3.3 Arc Length and Curvature

Big idea

A curve has properties that belong to the curve itself and properties that belong only to the way you happen to travel along it. Speed is in the second category: run along the same path twice as fast and every velocity vector doubles, but the path is unchanged. Length is in the first category, and so is bending. This section measures both, and the technical device that separates the two categories is reparametrizing by arc length.

Length comes first and comes cheaply. The speed $\|\mathbf{r}'(t)\|$ says how much distance is being covered per unit of parameter, so integrating the speed accumulates distance. That single integral covers plane curves and space curves at once, and it reduces to the familiar one-variable arc length formula when the curve is the graph of a function.

Curvature is the interesting quantity. The unit tangent vector has constant length, so it can only turn, never stretch, and the rate at which it turns is exactly how sharply the curve is bending. Measuring that rate against the parameter would let a fast parametrization inflate it, so the rate is measured against distance travelled instead. That makes curvature a property of the curve alone: a circle of radius $a$ has curvature $1/a$ no matter how you drive around it, and a straight line has curvature zero.

Bending also has a direction, and that gives the third idea here. At each point of a smooth curve there are three mutually perpendicular unit vectors: the tangent, the normal pointing the way the curve is turning, and the binormal perpendicular to both. This moving frame travels with the point and provides a coordinate system fitted to the curve rather than to the room, which is what the next section uses to split acceleration into a part that changes speed and a part that changes direction.

Decoder

Curvature is the magnitude of the rate of change of the unit tangent vector with respect to arc length, so it measures how quickly the direction of the curve turns per unit of distance travelled.

Two words carry the content. “Unit” means that the only thing the tangent vector can do as you move along is rotate, so its rate of change is pure turning with no contribution from speeding up. “Arc length” means the turning is measured per meter of curve rather than per unit of the parameter, which is what makes the answer independent of the parametrization.

In practice you almost never reparametrize by arc length to compute curvature, because solving for $t$ in terms of $s$ is usually impossible in closed form. The chain rule supplies the workaround. Since $ds/dt = \|\mathbf{r}'(t)\|$, dividing $\|\mathbf{T}'(t)\|$ by the speed converts a rate per unit parameter into a rate per unit distance, and a further computation turns that into the cross product formula, which needs only $\mathbf{r}'$ and $\mathbf{r}''$. Use whichever of the three formulas gives the least algebra for the curve in front of you.

The classic mistake is confusing curvature with the second derivative. For the graph of a function, $|f''|$ appears in the numerator of the curvature formula but is divided by $\left(1 + (f')^2\right)^{3/2}$, and ignoring that denominator gives the wrong answer everywhere except where the tangent is horizontal. A line with a steep slope has enormous $|f'|$ and zero curvature; the denominator is what reconciles those facts.

Definitions and results

Arc length. For a curve traced once by a smooth $\mathbf{r}$ on $[a,b]$,

$$ L = \int_a^b \|\mathbf{r}'(t)\| \, dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt $$

Dropping the third term gives the plane-curve version.

The arc-length function. Measuring from the starting value $a$,

$$ s(t) = \int_a^t \|\mathbf{r}'(u)\| \, du, \qquad \frac{ds}{dt} = \|\mathbf{r}'(t)\| $$

Since the speed is positive on a smooth curve, $s$ is strictly increasing and can in principle be inverted.

Arc-length parametrization. Substituting $t$ as a function of $s$ gives a parametrization in which the parameter is distance travelled. Its defining property is that the speed is identically $1$, so $\|d\mathbf{r}/ds\| = 1$, and it is the natural parametrization for stating results about the curve itself.

Unit tangent vector. As before, $\mathbf{T}(t) = \mathbf{r}'(t) / \|\mathbf{r}'(t)\|$, defined wherever the curve is smooth.

Curvature. The curvature is

$$ \kappa = \left\| \frac{d\mathbf{T}}{ds} \right\| = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{r}'(t)\|} = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} $$

All three expressions give the same nonnegative number. A line has $\kappa = 0$ and a circle of radius $a$ has $\kappa = 1/a$ at every point.

Curvature of a graph. For $y = f(x)$ with $f$ twice differentiable,

$$ \kappa(x) = \frac{|f''(x)|}{\left(1 + \left(f'(x)\right)^2\right)^{3/2}} $$

Radius of curvature and the osculating circle. Where $\kappa \ne 0$, the radius of curvature is $\rho = 1/\kappa$. The osculating circle at a point has that radius, lies in the plane of $\mathbf{T}$ and $\mathbf{N}$, and has its center a distance $\rho$ from the point in the direction of $\mathbf{N}$. It is the circle that best matches the curve there.

Principal unit normal. Where $\mathbf{T}'(t) \ne \mathbf{0}$,

$$ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} $$

Because $\mathbf{T}$ has constant length, $\mathbf{T}'$ is orthogonal to $\mathbf{T}$, so $\mathbf{N}$ is a unit vector perpendicular to the direction of travel, pointing toward the concave side of the curve.

