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Arc Length and Curvature · Tutorial 199 of 999

Radius of Curvature in Circles and Helices

Connect curvature to the radius of the circle that locally matches a curve, then calculate both quantities for circles and helices.

Intermediate 9 min read

What You'll Learn

  • Define radius of curvature and identify when it is finite.
  • Explain why a circle of radius R has curvature 1/R.
  • Calculate curvature and radius of curvature for a circular helix.
  • Evaluate a helix at a specified point using parameter derivatives.
  • Distinguish a helix’s radius of curvature from its distance to the axis.

From Turning to Radius

A curve’s curvature measures how quickly its direction changes as we move along the curve. A circle offers a direct geometric interpretation of that change: a small circle has a sharply bending path, while a large circle bends more gently. The quantity that expresses this local bending as a length is the radius of curvature.

For circles, this length is exactly the ordinary radius. For a helix, which winds around an axis while also moving along it, the radius of curvature is generally not the radius of the helix’s circular motion. The distinction is important: curvature describes the curve’s local bending in space, not simply its distance from an axis.

Definition: At a point where a regular, twice differentiable curve has positive curvature \(\kappa\), its radius of curvature is \(\rho=1/\kappa\). It is the radius of a circle having the same curvature at that point. If \(\kappa=0\), there is no finite radius of curvature; one may describe the limiting radius as infinite.

Here, “regular” means that the velocity vector is nonzero. For a parametrized space curve \(\mathbf{r}(t)\), this requires \(\mathbf{r}'(t)\ne\mathbf{0}\). Under this condition, the curvature can be computed from parameter derivatives by

$$ \kappa(t)= \frac{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|} {\|\mathbf{r}'(t)\|^3}, $$

provided the curve is twice differentiable. Thus, when this curvature is positive,

$$ \rho(t)= \frac{\|\mathbf{r}'(t)\|^3} {\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|}. $$

The reciprocal relationship does not require the curve to be a circle. It defines the radius of the circle that matches the curve’s bending at the point in question.

Why a Circle’s Curvature Radius Is Its Geometric Radius

Consider a circle of radius \(R>0\), parametrized by its angle \(\theta\): \(\mathbf{r}(\theta)=(R\cos\theta,R\sin\theta)\). Its speed is \(R\), so an increase \(d\theta\) in angle corresponds to an arc-length increase \(ds=R\,d\theta\). Its unit tangent is \(\mathbf{T}(\theta)=(-\sin\theta,\cos\theta)\).

The tangent turns through the same angle as the circle’s radius. Since the tangent has unit length, the magnitude of its derivative with respect to \(\theta\) is \(1\). Curvature is the magnitude of the tangent’s change per unit arc length, giving

$$ \kappa = \left\|\frac{d\mathbf{T}}{ds}\right\| = \left\|\frac{d\mathbf{T}}{d\theta}\right\| \left|\frac{d\theta}{ds}\right| = 1\cdot\frac{1}{R} = \frac{1}{R}. $$

Taking the reciprocal yields \(\rho=R\). This calculation explains the geometric meaning of the reciprocal: a circle of radius \(R\) turns through one unit of angle over arc length \(R\), so its curvature is \(1/R\).

Takeaway: Curvature has units of inverse length, while radius of curvature has units of length. Increasing the radius of a circle decreases its curvature in exactly the reciprocal proportion.

Worked Example: A Circle of Radius Five

Worked Example: A Circle of Radius Five

Find the curvature and radius of curvature of \(\mathbf{r}(t)=(5\cos t,5\sin t)\) at \(t=\pi/3\).

