13.4 Curvature and Normal Vectors of a Curve

Ulrich Hoensch MAT275 Rocky Mountain College Billings, MT 59102 Curvature

Deﬁnition 13.7

Let C be a curve that is parametrized by arc length (i.e. traced out in unit speed by u(s), s ∈ [0, L]). Then the curvature of C at s is

0 κ(s) = T (s) .

Remark: The curvature κ (“kappa”) measures the magnitude of the change in direction of the curve if it is traversed in unit speed. Large curvature

Small curvature Computation of Curvature

I If a curve C is traced out by r(t), t ∈ [a, b], then the curvature at t is |T0(t)| κ(t) = , |r0(t)| where T(t) = r0(t)/|r0(t)| is the unit tangent vector. 2 I If C is a curve in R that is traced out by r(t) = hx(t), y(t)i, then |x0(t)y 00(t) − y 0(t)x00(t)| κ(t) = . ((x0(t))2 + (y 0(t))2)3/2 3 I If C is a curve in R that is traced out by r(t), then |r0(t) × r00(t)| κ(t) = . |r0(t)|3 Examples of Curvature Calculation

1. If C is a line parametrized by r(t) = r0 + tv, then the unit tangent vector is r0(t)/|r0(t)| = v/|v|. Since the unit tangent vector is constant and thus does not change direction, we have that the curvature of any line is zero. 2 2 2 2 2. If C : x + y = r is a circle in R , then we may parametrize C using r(t) = hr cos t, r sin ti, t ∈ [0, 2π]. Now, x(t) = r cos t y(t) = r sin t x0(t) = −r sin t y 0(t) = r cos t x00(t) = −r cos t y 00(t) = −r sin t. So, x0(t)y 00(t) − y 0(t)x00(t) = r 2 sin2 t + r 2 cos2 t = r 2, (x0(t))2 + (y 0(t))2 = r 2 sin2 t + r 2 cos2 t = r 2, and so |x0(t)y 00(t) − y 0(t)x00(t)| r 2 r 2 1 κ = = = = . ((x0(t))2 + (y 0(t))2)3/2 (r 2)3/2 r 3 r Thus, the curvature of a circle is the reciprocal of its radius. Examples of Curvature Calculation 3. Suppose H is the helix parametrized by r(t) = ha cos t, a sin t, bti, a, b > 0. Then, r0(t) = h−a sin t, a cos t, bi r00(t) = h−a cos t, −a sin t, 0i, and

i j k 0 00 r (t) × r (t) = −a sin t a cos t b

−a cos t −a sin t 0 = hab sin t, −ab cos t, a2 sin2 t + a2 cos2 ti = hab sin t, −ab cos t, a2i.

Deﬁnition 13.8

If T(t) is the unit tangent vector of a curve parametrized by r(t), then we deﬁne the unit normal vector N(t) as

T0(t) N(t) = . |T0(t)|

So, N(t) indicates the direction of change of T(t).

Results. 1. N(t) ⊥ T(t), hence the name unit normal vector. Also, N(t) always points to the “inside” of the curve. 2. If the curve is parametrized by arc length, then 1 N = T0(s). κ The Unit Normal Vector N The Unit Normal Vector N Proof of the Results. 1. We have |T(t)| = 1 for all t, so T(t) · T(t) = 12 = 1 for all t. Diﬀerentiating both sides of the previous equation and using the product rule, we obtain

T0(t) · T(t) + T(t) · T0(t) = 0 2T0(t) · T(t) = 0 T0(t) · T(t) = 0 T0(t) · T(t) = 0 |T0(t)| N(t) · T(t) = 0.

2. Since κ = |T0(s)|, we have

T0(s) T0(s) N = = . |T0(s)| κ Osculating Circle Deﬁnition 13.9

Suppose P is a point of on a curve. The osculating circle (or circle of curvature) at P is the circle C with the following two properties. I C is tangent to the curve at P; I C has the same curvature as the curve at P; this means the radius of C is 1/κ, where κ is the curvature at P. Practice Problems for Section 13.4

p.751-752: 3, 7, 9, 11, 13, 29.