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Arc Length and Curvature · Tutorial 192 of 999

Curvature as Turning per Unit Distance

Learn how curvature measures the change in a curve’s direction per unit of distance traveled, and how that interpretation applies to circles and nearly straight curves.

Intermediate 10 min read

What You'll Learn

  • Define curvature using the derivative of the unit tangent with respect to arc length.
  • Explain why curvature measures the infinitesimal turning angle per unit distance.
  • Compute curvature directly for unit-speed lines and circles.
  • Relate the curvature of a circle to its radius.
  • Recognize how small curvature corresponds to a nearly straight curve.
  • Distinguish a derivative with respect to arc length from one with respect to an arbitrary parameter.

Turning Measured Along a Curve

A curve can change direction quickly or slowly as we travel along it. Curvature gives a precise measure of that change: it tells us how much the unit tangent turns for each unit of distance traveled. A straight line has no turning and therefore has zero curvature. A circle turns continuously, with tighter circles turning more sharply than larger ones.

The key is to measure turning against arc length, not simply against whichever parameter happens to describe the curve. Arc length records distance traveled along the curve. Measuring the change in direction per unit of that distance gives curvature a direct geometric meaning.

Definition of Curvature

Let a smooth regular curve be parametrized by arc length \(s\), and write its position as \(\mathbf{r}(s)\). Because \(s\) measures distance, the curve has unit speed: \(\left\lVert \frac{d\mathbf{r}}{ds}\right\rVert=1\). Its unit tangent vector is therefore \(\mathbf{T}(s)=\frac{d\mathbf{r}}{ds}\).

Definition: The curvature of a smooth regular curve at a point is the magnitude of the rate of change of its unit tangent with respect to arc length: \(\kappa(s)=\left\lVert\frac{d\mathbf{T}}{ds}\right\rVert\).

Curvature is nonnegative because it is a magnitude. If \(\mathbf{T}\) does not change at a point, then \(\kappa=0\) there. If \(\mathbf{T}\) changes rapidly over a short distance, then curvature is large. When distances are measured in units of length, curvature has units of inverse length: for example, \(\mathrm{m}^{-1}\) if distances are measured in metres.

This definition requires a unit tangent that is differentiable with respect to arc length at the point in question. Earlier material on unit tangents and arc-length parametrization explains how to describe a regular curve using arc length. Here, the essential point is that the derivative in the definition is specifically with respect to distance along the curve.

Why Curvature Is Turning per Unit Distance

To make the interpretation precise, compare the unit tangent at \(s\) with the unit tangent a short distance \(h\) later. Let \(\delta(h)\) be the angle between these two unit vectors. The endpoints of the vectors lie on the unit circle in the plane they span, so the chord between them has length \(2\sin(\delta(h)/2)\). Thus

$$ \left\lVert\mathbf{T}(s+h)-\mathbf{T}(s)\right\rVert = 2\sin\left(\frac{\delta(h)}{2}\right). $$

Since \(\mathbf{T}\) is differentiable, the magnitude of its change per unit arc length tends to \(\left\lVert\frac{d\mathbf{T}}{ds}\right\rVert\) as \(h\) tends to zero. Also, \(\delta(h)\) tends to zero, and \(\frac{2\sin(\delta/2)}{\delta}\) tends to \(1\) as \(\delta\) tends to zero. Combining these facts gives

$$ \lim_{h\to 0}\frac{\delta(h)}{|h|} = \left\lVert\frac{d\mathbf{T}}{ds}\right\rVert = \kappa(s). $$

The limit says that curvature is the infinitesimal angle through which the tangent turns, divided by the distance traveled. The angle is measured in radians, as is standard when relating angles to arc length. This argument also covers the case \(\frac{d\mathbf{T}}{ds}=\mathbf{0}\): the limiting turning angle per unit distance is then zero.

Geometric interpretation: A curvature of \(0.2\ \mathrm{m}^{-1}\) means that, locally, the tangent’s turning angle changes by about \(0.2\) radians for each metre traveled. The definition gives the exact limiting rate at a point.

