13.11 13.4: Curvature

Curvature is a measure of how much a curve bends. It is used to study geometric properties of curves and motion along curves, and has applications in diverse areas such as roller coaster design (Figure 1), optics, eye surgery (see Exercise 60), and biochemistry (Figure 2).

Curvature is a key ingredient in roller coaster design.
Biochemists study the effect of the curvature of DNA strands on biological processes.

In Chapter 4, we used the second derivative f″(x) to measure the bending or concavity of the graph of y = f(x), so it might seem natural to take f″(x) as our definition of curvature. However, there are two reasons why this proposed definition will not work. First, f″(x) makes sense only for a graph y = f(x) in the plane, and our goal is to define curvature for curves in three-space. A more serious problem is that f″(x) does not truly capture the intrinsic curvature of a curve. A circle, for example, is symmetric, so its curvature ought to be the same at every point (Figure 3). But the upper semicircle is the graph of f(x) = (1 − x2)1/2 and the second derivative f″(x) = −(1 − x2)−3/2 does not have the same value at each point of the semicircle. We must look for a definition that depends only on the curve itself and not how it is oriented relative to the axes.

The second derivative of does not capture the curvature of the circle, which by symmetry should be the same at all points.

Consider a path with parametrization r(t) = 〈x(t), y(t), z(t)〉. We assume that r′(t) = 0 for all t in the domain of r(t). A parametrization with this property is called regular. At every point P along the path there is a unit tangent vector T = TP that points in the direction of motion of the parametrization. We write T(t) for the unit tangent vector at the terminal point of r(t):

For example, if r(t) = 〈t, t2, t3〉, then r′(t) = 〈1, 2t, 3t2〉, and the unit tangent vector at P = (1, 1, 1), which is the terminal point of r(1) = 〈1, 1, 1〉, is

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If we choose another parametrization, say r1(s), then we can also view T as function of s: T(s) is the unit tangent vector at the terminal point of r1(s).

Now imagine walking along a path and observing how the unit tangent vector T changes direction (Figure 4). A change in T indicates that the path is bending, and the more rapidly T changes, the more the path bends. Thus, would seem to be a good measure of curvature. However, depends on how fast you walk (when you walk faster, the unit tangent vector changes more quickly). Therefore, we assume that you walk at unit speed. In other words, curvature is the magnitude , where s is the parameter of an arc length parametrization. Recall that r(s) is an arc length parametrization if ∥r(s)∥ = 1 for all s.

The unit tangent vector varies in direction but not in length.

DEFINITION Curvature

Let r(s) be an arc length parametrization and T the unit tangent vector. The curvature at r(s) is the quantity (denoted by a lowercase Greek letter “kappa”)

Our first two examples illustrate curvature in the case of lines and circles.

EXAMPLE 1: A Line Has Zero Curvature

Compute the curvature at each point on the line r(t) = 〈x0, y0, z0〉 + tu, where ∥u∥ = 1.

Solution First, we note that because u is a unit vector, r(t) is an arc length parametrization. Indeed, r′(t) = u and thus ∥r′(t)∥ = ∥u∥ = 1. Thus we have T(t) = r′(t)/∥r′(t)∥ = r′(t) and hence T′(t) = r″(t) = 0 (because r′(t) = u is constant). As expected, the curvature is zero at all points on a line:

EXAMPLE 2: The Curvature of a Circle of Radius R Is 1/R

Compute the curvature of a circle of radius R.

Solution Assume the circle is centered at the origin, so that it has parametrization r(θ) = 〈R cos θ, R sin θ〉 (Figure 5). This is not an arc length parametrization if R ≠ 1. To find an arc length parametrization, we compute the arc length function:

The unit tangent vector at a point on a circle of radius R.

Example 2 shows that a circle of large radius R has small curvature 1/R. This makes sense because your direction of motion changes slowly when you walk at unit speed along a circle of large radius.

Thus s = Rθ, and the inverse of the arc length function is θ = g(s) = s/R. In Section 13.3, we showed that r1(s) = r(g(s)) is an arc length parametrization. In our case, we obtain

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The unit tangent vector and its derivative are

By definition of curvature,

This shows that the curvature is 1/R at all points on the circle.

