Section Arclength and Curvature. (1) Arclength, (2) Parameterizing Curves by Arclength, (3) Curvature, (4) Osculating and Normal Planes.

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1 Section 10.3 Arclength and Curvature (1) Arclength, (2) Parameterizing Curves by Arclength, (3) Curvature, (4) Osculating and Normal Planes. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 1 / 15

2 Arclength Let r(t) = f (t), g(t), h(t) be a smooth curve for t [a, b]. The arclength is b L = r (t) dt = a b a Partition [a, b] into n subintervals of equal size t = b a n. a = t 0 <... < t i < t i+1 <... < t n = b Over each interval [t i, t i+1 ], then length of the curve is approximately r (t i+1 ) r (t i ) r (t i ) t The total length is then lim n n r (t i ) t. i=1 [f (t)] 2 + [g (t)] 2 + [h (t)] 2 dt. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 2 / 15

3 Example A tower which is shaped like a cylinder is 30 feet high and has a radius of 10 feet. On the outside of the tower is a circular staircase which wraps around the tower 3 times from the base to the top of the tower. How long is the staircase? Solution: We can model the staircase using a helix curve r(t) = 10 cos(t), 10 sin(t), 30 6 t for t [0, 6]. r (t) = 10 sin(t), 10 cos(t), 5 r (t) = ( 10 sin(t)) 2 + (10 cos(t)) 2 + The length is 6 ( ) = r (t) dt, which equals feet. 0 MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 3 / 15

4 The Arclength Function The length of the portion of the curve r over the interval [a, t] is s(t) = t a r (τ) dτ Since s (t) = r (t) > 0, s(t) has an inverse, say t = t(s) and dt ds = 1 ds dt = 1 r (t) We can reparameterize the curve r(t) by its arclength s as r 0 (s) = r (t(s)). In particular, d r 0 (s) ds = d r(t(s)) dt(s) = r 1 (t) dt ds r (t) Thus, r 0 (s) is a unit tangent vector. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 4 / 15

5 Example: Reparameterize the curve by its arclength s. r(t) = e t sin(t), e t cos(t), e t where t [0, ) Solution: We first calculate the arclength function r (t) = e t (sin(t)+cos(t)), e t (cos(t) sin(t)), e t r (t) = 3e t So the arclength function s(t) = ( ) 3(e t 1) and has an inverse t = ln s Arclength parametrization for r(t) is r 0 (s) = ( ) ( ( )) ( ( )) s s s + 1 sin ln + 1, cos ln + 1, MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 5 / 15

6 Curvature For a smooth curve C defined by r(t), the unit tangent vector r (t) T (t) = r gives the instantaneous direction of the curve. (t) Question 1: Does the curve r(t) actually curve? If T (t) does not change direction, that is T (t) = 0, then the curve r(t) is a line. Thus, as long as T (t) 0, the curve r(t) indeed curves (bends and/or twists). Question 2: How sharply does the curve r(t) curve? That is, what is the rate of change of T (t)? The curvature of a curve r(t) at t is (s = s(t) is the arclength of r(t)). d T (s) κ(t) = ds MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 6 / 15

7 Curvature The curvature of a curve r(t) at t is d T (s) κ(t) = ds where s = s(t) is the arclength of r(t). Recall that s(t) = We have, in terms of the original parameter t, d T dt κ(t) = ds = T (t) r (t) Another formula, dt κ(t) = r (t) r (t) r (t) 3 t a r (τ) dτ. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 7 / 15

8 Example Find the curvature of a circle of radius a. Solution: We parameterize the circle as r(t) = a cos(t), a sin(t), 0. r (t) = a sin(t), a cos(t), 0 r (t) = a T (t) = sin(t), cos(t), 0 T (t) = cos(t), sin(t), 0 T (t) = 1 Therefore, κ(t) = 1 a. An osculating circle along the curve C at the point P is the circle which best fits inside the curve near P. The curvature at P is the reciprocal of the radius of the osculating circle. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 8 / 15

9 Curvature in R 2 For a planar curve y = f (x) we can take the parametrization r(t) = t, f (t), 0. r (t) = 1, f (t), 0 = i + f (t) j r (t) = 0, f (t), 0 = f (t) j r (t) r (t) = f (t) k r (t) = 1 + [f (t)] 2 r (t) r (t) = f (t) Thus, the curvature of a planar curve y = f (x) is κ(x) = f (x) (1 + f (x) 2 ) 3 2 MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 9 / 15

10 Normal and Binormal Vectors Question 3: We have established the curvature, κ(t), for the curve r(t). What is the direction in which the curve is bending? Answer: The direction in which the curve bends is in the same direction as T, and as we are only interested in direction, we take the unit vector: T N(t) (t) = T (t) Since T (t) = 1 it follows that T (t) T (t) = 1. Therefore, T (t) T (t) = 0 or T (t) T (t), which implies N (t) T (t) N (t) is the unit normal vector of the curve. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 10 / 15

11 The unit binormal vector of the curve is the vector B (t) = T (t) N (t) N (t) is in the direction which the curve bends at t and is orthogonal to the curve. N (t) points to the concave side of C. The osculating plane is orthogonal to the curve and contains the unit tangent and unit normal vectors. The unit binormal vector is the torsion or twisting direction of the curve. The normal plane is orthogonal to the curve and contains the unit normal and unit binormal vectors. MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 11 / 15

12 For the helix r(t) = cos(t), sin(t), t, (i) Find the unit normal and unit binormal vectors of r(t). (ii) Find the normal plan and osculating plane at the point P where t = 4 r (t) = sin(t), cos(t), 1 r (t) = 2 T (t) = 1 2 sin(t), cos(t), 1 T (t) = 1 2 cos(t), sin(t), 0 T (t) = 1 N (t) = cos(t), sin(t), 0 2 B (t) = T (t) N (t) = 1 2 sin(t), cos(t), 1 MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 12 / 15

13 The osculating plane passes through r ( ) ( 2 4 = normal vector B ( ) 4 = 1 2, 1 2, , 2 2, 4 x y + 2z = 4 The normal plane passes through r ( ) ( 2 ) 4 = 2, 2 2, 4 normal vector T ( ) 4 = 1, 1, 2. x + y + 2 2z = 4 ) and has a and has a MATH 127 (Section 10.3) Arclength and Curvature The University of Kansas 13 / 15