Worked Example: Curvature of a Unit-Speed Line

Worked Example: Curvature of a Unit-Speed Line

Consider the line \(\mathbf{r}(s)=\left(\frac{3s}{5},\frac{4s}{5}\right)\). Determine its curvature.

Step 1: Find the unit tangent. Differentiate the position with respect to \(s\):

$$ \mathbf{T}(s)=\frac{d\mathbf{r}}{ds} =\left(\frac35,\frac45\right). $$

Its magnitude is \(\sqrt{(3/5)^2+(4/5)^2}=\sqrt{9/25+16/25}=1\), so \(s\) is indeed arc length and this is a unit tangent.

Step 2: Differentiate the tangent.

$$ \frac{d\mathbf{T}}{ds}=(0,0), \qquad \kappa(s)=\left\lVert(0,0)\right\rVert=0. $$

Check: The tangent vector is constant, so its direction never changes as the point moves along the line. That agrees with the zero curvature obtained from the definition.

Curvature of a Circle

A circle is a useful model because its direction changes uniformly as one travels around it. For a circle of radius \(R>0\), use arc length \(s\) to write

$$ \mathbf{r}(s)=\left(R\cos\left(\frac{s}{R}\right),R\sin\left(\frac{s}{R}\right)\right). $$

Differentiating gives the tangent. Its magnitude is \(1\), confirming that \(s\) is arc length. Differentiating once more gives the rate at which the tangent changes:

$$ \mathbf{T}(s)=\left(-\sin\left(\frac{s}{R}\right),\cos\left(\frac{s}{R}\right)\right), \qquad \frac{d\mathbf{T}}{ds} = \left(-\frac1R\cos\left(\frac{s}{R}\right),-\frac1R\sin\left(\frac{s}{R}\right)\right). $$

The magnitude of this tangent derivative is \(1/R\), because the squared sine and cosine terms sum to \(1\). Thus a circle’s curvature is constant and inversely proportional to its radius.

Theorem: A circle of radius \(R\) has curvature \(\kappa=\frac1R\). A smaller radius means greater curvature; a larger radius means smaller curvature.

Worked Example: Comparing Two Circles

Worked Example: Comparing Two Circles

Find and compare the curvatures of circles with radii \(3\ \mathrm{m}\) and \(\frac12\ \mathrm{m}\).

Step 1: Apply the circle result. For the circle with radius \(3\ \mathrm{m}\),

$$ \kappa_1=\frac{1}{3\ \mathrm{m}}=\frac13\ \mathrm{m}^{-1}. $$

For the circle with radius \(\frac12\ \mathrm{m}\),

$$ \kappa_2=\frac{1}{\frac12\ \mathrm{m}}=2\ \mathrm{m}^{-1}. $$

Step 2: Compare. The ratio is \(\frac{\kappa_2}{\kappa_1}=\frac{2}{1/3}=6\). Therefore the smaller circle has six times the curvature of the larger one.

Check: The smaller circle has a tighter bend, so it should turn more per unit distance. The computed curvatures, \(2\ \mathrm{m}^{-1}\) and \(\frac13\ \mathrm{m}^{-1}\), agree with that geometric comparison.

Nearly Straight Curves

A curve need not be exactly straight to have small curvature. Small curvature means its tangent changes slowly relative to the distance traveled. One way to see this is to describe a unit tangent whose direction angle changes only gradually.

For example, define a curve by specifying its tangent for \(0\leq s\leq \frac{1}{10}\):

$$ \mathbf{T}(s)=(\cos(s^2),\sin(s^2)), \qquad \mathbf{r}(s)=\int_0^s\mathbf{T}(u)\,du. $$

The integral defines the position, and the fundamental theorem of calculus gives \(\frac{d\mathbf{r}}{ds}=\mathbf{T}(s)\). Since \(\lVert\mathbf{T}(s)\rVert=\sqrt{\cos^2(s^2)+\sin^2(s^2)}=1\), this is a unit-speed curve. Its tangent derivative and curvature are

$$ \frac{d\mathbf{T}}{ds} = (-2s\sin(s^2),2s\cos(s^2)), \qquad \kappa(s)=\sqrt{4s^2\sin^2(s^2)+4s^2\cos^2(s^2)}=2s. $$

On the given interval, curvature rises from \(0\) to \(\frac15\) in inverse units of the distance used for \(s\). The tangent angle is \(s^2\), so from \(s=0\) to \(s=\frac{1}{10}\) it changes by \(\frac{1}{100}\) radians. This small change over a short distance illustrates why the curve is nearly straight on that interval.

