Unit Normal Vector

We noted above that T′(t) and T(t) are orthogonal. The unit vector in the direction of T′(t), assuming it is nonzero, is called the unit normal vector and is denoted N(t) or simply N:

Furthermore, ∥T′(t)∥ = υ(t)κ(t) by Eq. (2), so we have

Intuitively, N points the direction in which the curve is turning (see Figure 11). This is particularly clear for a plane curve. In this case, there are two unit vectors orthogonal to T (Figure 10), and of these two, N is the vector that points to the “inside” of the curve.

For a plane curve, the unit normal vector points in the direction of bending.

EXAMPLE 5: Unit Normal to a Helix

Find the unit normal vector at to the helix r(t) = 〈cos t, sin t, t〉.

Solution The tangent vector r′(t) = 〈− sin t, cos t, 1〉 has length , so

Hence, (Figure 11).

Unit tangent and unit normal vectors at on the helix in Example 5.

We conclude by describing another interpretation of curvature in terms of the osculating or “best-fitting circle” circle. Suppose that P is a point on a plane curve where the curvature κP is nonzero. The osculating circle, denoted OscP, is the circle of radius R = 1/κP through P whose center Q lies in the direction of the unit normal N (Figure 12). In other words, the center Q is determined by

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The center Q of the osculating circle at P lies at a distance from P in the normal direction.

Among all circles tangent to at P, OscP “best fits” the curve (Figure 13; see also Exercise 71). We refer to R = 1/κP as the radius of curvature at P. The center Q of OscP is called the center of curvature at P.

Among all circles tangent to the curve at P, the osculating circle is the “best fit” to the curve.

EXAMPLE 6

Parametrize the osculating circle to y = x2 at .

Solution Let f(x) = x2. We use the parametrization

and proceed by the following steps.

Step 1. Find the radius.

Apply Eq. (5) to f(x) = x2 to compute the curvature:

The osculating circle has radius .

Step 2. Find N at .

For a plane curve, there is an easy way to find N without computing T′. The tangent vector is r′(x) = 〈1, 2x〉, and we know that 〈2x, −1〉 is orthogonal to r′(x) (because their dot product is zero). Therefore, N(x) is the unit vector in one of the two directions ± 〈2x, −1〉. Figure 14 shows that the unit normal vector points in the positive y- direction (the direction of bending). Therefore,

The osculating circle of y = x2 at has center Q and radius .

Step 3. Find the center Q.

Apply Eq. (8) with :

Step 4. Parametrize the osculating circle.

The osculating circle has radius , so it has parametrization

If a curve lies in a plane, then this plane is the osculating plane. For a general curve in three-space, the osculating plane varies from point to point.

To define the osculating circle at a point P on a space curve , we must first specify the plane in which the circle lies. The osculating plane is the plane through P determined by the unit tangent TP and the unit normal NP at P (we assume that T′ ≠ 0, so N is defined). Intuitively, the osculating plane is the plane that “most nearly” contains the curve near P (see Figure 15). The osculating circle is the circle in the osculating plane through P of radius R = 1/κP whose center is located in the normal direction NP from P. Equation (8) remains valid for space curves.

Osculating circles to r(t) = 〈cos t, sin t, sin 2t〉.

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