Differentiate each component:

$$ \mathbf{r}'(t)=(-5\sin t,5\cos t), \qquad \mathbf{r}''(t)=(-5\cos t,-5\sin t). $$

At \(t=\pi/3\), \(\sin(\pi/3)=\sqrt{3}/2\) and \(\cos(\pi/3)=1/2\). Therefore, the point and derivatives are

$$ \mathbf{r}\left(\frac{\pi}{3}\right) = \left(\frac{5}{2},\frac{5\sqrt{3}}{2}\right), \quad \mathbf{r}'\left(\frac{\pi}{3}\right) = \left(-\frac{5\sqrt{3}}{2},\frac{5}{2}\right), \quad \mathbf{r}''\left(\frac{\pi}{3}\right) = \left(-\frac{5}{2},-\frac{5\sqrt{3}}{2}\right). $$

The speed is \(\|\mathbf{r}'(\pi/3)\|=\sqrt{75/4+25/4}=5\). The two-dimensional determinant corresponding to the cross-product magnitude is

$$ \left| \left(-\frac{5\sqrt{3}}{2}\right) \left(-\frac{5\sqrt{3}}{2}\right) - \left(\frac{5}{2}\right)\left(-\frac{5}{2}\right) \right| = \left|\frac{75}{4}+\frac{25}{4}\right| = 25. $$

Thus \(\kappa=25/5^3=1/5\), and \(\rho=1/(1/5)=5\). As a check, the parametrization describes a circle of geometric radius \(5\), so the calculated radius of curvature agrees with that radius.

Curvature and Radius of a Circular Helix

A circular helix of radius \(a>0\) and vertical advance \(b\) per unit angular parameter can be written as

$$ \mathbf{r}(t)=(a\cos t,a\sin t,bt). $$

The first two coordinates trace a circle of radius \(a\), while the third coordinate changes steadily with \(t\). The parameter \(b\) may be positive, negative, or zero; the usual nonplanar helix has \(b\ne0\). Because \(a>0\), this parametrization is regular for every \(t\).

Differentiate and compute the speed:

$$ \mathbf{r}'(t)=(-a\sin t,a\cos t,b), \qquad \mathbf{r}''(t)=(-a\cos t,-a\sin t,0), $$
$$ \|\mathbf{r}'(t)\| = \sqrt{a^2\sin^2t+a^2\cos^2t+b^2} = \sqrt{a^2+b^2}. $$

The speed is constant, but the velocity direction changes. The cross product is

$$ \mathbf{r}'(t)\times\mathbf{r}''(t) = (ab\sin t,-ab\cos t,a^2). $$

Its magnitude is \(\sqrt{a^2b^2\sin^2t+a^2b^2\cos^2t+a^4}=a\sqrt{a^2+b^2}\), since \(a>0\). Substitution into the curvature formula gives the following result.

Result: For \(\mathbf{r}(t)=(a\cos t,a\sin t,bt)\), with \(a>0\), \[ \kappa(t)=\frac{a}{a^2+b^2}, \qquad \rho(t)=\frac{a^2+b^2}{a}. \] Both values are constant along the helix.

The radius \(a\) measures distance from the helix to its axis. The radius of curvature \((a^2+b^2)/a\) measures local bending. If \(b=0\), the curve is a circle in a plane and the formula reduces to \(\rho=a\). If \(b\ne0\), then \(\rho=a+b^2/a>a\): the helix bends less sharply than the circle obtained by looking only at its projection onto the horizontal plane.

Worked Example: A Helix at a Specified Point

Worked Example: A Helix at a Specified Point

For \(\mathbf{r}(t)=(3\cos t,3\sin t,4t)\), find the point, curvature, and radius of curvature at \(t=0\).

The point is \(\mathbf{r}(0)=(3,0,0)\). The derivatives there are obtained from

$$ \mathbf{r}'(t)=(-3\sin t,3\cos t,4), \qquad \mathbf{r}''(t)=(-3\cos t,-3\sin t,0), $$
$$ \mathbf{r}'(0)=(0,3,4), \qquad \mathbf{r}''(0)=(-3,0,0). $$

The speed is \(\sqrt{0^2+3^2+4^2}=5\). The cross product and its magnitude are

$$ \mathbf{r}'(0)\times\mathbf{r}''(0) = (0,-12,9), \qquad \|(0,-12,9)\| = \sqrt{144+81} = 15. $$

Therefore, \(\kappa(0)=15/5^3=15/125=3/25\), and \(\rho(0)=25/3\). Check the result with the helix formula: \(a=3\), \(b=4\), so \(\kappa=3/(9+16)=3/25\) and \(\rho=(9+16)/3=25/3\). The direct calculation and the general formula agree.