In practice, it is often impossible to find an arc length parametrization explicitly. Fortunately, we can compute curvature using any regular parametrization r(t). To derive a formula, we need the following two results.

First is the fact that T(t) and T′(t) are orthogonal (see the marginal note). Second, arc length s is function s(t) of time t, so the derivatives of T with respect to t and s are related by the Chain Rule. Denoting the derivative with respect to t by a prime, we have

REMINDER

To prove that T(t) and T′(t) are orthogonal, note that T(t) is a unit vector, so T(t) · T(t) = 1. Differentiate using the Product Rule for Dot Products:

This shows that T(t) · T′(t) = 0

where is the speed of r(t). Since curvature is the magnitude , we obtain

THEOREM 1: Formula for Curvature

If r(t) is a regular parametrization, then the curvature at r(t) is

To apply Eq. (3) to plane curves, replace r(t) = 〈x(t), y(t)〉 by r(t) = 〈x(t), y(t), 0〉 and compute the cross product.

Proof

Since υ(t) = ∥r′(t)∥, we have r′(t) = υ(t)T(t). By the Product Rule,

r″(t) = υ′(t)T(t) + υ(t)T′(t)

Now compute the following cross product, using the fact that T(t) × T(t) = 0:

Because T(t) and T′(t) are orthogonal,

REMINDER

By Theorem 1 in Section 12.4,

∥v × w∥ = ∥v∥ ∥w∥ sin θ

where θ is the angle between v and w.

Eq. (4) yields ∥r′(t) × r″(t)∥ = υ(t)2∥T′(t)∥. Using Eq. (2), we obtain

∥r′(t) × r″(t)∥ = υ(t)2∥T′(t)∥ = υ(t)3κ(t) = ∥r′(t)∥3κ(t)

This yields the desired formula.

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EXAMPLE 3: Twisted Cubic Curve

Calculate the curvature κ(t) of the twisted cubic r(t) = 〈t, t2, t3〉. Then plot the graph of κ(t) and determine where the curvature is largest.

Solution The derivatives are

The parametrization is regular because r′(t) ≠ 0 for all t, so we may use Eq. (3):

The graph of κ(t) in Figure 6 shows that the curvature is largest at t = 0. The curve r(t) is illustrated in Figure 7. The plot is colored by curvature, with large curvature represented in blue, small curvature in green.

Graph of the curvature κ(t) of the twisted cubic r(t) = 〈t, t2, t3〉.
Graph of the twisted cubic r(t) = 〈t, t2, t3〉 colored by curvature.

In the second paragraph of this section, we pointed out that the curvature of a graph y = f(x) must involve more than just the second derivative f″(x). We now show that the curvature can be expressed in terms of both f″(x) and f′(x).

THEOREM 2: Curvature of a Graph in the Plane

The curvature at the point (x, f (x)) on the graph of y = f(x) is equal to

Proof

The curve y = f(x) has parametrization r(x) = 〈x, f(x)〉. Therefore, r′(x) = 〈1, f′(x)〉 and r″(x) = 〈0, f″(x)〉. To apply Theorem 1, we treat r′(x) and r″(x) as vectors in R3 with z-component equal to zero. Then

Since , Eq. (3) yields

CONCEPTUAL INSIGHT

Curvature for plane curves has a geometric interpretation in terms of the angle of inclination, defined as the angle θ between the tangent vector and the horizonal (Figure 8). The angle θ changes as the curve bends, and we can show that the curvature κ is the rate of change of θ as you walk along the curve at unit speed (see Exercise 61).

The angle θ changes as the curve bends.

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EXAMPLE 4

Compute the curvature of f(x) = x3 − 3x2 + 4 at x = 0, 1, 2, 3.

Solution We apply Eq. (5):

We obtain the following values:

Figure 9 shows that the graph bends more where the curvature is large.

Graph of f(x) = x3 − 3x2 + 4 and the curvature κ(x).