin for “kissing”) is the circle that best approximates the curve locally. It lies in the plane spanned by $\mathbf{T}$ and $\mathbf{N}$ (the principal normal), and its radius is $\rho = 1/\kappa$. A tightly bending curve has large $\kappa$ and a small osculating circle.
Worked Example
Find the curvature of the helix $\mathbf{r}(t) = \langle \cos t,\, \sin t,\, t \rangle$. (Stewart 13.3, Example 3.)
Step 1. Compute the needed derivatives. $$\mathbf{r}'(t) = \langle -\sin t,\, \cos t,\, 1 \rangle, \qquad \mathbf{r}''(t) = \langle -\cos t,\, -\sin t,\, 0 \rangle.$$
Step 2. Compute the cross product. $$\mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin t & \cos t & 1 \\ -\cos t & -\sin t & 0 \end{vmatrix}$$ $$= \mathbf{i}(\cos t \cdot 0 - 1 \cdot (-\sin t)) - \mathbf{j}((-\sin t)(0) - 1(-\cos t)) + \mathbf{k}((-\sin t)(-\sin t) - \cos t(-\cos t))$$ $$= \langle \sin t,\, -\cos t,\, \sin^2 t + \cos^2 t \rangle = \langle \sin t,\, -\cos t,\, 1 \rangle.$$
Step 3. Compute the magnitudes. $$|\mathbf{r}' \times \mathbf{r}''| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}.$$ $$|\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}.$$
Step 4. Apply the formula. $$\kappa = \frac{\sqrt{2}}{(\sqrt{2})^3} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}.$$
The curvature is the constant $\kappa = 1/2$. The helix bends at a uniform rate, and its osculating circle has radius $\rho = 2$.
confusing $\kappa$ with $|{\bf r}''(t)|$. Acceleration $\mathbf{r}''(t)$ is NOT the same as curvature. Curvature measures the geometric bending of the path, independent of how fast a particle traverses it. If you speed up along a circular track, the curvature of the track does not change, but $|\mathbf{r}''(t)|$ does. The vector formula $\kappa = |\mathbf{r}' \times \mathbf{r}''|/|\mathbf{r}'|^3$ corrects for the speed of traversal.
Common Misconceptions
the curvature of a curve equals the magnitude of the acceleration vector $|\mathbf{r}''(t)|$.
This is the rate-as-fixed-number error. Curvature $\kappa = |\mathbf{r}' \times \mathbf{r}''|/|\mathbf{r}'|^3$ measures the geometric bending of the path per unit arc length, independently of how fast the parameter increases. The magnitude $|\mathbf{r}''(t)|$ depends on the parameterization speed; reparameterizing the same curve at twice the speed quadruples $|\mathbf{r}''|$ without changing the geometry. The formula for $\kappa$ divides by $|\mathbf{r}'|^3$ precisely to remove this speed dependence.
Leveled Practice
Problem 1. Find the curvature of the parabola $y = x^2$ at $x = 0$.
Show answer
Use the scalar formula with $y' = 2x$ and $y'' = 2$.
At $x = 0$: $\kappa = \dfrac{|2|}{(1 + 0)^{3/2}} = 2$.
The osculating circle at the vertex has radius $\rho = 1/2$.
Problem 2. Find the curvature of the circle $\mathbf{r}(t) = \langle a\cos t, a\sin t, 0\rangle$ ($a > 0$).
Show answer
$\mathbf{r}' = \langle -a\sin t, a\cos t, 0\rangle$, $\mathbf{r}'' = \langle -a\cos t, -a\sin t, 0\rangle$.
$\mathbf{r}' \times \mathbf{r}'' = \langle 0, 0, a^2\sin^2 t + a^2\cos^2 t\rangle = \langle 0,0,a^2\rangle$.
$\kappa = \dfrac{a^2}{(a)^3} = \dfrac{1}{a}$.
This confirms that a circle of radius $a$ has constant curvature $1/a$.