1.1. Curves Curves

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1 1.1. Curve Curve Note. The hitorical note in thi ection are baed on Morri Kline Mathematical Thought From Ancient to Modern Time, Volume 2, Oxford Univerity Pre (1972, Analytic and Differential Geometry in the Eighteenth Century (Chapter 23. Page number in the hitorical note refer to thi reference. Image are from the MacTutor Hitory of Mathematic archive. Note. We define the curvature of a path (or curve in 3-dimenional pace in thi ection. We inpire our definition by appealing to phyical intuition baed on velocitie and acceleration. Therefore, there i a need to employ the technique of calculu. The term differential geometry wa firt ued by Luigi Bianchi ( in See page 554. Note. Iaac Newton ( tudied the curvature of plane curve (that i, curve in a plane. He publihed hi reult in Geometricia Analytica (in 1736 though hi work eem to date to He doe o by introducing a center of a the limiting point of the interection of normal curve near a given point. He define the curvature at the point a reciprocal of the radiu of the reulting oculating circle (a term ued by Gottfried Leibniz [ ] in a paper in Newton calculated the curvature of everal curve and a a reult duplicated ome earlier work of Chritiaan Huygen ( See page 556 and 557. Iaac Newton Gottfried Leibniz

2 1.1. Curve 2 Note. The tudy of curve in 3-dimenional pace (a oppoed to thoe retricted to a plane a dicued above wa extenively tudied by Alexi-Claude Clairaut ( He gave expreion for the arc length of a pace curve and the quadrature of certain area on urface. However, very little had been done in the theory of urface by See page 557. Alexi Clairaut Leonhard Euler Note. The next major tep in the tudy of curve in 3-dimenional pace i due to Leonhard Euler ( He included much of thi work in hi Mechanica (publihed in 1736 and preented more reult in hi Theoria Motu Corporum Solidorum eu Rigidorum (publihed in In thi econd book Euler derived the formula for the radial and normal component of acceleration in polar coordinate for a plane curve f(t. Thee are given in Calculu 3 and ued in the proof of Kepler Firt Law; ee my online note for Velocity and Acceleration in Polar Coordinate at faculty.etu.edu/gardnerr/2110/note-12e/c136.pdf. Euler tarted to explore curve in 3-dimenional curve in 1774 and preented a paper with hi reult in See page 558. Note. A we do in thi ection, Euler repreented pace curve by the parametric equation x = x(, y = y(, z = z(, where i arc length. He define the oculating plane aociated with a curve in pace, a we will do (though the term oculating plane i due to Johann Bernoulli, See page 558 and 559.

3 1.1. Curve 3 Note. Alexi Clairaut introduced the idea that a curve in 3-dimenional pace ha two curvature, which we hall call curvature and torion. Torion wa formulated explicitly and analytically by Michel-Ange Lancret ( Lancret alo called attention to three principal direction aociated with each point on a curve in 3-dimenional pace in hi Mémoire de Mathématique et de Phyique Préenté à l Académie Royal de Science, par Diver Sçavan, et Lu dan e Aemblée, 1 (1806, Thee direction correpond to our unit tangent vector T, principal normal vector N, and binormal vector B. Michel-Ange Lancret (from Wikipedia Augutin Loui Cauchy Note. Augutin Loui Cauchy ( in hi Leçon ur le application du calcul infinitéimal à la géométrie (publihed in 1826 clarifie much of the theory of curve in 3-dimenion. He alo clean up many of the conceptual problem with calculu involving the vague concept of infiniteimal and differential. In connection with curve, he point out that when one write d 2 = dx 2 + dy 2 + dz 2 one hould mean ( 2 d = dt ( 2 dx + dt ( 2 dy + dt ( 2 dx. dt Cauchy development of the geometry of curve i practically modern. See page 560.