Worked Example: Curvature at a Point on a Nearly Straight Curve

For the curve just defined, find its curvature at \(s=\frac{1}{20}\), and state the tangent’s total angle change from \(s=0\) to \(s=\frac{1}{20}\).

Step 1: Evaluate curvature. Since \(\kappa(s)=2s\),

$$ \kappa\left(\frac{1}{20}\right) = 2\left(\frac{1}{20}\right) = \frac{1}{10}. $$

Step 2: Find the tangent angle change. The tangent direction angle at \(s\) is \(s^2\). It is \(0\) at the starting point and \(\left(\frac{1}{20}\right)^2=\frac{1}{400}\) at the specified point. The angle change is therefore \(\frac{1}{400}\) radians.

Check: The curvature is the rate of angle change with respect to \(s\): differentiating the angle \(s^2\) gives \(2s\), which at \(s=\frac{1}{20}\) is \(\frac{1}{10}\). The angle change is small over this short distance, consistent with the curve being nearly straight there.

A Common Mistake: Differentiating with Respect to the Wrong Variable

The definition uses \(\frac{d\mathbf{T}}{ds}\), where \(s\) is arc length. A derivative with respect to a different parameter is not automatically the same quantity. For instance, consider the circle \(\mathbf{r}(t)=(3\cos t,3\sin t)\). A mistaken calculation might be

$$ \mathbf{r}''(t)=(-3\cos t,-3\sin t), \qquad \left\lVert\mathbf{r}''(t)\right\rVert=3, \qquad \text{so “}\kappa=3\text{.”} $$

This is wrong: it uses the second derivative of position with respect to \(t\), not the derivative of the unit tangent with respect to arc length. In fact, \(\mathbf{r}'(t)=(-3\sin t,3\cos t)\) has magnitude \(3\), so \(t\) is not arc length. The circle has radius \(3\), and its curvature is \(\frac13\), not \(3\).

To correct the calculation, write the same circle in terms of arc length \(s=3t\): \(\mathbf{r}(s)=(3\cos(s/3),3\sin(s/3))\). Its unit tangent is \(\mathbf{T}(s)=(-\sin(s/3),\cos(s/3))\), so

$$ \frac{d\mathbf{T}}{ds} = \left(-\frac13\cos\left(\frac{s}{3}\right),-\frac13\sin\left(\frac{s}{3}\right)\right), \qquad \kappa=\left\lVert\frac{d\mathbf{T}}{ds}\right\rVert=\frac13. $$

Correction to remember: Curvature is defined using the unit tangent’s rate of change per unit arc length. Do not replace that derivative with a derivative in an arbitrary parameter without accounting for the difference.

Why Curvature Matters

Curvature distinguishes curves that may travel through similar locations but bend in different ways. It quantifies local bending without depending on a verbal description such as “sharp” or “gentle.” The circle result gives a practical reference: radius and curvature vary inversely. For more general curves, the definition still applies even when curvature changes from point to point.

When interpreting a curvature value, keep track of its units and scale. A large curvature means substantial turning over a short distance; a small curvature means the direction changes slowly. Zero curvature at a point means the tangent has zero instantaneous rate of change there. It does not, by itself, assert that the entire curve is a straight line.

Check Your Understanding

Use the definition and examples above to answer these questions.

  1. State the definition of curvature for a curve parametrized by arc length.
  2. Why does the magnitude of the unit tangent’s derivative measure turning per unit distance?
  3. What is the curvature of a circle of radius \(5\ \mathrm{m}\), including units?
  4. What does small curvature indicate about the change in tangent direction as distance is traveled?
  5. Why is the magnitude of the second derivative of \((3\cos t,3\sin t)\) not the curvature of that circle?