A road can be gently curved or tightly bent. Curvature puts a number on that: how fast the direction of travel turns per unit of distance. It builds on the arc length you just met, because "per unit of distance" means per unit of $s$, not per unit of time.
Key idea
Unit tangent $\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$. Curvature
$$\kappa = \left|\frac{d\mathbf{T}}{ds}\right| = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.$$
For a graph $y = f(x)$: $\kappa = \dfrac{|f''(x)|}{\left(1 + f'(x)^2\right)^{3/2}}$. The unit normal $\mathbf{N} = \dfrac{\mathbf{T}'}{|\mathbf{T}'|}$ points toward the inside of the bend.
Why it works
$\mathbf{T}$ only carries direction (length 1), so its rate of change measures turning. But $\mathbf{T}'(t)$ is a rate per unit time: drive the same road twice as fast and it doubles. Dividing by the speed $\frac{ds}{dt} = |\mathbf{r}'(t)|$ converts it to a rate per unit distance, a property of the road alone.
Check on a circle of radius $R$: $\mathbf{r} = \langle R\cos t, R\sin t \rangle$ has speed $R$ and $\mathbf{T} = \langle -\sin t, \cos t \rangle$, so $|\mathbf{T}'| = 1$ and $\kappa = \frac{1}{R}$. Small circles bend sharply. Also, since $\mathbf{T} \cdot \mathbf{T} = 1$, differentiating gives ${}2\mathbf{T} \cdot \mathbf{T}' = 0$: $\mathbf{T}'$ is perpendicular to $\mathbf{T}$, which is why $\mathbf{N}$ is a normal.
Check on a circle of radius $R$: $\mathbf{r} = \langle R\cos t, R\sin t \rangle$ has speed $R$ and $\mathbf{T} = \langle -\sin t, \cos t \rangle$, so $|\mathbf{T}'| = 1$ and $\kappa = \frac{1}{R}$. Small circles bend sharply. Also, since $\mathbf{T} \cdot \mathbf{T} = 1$, differentiating gives ${}2\mathbf{T} \cdot \mathbf{T}' = 0$: $\mathbf{T}'$ is perpendicular to $\mathbf{T}$, which is why $\mathbf{N}$ is a normal.
Worked examples
Worked example: A unit tangent
Find $\mathbf{T}(1)$ for $\mathbf{r}(t) = \left\langle t, t^2, \frac{2}{3}t^3 \right\rangle$.
- $\mathbf{r}'(t) = \langle 1, 2t, 2t^2 \rangle$, so $\mathbf{r}'(1) = \langle 1, 2, 2 \rangle$.
- Its length is ${}3$.
- $\mathbf{T}(1) = \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle$, length 1.
Answer: $\left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle$
Worked example: Curvature of a helix
Find the curvature of $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangle$.
- $\mathbf{r}' = \langle -3\sin t, 3\cos t, 4 \rangle$, speed ${}5$; $\mathbf{T} = \frac{1}{5}\langle -3\sin t, 3\cos t, 4 \rangle$.
- $\mathbf{T}' = \frac{1}{5}\langle -3\cos t, -3\sin t, 0 \rangle$, length $\frac{3}{5}$.
- $\kappa = \frac{3/5}{5} = \frac{3}{25}$: less than $\frac{1}{3}$, the curvature of the circle it winds around, because the climb straightens it.
Answer: $\frac{3}{25}$
Worked example: Curvature of a graph
Find the curvature of $y = x^2$ at $x = 0$ and at $x = 1$.
- $y' = 2x$, $y'' = 2$.
- At $x = 0$: $\kappa = \frac{2}{(1 + 0)^{3/2}} = 2$.
- At $x = 1$: $\kappa = \frac{2}{(1 + 4)^{3/2}} = \frac{2}{5\sqrt{5}} = \frac{2\sqrt{5}}{25} \approx 0.18$: the parabola straightens out.
Answer: ${}2$ and $\frac{2\sqrt{5}}{25}$
Worked example: The cross product formula
Find $\kappa$ for $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ at $t = 0$.
- $\mathbf{r}'(0) = \langle 1, 0, 0 \rangle$ and $\mathbf{r}''(0) = \langle 0, 2, 0 \rangle$.
- $\mathbf{r}' \times \mathbf{r}'' = \langle 0, 0, 2 \rangle$, length ${}2$; $|\mathbf{r}'|^3 = 1$.
- $\kappa = 2$.
Answer: ${}2$
Common mistake
A "unit" tangent that is not unit length. $\mathbf{T}$ is $\mathbf{r}'$ divided by its length. Skipping the division gives a vector whose length is the speed, and every formula built on it (curvature, $\mathbf{N}$) comes out wrong. Check: $|\mathbf{T}| = 1$. Likewise, $\kappa = |\mathbf{T}'(t)|$ forgets the second division by $|\mathbf{r}'(t)|$; test your formula on a circle of radius $R$, which must give $\frac{1}{R}$.