4 1.1. Curve 4 Note. We now turn our attention to Faber text. In thi ection, we encounter mot of the idea mentioned above in the hitorical urvey. We motivate the idea and definition by appealing to phyical intuition baed on the poition, velocity, and acceleration experienced by a particle a it follow a curve in 3-dimenional pace R 3 = E 3. Definition. A curve in E 3 i a vector valued function of the parameter t: α(t = (x(t,y(t,z(t. Note. We aume the function x(t, y(t, and z(t have continuou econd derivative. Definition. The derivative vector of curve α i α (t = (x (t,y (t,z (t. If α(t i the poition of a particle at time t, then α (t i the velocity vector of the particle and α (t i the acceleration vector of the particle. The peed of the particle i the calar function α (t. Note. According to Newton Second Law of motion, the force acting on a particle of ma m and poition α(t i F(t = m α (t. Definition. The length (or arclength of the curve α(t for t [a,b] i S = b a α (t dt.

5 1.1. Curve 5 Note. If β(t i a curve for t [a,b], then β can be written a a function of arclength (which we will denote α( a follow. Firt, S(t = t a β (t dt (that i, S(t i an antiderivative of peed which atifie S(a = 0. Therefore S i a one to one function and S 1 exit. S 1 give the time at which the particle ha travelled along β(t a (gro ditance. So we denote thi a t = S 1 (. Second, we make the ubtitution for t: β(t = β(s 1 ( α(. However, it may be algebraically impoible to calculate t = S 1 ( (ee page 11, number 5. Recall. If f i differentiable on an interval I and f i nonzero on I, then f 1 exit (i.e. f i one-to-one on I on f(i and f 1 i differentiable on I. In addition, ( df 1 1 = ( dx df x=a x=f(a dx or f 1 (f(a = 1 f (a. Note. If β(t i parameterized a α( a above, then and d α d = d β ds 1 ds 1 d β(t = β(s 1 ( = α( = β (S 1 1 ( S (S 1 ( = β 1 (t S (t = β (t β (t. Notice d α d = α ( i a unit vector in the direction of the velocity vector of β(t.

6 1.1. Curve 6 Definition. If α( i a curve parameterized in term of arclength, then the unit tangent vector of α( i α ( = T(. ( α( i called a unit peed curve ince α ( = 1. Example 3 (page 6. Conider the circular helix β(t = (a cot,ain t,bt (ee Figure I-3, page 6. Parameterize β(t in term of arclength α( and calculate T(. Solution. We have β (t = ( a int,aco t,b. With S(t the total arclength travelled by a particle along the helix at time t, we have S (t = β (t = a 2 + b 2. Therefore, S(t = t a 2 + b 2 (taking S(0 = 0. Hence t = S 1 ( = and Alo, = ( α( = β(t = β(s 1 ( = β ( a co T( = α ( = ( (,ain, 1 ( a in b. ( (,aco,b. Notice that T = β (t β (t = β (S 1 ( β (S 1 (.

7 1.1. Curve 7 Note. T( alway ha unit length. The only way T( can change i in direction. Notice that thi correpond to a change in the direction of travel of a particle along the path α(. Since T( T( = 1, we have T ( T( = 0 (by the product rule and o T ( = α ( i orthogonal to T( = α (. Definition. The curvature of α( (denoted k( i k( = T ( = α (. If T ( 0 (and therefore curvature i nonzero then the unit vector in the direction of T ( i the principal normal vector, denoted N(. Notice. N( = T ( T ( = T ( k(. Example 3, page 6 (cont.. Calculate the curvature k( and principal normal vector N( for the helix β(t = (a cot,aint,bt. Solution. From above, we have T ( = α ( = 1 a 2 + b 2 ( a co ( (,ain and o k( = T ( = a a 2 (a contant. Now + b2 T N( = ( ( ( ( k( = co,in Notice that in term of t, N(t = (co t,in t,0.,0,0 if a > 0.