Binormal and the TNB frame. The binormal is $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$. The three vectors $\mathbf{T}$, $\mathbf{N}$, $\mathbf{B}$ are mutually orthogonal unit vectors forming a right-handed frame that moves with the point. The plane through the point spanned by $\mathbf{T}$ and $\mathbf{N}$ is the osculating plane, and $\mathbf{B}$ is normal to it.

Worked examples

Arc length of a helix, and its arc-length parametrization

Let $\mathbf{r}(t) = \langle 3\cos t, \; 3\sin t, \; 4t \rangle$. Then $\mathbf{r}'(t) = \langle -3\sin t, \; 3\cos t, \; 4 \rangle$ and

$$ \|\mathbf{r}'(t)\| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5 $$

The speed is constant, so one full turn from $t = 0$ to $t = 2\pi$ has length

$$ L = \int_0^{2\pi} 5\,dt = 10\pi $$

Because the speed is constant, the arc-length function is simply $s = 5t$, and solving gives $t = s/5$. Substituting,

$$ \mathbf{r}(s) = \left\langle 3\cos\frac{s}{5}, \; 3\sin\frac{s}{5}, \; \frac{4s}{5} \right\rangle $$

Check that this really is parametrized by arc length: its derivative is $\left\langle -\tfrac{3}{5}\sin\tfrac{s}{5}, \; \tfrac{3}{5}\cos\tfrac{s}{5}, \; \tfrac{4}{5} \right\rangle$, whose magnitude is $\sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = 1$.

Curvature of the helix, two ways

Continue with the same helix. Using the cross product formula, $\mathbf{r}''(t) = \langle -3\cos t, \; -3\sin t, \; 0 \rangle$, so

$$ \mathbf{r}' \times \mathbf{r}'' = \langle 12\sin t, \; -12\cos t, \; 9 \rangle $$

whose magnitude is $\sqrt{144 + 81} = 15$. Dividing by $\|\mathbf{r}'\|^3 = 125$,

$$ \kappa = \frac{15}{125} = \frac{3}{25} $$

Now confirm it with the tangent-vector formula. Since $\mathbf{T}(t) = \tfrac{1}{5}\langle -3\sin t, 3\cos t, 4 \rangle$, differentiating gives $\mathbf{T}'(t) = \tfrac{1}{5}\langle -3\cos t, -3\sin t, 0 \rangle$, with $\|\mathbf{T}'(t)\| = \tfrac{3}{5}$. Dividing by the speed $5$ gives $\tfrac{3}{25}$ again.

The radius of curvature is $\tfrac{25}{3} \approx 8.33$, larger than the radius $3$ of the cylinder the helix winds around. That is the right comparison: a flat circle of radius $3$ would have curvature $\tfrac{1}{3}$, and the helix bends less than that because part of its motion is a straight climb.

The moving frame on the helix

From the previous example, $\mathbf{T}'(t) = \tfrac{1}{5}\langle -3\cos t, -3\sin t, 0 \rangle$ and $\|\mathbf{T}'(t)\| = \tfrac{3}{5}$, so

$$ \mathbf{N}(t) = \langle -\cos t, \; -\sin t, \; 0 \rangle $$

This is horizontal and points from the moving point straight in toward the axis of the cylinder, which is the direction the helix is turning. The binormal is

$$ \mathbf{B}(t) = \mathbf{T} \times \mathbf{N} = \left\langle \tfrac{4}{5}\sin t, \; -\tfrac{4}{5}\cos t, \; \tfrac{3}{5} \right\rangle $$

Check the frame. The binormal has magnitude $\sqrt{\tfrac{16}{25} + \tfrac{9}{25}} = 1$. It is orthogonal to $\mathbf{T}$, since $\tfrac{1}{5}\left[ (-3\sin t)\left(\tfrac{4}{5}\sin t\right) + (3\cos t)\left(-\tfrac{4}{5}\cos t\right) + 4 \cdot \tfrac{3}{5} \right] = \tfrac{1}{5}\left[ -\tfrac{12}{5} + \tfrac{12}{5} \right] = 0$, and it is orthogonal to $\mathbf{N}$ because its horizontal part is perpendicular to $\langle -\cos t, -\sin t \rangle$ and $\mathbf{N}$ has no vertical part.

Curvature of a parabola

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

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

At the vertex, $\kappa(0) = 2$ and the radius of curvature is $\tfrac{1}{2}$: near the origin the parabola hugs a circle of radius one half centered at $\left(0, \tfrac{1}{2}\right)$. At $x = 1$ the curvature has already dropped to $2 / 5^{3/2} \approx 0.179$, a radius of about $5.59$, and it tends to $0$ as $|x|$ grows because the parabola becomes ever more nearly straight. The second derivative is constant throughout, so every bit of that variation comes from the denominator.

Practice

Start with length. These ask for the arc length of plane and space curves, the arc-length function, and parametrizations in which the parameter is distance travelled.

Then bending. These cover curvature by each of its formulas, the radius of curvature, and the unit tangent, principal normal and binormal vectors.

Quiz

Five items on arc length integrals, the arc-length parameter, computing curvature and the radius of curvature, and the TNB frame.