Arc length measures how far we travel along a curve. Curvature measures how quickly our direction changes as we travel. A tight bend has greater curvature than a gentle bend, regardless of how fast we move through it.

Turning per unit distance

Let $\mathbf r(t)$ be a regular curve: its velocity $\mathbf r'(t)$ is nonzero throughout the interval we are considering. Assume that $\mathbf r$ is twice continuously differentiable, so its first and second derivatives exist and are continuous.

The unit tangent vector is

$$\mathbf T(t)=\frac{\mathbf r'(t)}{\lVert\mathbf r'(t)\rVert}.$$

It records the direction of motion without the speed. Since $\lVert\mathbf T(t)\rVert=1$, any change in $\mathbf T$ is a change of direction.

How the tangent direction changes

The quantity $\lVert\mathbf T'(t)\rVert$ measures how fast that direction changes per unit of the parameter $t$. That still depends on how quickly we move along the curve. To measure the bending of the curve itself, we need change per unit of arc length instead.

Write $s$ for arc length and $\kappa$ (read “kappa”) for curvature. The definition is

$$\boxed{\kappa=\left\lVert\frac{d\mathbf T}{ds}\right\rVert.}$$

If the curve is already parametrized by arc length, we can differentiate its unit tangent directly with respect to $s$. Curvature is nonnegative and has units of inverse length.

Using an arbitrary parameter

Finding an explicit arc-length parametrization can be difficult. We can avoid that step by using the parameter we already have. Recall that the arc-length function satisfies

$$s(t)=\int_{t_0}^t\lVert\mathbf r'(u)\rVert\,du, \qquad \frac{ds}{dt}=\lVert\mathbf r'(t)\rVert,$$

where $t_0$ is a fixed starting parameter and $u$ is a dummy integration variable. The chain rule connects the two rates of change:

$$\frac{d\mathbf T}{dt} =\frac{d\mathbf T}{ds}\frac{ds}{dt} =\frac{d\mathbf T}{ds}\lVert\mathbf r'(t)\rVert.$$

Taking norms and dividing by the positive speed gives

$$\boxed{ \kappa(t)=\frac{\lVert\mathbf T'(t)\rVert}{\lVert\mathbf r'(t)\rVert}. }$$

Here $\kappa(t)$ means the curvature at the point reached at parameter $t$. Dividing by speed removes the effect of moving through the curve faster or slower. The denominator is the length of the velocity, not the length of the position vector.

Circles and lines

A circle of radius a

For a fixed $a>0$, consider

$$\mathbf r(t)=\langle a\cos t,a\sin t\rangle.$$

This traces a circle of radius $a$ centered at the origin.

A circle has the same curvature everywhere

First find the velocity and its length:

$$\mathbf r'(t)=\langle-a\sin t,a\cos t\rangle, \qquad \lVert\mathbf r'(t)\rVert =\sqrt{a^2\sin^2t+a^2\cos^2t}=a.$$

Dividing by $a$ gives the unit tangent. Differentiating that vector gives

$$\mathbf T(t)=\langle-\sin t,\cos t\rangle, \qquad \mathbf T'(t)=\langle-\cos t,-\sin t\rangle.$$

Its length is $\lVert\mathbf T'(t)\rVert=\sqrt{\cos^2t+\sin^2t}=1$. Therefore

$$\boxed{\kappa(t)=\frac1a.}$$

The curvature is the same everywhere on the circle. A smaller circle bends more sharply, so its curvature is larger.

A straight line

Let

$$\mathbf r(t)=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle, \qquad a^2+b^2+c^2>0.$$

The point $(x_0,y_0,z_0)$ lies on the line, and the nonzero vector $\langle a,b,c\rangle$ gives its direction. Then

$$\mathbf r'(t)=\langle a,b,c\rangle, \qquad \mathbf T(t)=\frac{\langle a,b,c\rangle}{\sqrt{a^2+b^2+c^2}}.$$

The unit tangent is constant, so $\mathbf T'(t)=\mathbf0$ and

$$\boxed{\kappa(t)=0.}$$

A line does not turn. Notice that zero curvature is allowed even though zero velocity is not allowed in the curvature formula.

Computing curvature with a cross product

Finding $\mathbf T$ and then differentiating it can be cumbersome, especially when speed is not constant. For a regular curve in $\mathbb R^3$, an equivalent formula uses the first two derivatives directly:

$$\boxed{ \kappa(t)= \frac{\lVert\mathbf r'(t)\times\mathbf r''(t)\rVert} {\lVert\mathbf r'(t)\rVert^3}. }$$

The symbol $\times$ denotes the cross product. Take the norm of that cross product in the numerator, and cube the speed in the denominator.

Example: a polynomial space curve

Take

$$\mathbf r(t)=\langle t,t^2,t^3\rangle.$$

Differentiate twice:

$$\mathbf r'(t)=\langle1,2t,3t^2\rangle, \qquad \mathbf r''(t)=\langle0,2,6t\rangle.$$

Using the coordinate unit vectors $\mathbf i$, $\mathbf j$, and $\mathbf k$, the determinant expansion gives

$$\begin{aligned} \mathbf r'(t)\times\mathbf r''(t) &=\begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 1&2t&3t^2\\ 0&2&6t \end{vmatrix}\\ &=(12t^2-6t^2)\mathbf i-6t\mathbf j+2\mathbf k\\ &=\langle6t^2,-6t,2\rangle. \end{aligned}$$

Thus

$$\lVert\mathbf r'(t)\times\mathbf r''(t)\rVert =\sqrt{36t^4+36t^2+4}, \qquad \lVert\mathbf r'(t)\rVert=\sqrt{1+4t^2+9t^4}.$$

Put these into the formula:

$$\boxed{ \kappa(t)=\frac{\sqrt{36t^4+36t^2+4}}{(1+4t^2+9t^4)^{3/2}}. }$$

This is valid for every real $t$: the first component of $\mathbf r'(t)$ is always $1$, so the velocity never vanishes.

