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 2 [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 2 [a; b], then the curvature at t is jT0(t)j κ(t) = ; jr0(t)j where T(t) = r0(t)=jr0(t)j 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 jx0(t)y 00(t) − y 0(t)x00(t)j 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 jr0(t) × r00(t)j κ(t) = : jr0(t)j3 Examples of Curvature Calculation 1. If C is a line parametrized by r(t) = r0 + tv, then the unit tangent vector is r0(t)=jr0(t)j = v=jvj. 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 2 [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 jx0(t)y 00(t) − y 0(t)x00(t)j 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: 0 00 p 2 2 2 2 2 2 4 So,p jr (t) × r (t)pj = a b sin t + a b cos t + a = a4 + a2b2 = a a2 + b2. Since p p jr0(t)j = a2 sin2 t + a2 cos2 t + b2 = a2 + b2, the curvature of the helix is p jr0(t) × r00(t)j a a2 + b2 a κ = = p = : jr0(t)j3 ( a2 + b2)3 a2 + b2 The Unit Normal Vector N 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) = : jT0(t)j 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 jT(t)j = 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 jT0(t)j N(t) · T(t) = 0: 2. Since κ = jT0(s)j, we have T0(s) T0(s) N = = : jT0(s)j 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..