8 1.1. Curve 8 That i, N(t i a vector that point from the particle at β(t = (a cot,ain t,bt back to the z axi (that i, N(t i a unit vector from β(t to (0,0,bt. Note. If we take b = 0 in Example 3, we jut get β(t to trace out a circle of radiu a a in the xy plane. The curvature of thi circle i k( = a 2 + b = 1 2 a. Therefore, circle of mall radiu have large curvature and circle of large radiu have mall curvature (and the curvature of a traight line i 0. See Figure I-4. 1 Definition. For a given value of, the circle of radiu which i tangent to α k( and which lie in the plane of T( and N( i the oculating circle of α at point α(. The center of the oculating circle i the center of curvature of α at point α(, denoted c(. The plane containing the oculating circle i the oculating plane. See Figure I-6. Note. c( i calculated by going from point α( a ditance N(. That i, c( = α( + 1 k( N(. 1 k( in the direction Example (p. 13, # 12. Conider the helix above parameterized in term of : ( ( ( b α( = a co,ain,. Find c( and how that it i alo a helix. Solution. The center of curvature i c( = α( + 1 k( N(,

9 1.1. Curve 9 a where, from Example 3, k( = a 2 + b and 2 ( ( ( N( = co, in,0. So c( = = ( ( a co a2 + b ( 2 a in a2 + b ( ( 2 b2 a co a a, b2 a in ( co, a2 + b ( 2 in, a2 + b ( 2, b b. If we let A = b2 a and t = which i a circular helix. then a2 + b2, c(t = (A cot,ain t,bt, Note. The curvature k( of a curve α( give an idea of how a curve twit but doe not provide a complete decription of the curve gyration (a the text put it; ee page 8. There i information in how the oculating plane tilt a varie. Recall. A plane in E 3 i determined by a point (x 0,y 0,z 0 and a normal vector n = (A, B, C (we will not notationally ditinguih between point and vector. If (x,y,z i a point in the plane, then a vector from (x 0,y 0,z 0 to (x,y,z i perpendicular to n and o n (x x 0,y y 0,z z 0 = (A,B,C (x x 0,y y 0,z z 0 = A(x x 0 + B(y y 0 + C(z z 0 = 0.

10 1.1. Curve 10 Thi can be rearranged a Ax + By + Cz = D for ome contant D. Notice that twiting of the plane would be reflected in change in the direction of the normal vector. Example. Find the equation of the plane through the point (1,2,3, ( 2,3,3 and (1, 2, 4. Solution. The vector a = (1 ( 2,2 3,3 3 = (3,1,0 and b = (1 1,2 2,4 3 = (0,0,1 both lie in the deired plane. Recall that in E 3, a and b are both orthogonal to a b (provided a i not a calar multiple of b - See Appendix A for more detail. So we can take n = a b a a normal vector for the deired plane. i j k n = a b = = i 3 j + 0 k = (1, 3, So the deired plane atifie 1(x 1 3(y 2 + 0(z 3 = 0 or x 3y = 5. Expreed parametrically, x = 5 + 3t y = t z = z (a free variable. (Again, we make no notational ditinction between a vector and a point. There i a difference, though: point have location, but vector don t [nonzero vector have a length and a direction, but no poition]. Definition. The binormal vector i B = T N.

11 1.1. Curve 11 Note. B i orthogonal to both T and N (and therefore to the oculating plane. The derivative of B i B = ( T N = T N + T N (where repreent derivative with repect to whatever the variable of parameterization i. Since T = k N, T N = 0 and o B = T N. Since N i a unit vector, N i perpendicular to N (a argued above for T. Alo, T i perpendicular to N ince N = (1/k T. So both T and N are perpendicular to N and o B = T N i a multiple of N (ince we are in 3-dimenion, ay B = τ N. (Notice that if T and N are multiple of each other, then τ = 0. Definition. The torion of α at α( i the function τ( where B ( = τ( N(. Note. The torion meaure the twiting (or turning of the oculating plane and therefore decribe how much α depart from being a plane curve (a the text ay - ee page 9. Example. Calculate B, B and τ( for the helix above. Solution. We have T( = 1 ( a in ( (,aco,b and ( N( = co ( (,in,0, o B = T N = 1 i j k ( ( a in a2 a co +b 2 a2 b ( ( +b 2 co a2 in +b 2 a2 0 +b 2

