Curvature of Plane Curves

How sharply does a curve bend at each point? That single question opens the door to all of differential geometry. A perfectly straight line has zero bending; a tight circle bends hard everywhere; a free-form curve might bend gently here, sharply there. Curvature $\kappa$ is the scalar quantity that pins down "how sharply" at every point, and once you can compute it you have unlocked everything from highway transition design to Einstein's field equations.

The arc-length picture

Imagine driving along a curve at constant unit speed — the parameter is arc length $s$. The position vector $\mathbf{r}(s)$ has unit tangent $\mathbf{T}(s) = \mathbf{r}'(s)$, and the only way the tangent direction can change is by rotating (its length stays $1$). The rate of rotation is the curvature: $\kappa(s) = |\mathbf{T}'(s)| = |\mathbf{r}''(s)|$. Geometrically, $\kappa$ is the magnitude of the acceleration when you traverse the curve at unit speed — pure centripetal acceleration, no tangential component.

The general parametrization formula

Real curves rarely come arc-length-parametrized, so we need a formula that works for any smooth $\mathbf{r}(t) = (x(t), y(t))$:

The numerator is a 2×2 determinant measuring how much $\mathbf{r}'$ rotates per unit $t$; the denominator $|\mathbf{r}'(t)|^3$ is the speed-cubed correction that turns "per unit $t$" into "per unit arc length." When the curve is the graph of $y = f(x)$, set $x = t, y = f(t)$ and the formula collapses to:

The grapher above lets you sketch $y = f(x)$ and read off where the curve bends hard versus softly. Notice the $(1 + f'^2)^{3/2}$ in the denominator: at a steep stretch (large $|f'|$) the curve must bend further per unit $x$ just to register the same change in tangent direction, so the denominator damps $\kappa$. The numerator $|f''|$ alone would be misleading.

The osculating circle

At any point where $\kappa \ne 0$, there is a unique circle that matches the curve to second order: same point, same tangent, same curvature. Its radius is the radius of curvature $R = 1/\kappa$, and its centre lies along the inward normal at distance $R$. This osculating circle (Latin osculari, "to kiss") is the curve's best circular approximation locally. A straight line has $\kappa = 0$ and infinite radius of curvature — the osculating "circle" degenerates to the line itself.

Signed curvature and turning number

For a plane curve traversed in a chosen direction, you can keep the sign of $x'y'' - y'x''$ rather than its absolute value. The result is signed curvature $\kappa_s$: positive when the curve turns left, negative when it turns right. For any simple closed curve traversed counterclockwise, $\int_C \kappa_s, ds = 2\pi$ — the total turning is exactly one full revolution. This is the planar prototype of the Gauss-Bonnet theorem you will see in §7.4.

Pause and think: The parabola $y = x^2$ has curvature $\kappa(x) = \dfrac{2}{(1 + 4x^2)^{3/2}}$. Where is the curvature largest? Why does it shrink as $|x|$ grows, even though the parabola "obviously" keeps bending? (Hint: imagine driving along the parabola at unit speed — how much does your heading change per metre travelled when you are far from the vertex?)

Try it

• Predict first, then compute: what is the curvature of the unit circle $x^2 + y^2 = 1$ at every point? Verify with the parametrization $\mathbf{r}(t) = (\cos t, \sin t)$ and the determinant formula.
• For the curve $y = e^x$, find $\kappa(0)$ and the radius of curvature there. Then take $x \to +\infty$: what happens to $\kappa(x)$, and why does that match the visual intuition that the exponential "flattens out into a vertical line"?
• Use the formula to show that $y = \sin x$ has $\kappa(0) = 0$. At which $x$ values is $\kappa$ maximised? (Hint: differentiate the curvature expression, or notice the curve is steepest at zero crossings.)
• True or false: if a smooth curve has $\kappa(t) > 0$ everywhere, then the curve cannot cross itself. Find a counterexample using a small lemniscate or a figure-eight.

A trap to watch for

The formula $\kappa = |f''|/(1+f'^2)^{3/2}$ tempts beginners to read $|f''|$ as "the curvature" and treat the denominator as a nuisance correction. That is dangerously wrong. At a steep stretch (say $y = x^{10}$ near $x = 1$), $|f''|$ is enormous but the curve is still nearly straight in absolute terms — it just looks bent on a plot because the $y$-axis is compressed. The denominator $(1+f'^2)^{3/2}$ is the chain-rule correction that converts second derivatives in $x$ into rates of tangent rotation per unit arc length, which is the geometric quantity. Always trust the full formula, never the numerator alone.

What you now know

You can compute the curvature of any smooth plane curve, interpret it geometrically as the rate of rotation of the unit tangent (or as the reciprocal radius of the osculating circle), and locate the role of curvature in road engineering, relativity, and graphics. The next section extends the story to three dimensions, where bending alone is no longer enough — a space curve can also twist, captured by torsion.

References

• Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, ch. 7.
• do Carmo, M. P. (2016). Differential Geometry of Curves and Surfaces (2nd ed.). Dover, ch. 1.
• Pressley, A. (2010). Elementary Differential Geometry (2nd ed.). Springer, ch. 1-2.
• Spivak, M. (1999). A Comprehensive Introduction to Differential Geometry (3rd ed., Vol. 2). Publish or Perish, ch. 1.
• Misner, C. W., Thorne, K. S., Wheeler, J. A. (2017). Gravitation. Princeton University Press, ch. 8 (curvature as gravity).