Curvature of a graph

For a graph $y=f(x)$, we can use $x$ itself as the parameter. Assume that $f$ is twice continuously differentiable. View the graph in the $z=0$ plane:

$$\mathbf r(x)=\langle x,f(x),0\rangle.$$

Then

$$\mathbf r'(x)=\langle1,f'(x),0\rangle, \qquad \mathbf r''(x)=\langle0,f''(x),0\rangle,$$

and

$$\lVert\mathbf r'(x)\rVert=\sqrt{1+(f'(x))^2}.$$

The cross product has only a third component:

$$\mathbf r'(x)\times\mathbf r''(x) =\begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 1&f'(x)&0\\ 0&f''(x)&0 \end{vmatrix} =f''(x)\mathbf k.$$

Its norm is $|f''(x)|$, giving

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

Keep the absolute value: curvature measures the amount of bending and cannot be negative. The denominator is always positive. This formula applies where the curve is represented by a twice differentiable graph $y=f(x)$; a general plane curve need not have one such representation everywhere.

Example: a parabola

For $y=x^2$, the derivatives are $f'(x)=2x$ and $f''(x)=2$. Therefore

$$\boxed{\kappa(x)=\frac{2}{(1+4x^2)^{3/2}}.}$$

The denominator is smallest at $x=0$, so curvature is greatest at the vertex:

$$\kappa(0)=2, \qquad \lim_{x\to\pm\infty}\kappa(x)=0.$$

The parabola bends most sharply at its vertex

Farther from the vertex, the tangent line becomes steeper, but its direction changes less per unit distance traveled. A large slope does not necessarily mean large curvature.

The TNB frame

The unit tangent tells us the direction of travel. When that direction is changing, we can also identify the direction in which it turns.

Since $\mathbf T$ has constant length,

$$\mathbf T(t)\cdot\mathbf T(t)=1.$$

Differentiate using the dot-product rule:

$$2\mathbf T(t)\cdot\mathbf T'(t)=0.$$

So $\mathbf T'(t)$ is perpendicular to $\mathbf T(t)$. Wherever $\mathbf T'(t)\ne\mathbf0$, normalize it to get the principal unit normal, also called the unit normal:

$$\boxed{ \mathbf N(t)=\frac{\mathbf T'(t)}{\lVert\mathbf T'(t)\rVert}. }$$

The binormal is

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

These three vectors form the TNB frame:

• $\mathbf T$ points along the curve in the direction of motion.
• $\mathbf N$ points in the direction in which the unit tangent is changing.
• $\mathbf B$ is perpendicular to both, with its direction fixed by the cross-product order $\mathbf T\times\mathbf N$.

They are orthonormal: mutually perpendicular and each of length $1$. The length of $\mathbf B$ is $1$ because it is the cross product of two perpendicular unit vectors. The ordered frame is right-handed.

The nonzero condition matters. If $\mathbf T'(t)=\mathbf0$, then curvature is zero and the formulas above do not define $\mathbf N$ or $\mathbf B$ at that point. In particular, a straight line has a unit tangent and zero curvature, but its principal normal and binormal are not defined by these formulas.

Optional derivation of the cross-product formula

This derivation is not required; it explains why the computational formula agrees with turning per unit arc length.

Write $v(t)=\lVert\mathbf r'(t)\rVert$ for speed. To keep the expressions short, suppress the argument $t$ below. Since $\mathbf r'=v\mathbf T$, the product rule gives

$$\mathbf r''=v'\mathbf T+v\mathbf T'.$$

Take the cross product with $\mathbf r'$:

$$\begin{aligned} \mathbf r'\times\mathbf r'' &=(v\mathbf T)\times(v'\mathbf T+v\mathbf T')\\ &=vv'(\mathbf T\times\mathbf T)+v^2(\mathbf T\times\mathbf T')\\ &=v^2(\mathbf T\times\mathbf T'). \end{aligned}$$

The first term vanishes because a vector crossed with itself is zero. We also know that $\mathbf T$ is a unit vector perpendicular to $\mathbf T'$, so

$$\lVert\mathbf T\times\mathbf T'\rVert=\lVert\mathbf T'\rVert.$$

This remains true when $\mathbf T'=\mathbf0$: both sides are zero. Taking norms now gives

$$\lVert\mathbf r'\times\mathbf r''\rVert=v^2\lVert\mathbf T'\rVert.$$

Finally, use the definition $\kappa=\lVert\mathbf T'\rVert/v$ and the fact that $v>0$:

$$\boxed{ \kappa =\frac{\lVert\mathbf T'\rVert}{v} =\frac{\lVert\mathbf r'\times\mathbf r''\rVert}{v^3} =\frac{\lVert\mathbf r'\times\mathbf r''\rVert}{\lVert\mathbf r'\rVert^3}. }$$