12 1.1. Curve 12 and = 1 (b in B = ( b co a 2 + b 2 Therefore, ince B = τ N, we have τ( = ( (, b co ( (,in b a 2 + b 2.,a,0. Note. Notice that the torion i a contant in the previou example. Thi make ene ince the oculating plane tilt at a contant rate a a particle travel (uniformly up the helix (or pring. With b = 0, the helix i, in fact, a circle in the xy plane and o the oculating plane doe not change and τ = 0. Lemma. If τ = 0 at every point of a curve α, then α lie in a plane. Proof. Since B( = τ( N(, then with τ( = 0 we have B ( = 0 and o B( = B i a contant. Conider the plane containing α(0 with normal vector B. Let α( be an arbitrary point on curve α and define f( = ( α( α(0 B = α( B α(0 B. Then f ( = α ( B = T( B = 0 (ince B T. So f( i a contant with f(0 = ( α(0 α(0 B = 0. Therefore, f( = 0 for all and o f( = ( α( α(0 B = 0. So vector α( α(0 i orthogonal to B and point α( lie in the plane containing point α(0 and with normal vector B. Note. We will ee that the hape of a curve i completely determined by the curvature k( and torion τ(. Example (Exercie 14, page 14. Prove the Second Serret-Frenet Formula: N = k T + τ B. The Firt Serret-Frenet Formula i T = k N and the Third Serret-Frenet Formula i B = τ N. Proof. Since N, T, and B are mutually orthogonal, and each i a unit vector, we

13 1.1. Curve 13 can write α = ( α N N + ( α T T + ( α B B. Differentiating with repect to : α = ( α N + α N N + ( α N N +( α T + α T T + ( α T T + ( α B + α B B + ( α B B. So α = α + ( α N N + ( α k N T + ( α ( τ N B +( α N N + ( α Tk N + ( α B( τ N (1 uing the Firt and Third Serret-Frenet Formula. Now d d [1] = d [ N d 2] = d [ ] N N = 2N d N = 0. So N and N are orthogonal. Equating multiple of N in equation (1 (ince T and B are alo orthogonal to N: α ( N + k T τ B = 0. (2 So equation (2 implie either N = k T +τ B, or N +k T τ B i orthogonal to α. In the firt cae we are done, in the econd cae it mut be that α = a( N = a N (ince N, T and B are all orthogonal to N. Then α = a N + a N = T implie that T = a N and a N = 0. So a = 0 and a(a = a i contant. Therefore α( lie on a phere of radiu a. Now B = T N o But B B = T N + T N = (k N N + (a N N = 0. = τ N o τ = 0. Therefore, a commented in Exercie , α( i planar. So α( i a circle of radiu a. In thi cae, α = a N and k = 1/a. Since τ = 0, k T τ B = k T = 1 a T = 1 a (a N = N.

14 1.1. Curve 14 In either cae, the reult hold. (Note: The econd cae occur in the cae of circular motion: conider α(t + (a cot,ain t,0. Serret Frenet Frame (from Wikipedia Joeph Alfred Serret (from MacTutor Note. The Serret-Frenet formula were independently dicovered by two French mathematician, Jean Frédéric Frenet ( and Joeph Alfred Serret ( Frenet included thee idea in hi doctoral thei of Notice that the three Serret-Frenet formula are vector formula in 3-dimenion, and o actually correpond to nine (component formula. Frenet tated only ix of the calar formula. He publihed hi reult in a paper titled Sur quelque propriété de courbe à double courbure in the Journal de mathematique pure et applique in See the MacTutor Hitory of Mathematic biography of Frenet at mc.t-andrew.ac.uk/biographie/frenet.html. Note. Serret tated all nine of the (component formula in He alo publihed paper on number theory, calculu, the theory of function, group theory, mechanic, differential equation, and atronomy. See the MacTutor Hitory of Mathematic biography of Serret at graphie/serret.html. Revied: 6/25/2016