Worked Example: A Helix with a Faster Parameter

Worked Example: A Helix with a Faster Parameter

Find the curvature and radius of curvature of \(\mathbf{r}(t)=(4\cos 2t,4\sin 2t,6t)\) at \(t=\pi/4\).

At this parameter value, \(2t=\pi/2\), so the point is \((0,4,3\pi/2)\). Differentiating gives

$$ \mathbf{r}'(t)=(-8\sin 2t,8\cos 2t,6), \qquad \mathbf{r}''(t)=(-16\cos 2t,-16\sin 2t,0). $$

At \(t=\pi/4\), these vectors are \(\mathbf{r}'=(-8,0,6)\) and \(\mathbf{r}''=(0,-16,0)\). The speed is \(\sqrt{64+36}=10\), and

$$ \mathbf{r}'\times\mathbf{r}'' = (96,0,128), \qquad \|\mathbf{r}'\times\mathbf{r}''\| = \sqrt{96^2+128^2} = \sqrt{25600} = 160. $$

Hence \(\kappa=160/10^3=160/1000=4/25\), and \(\rho=25/4\). To check, use the angular variable \(u=2t\); the same geometric helix is \((4\cos u,4\sin u,3u)\), with \(a=4\) and \(b=3\). The formula gives \(\kappa=4/(16+9)=4/25\) and \(\rho=25/4\), matching the derivative calculation.

A Common Mistake: Confusing Two Radii

For the helix \((3\cos t,3\sin t,4t)\), it is tempting to report \(3\) as the radius of curvature because \(3\) is the distance from the helix to its axis. That is incorrect. The distance to the axis is \(a=3\), but the curvature is \(3/(3^2+4^2)=3/25\), so the radius of curvature is \(25/3\), not \(3\).

The error comes from treating the helix as if it were just its circular projection. The helix also moves in the vertical direction, which affects its speed and how rapidly its tangent turns per unit distance. The full space-curve calculation accounts for both effects.

Using the Relationship Reliably

For a circle, the geometric radius is already the radius of curvature. For a helix, first identify the parameters in its representation and use the helix formula, or calculate from derivatives if the parametrization is presented in another form. In either case, evaluate at the specified parameter value when the question asks for a particular point, and verify that the parametrization is regular.

1
Identify the curve.
For a circle, find its geometric radius. For a standard helix, identify \(a\) and \(b\).
2
Find curvature.
Use \(\kappa=1/R\) for a circle, the helix formula for a standard helix, or the parameter-derivative formula for a general parametrization.
3
Take the reciprocal.
When \(\kappa>0\), calculate \(\rho=1/\kappa\), keeping the result in units of length if units are given.
4
Check the interpretation.
For a circle, the result should equal its geometric radius. For a helix, do not substitute its distance to the axis for its radius of curvature.

Radius of curvature is useful because it turns an inverse-length measure of bending into a directly interpretable length. A small value means strong local bending; a large value means gentle bending. For a circle, that length describes the whole curve. For a helix, it describes the local curvature at each point, even though the helix itself does not lie in a single circle.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the following questions.

  1. What is the radius of curvature of a circle of radius \(7\), and what is its curvature?
  2. For a helix with \(a=2\) and \(b=3\), calculate its curvature and radius of curvature.
  3. For a standard circular helix, do curvature and radius of curvature change with the parameter \(t\)? Explain using the formulas.
  4. Why is the distance from a helix to its axis generally not its radius of curvature?
  5. A regular curve has curvature \(0\) at a point. Does it have a finite radius of curvature there under the reciprocal definition?