Notes on the geometry of curves, Math 210 John Wood

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1 Baic definition Note on the geometry of curve, Math 0 John Wood Let f(t be a vector-valued function of a calar We indicate thi by writing f : R R 3 and think of f(t a the poition in pace of a particle at time t; f define the orbit of the particle f(t + h f(t i the diplacement in pace during the time interval from t to t + h (/h(f(t + h f(t i the [diplacement]/[time interval] or average velocity f (t = lim (f(t + h f(t i the velocity h f (t i the peed T (t = f (t f (t i the direction vector or unit tangent vector It i defined whenever f (t 0 h 0 f (t = f (t T (t expree velocity a peed (a calar time direction (a vector (t = t t 0 f (t dt i the ditance traveled along the orbit during the time interval from t 0 to t, or the arclength of the curve f (t i the acceleration An example The helix i defined by f(t = (a co ωt, a in ωt, bt f (t = ( aω in ωt, aω co ωt, b f (t = a ω + b T (t = ( aω in ωt, aω co ωt, b a ω + b The ditance along the helix from f(0 to f(π/ω i π/ω 0 a ω + b dt = (π/ω a ω + b The diplacement i f(π/ω f(0 = (a, 0, πb/ω (a, 0, 0 = (0, 0, πb/ω The acceleration i f (t = ( aω co ωt, aω in ωt, 0 Notice that f (t = f (t (0, 0, ω For any orbit f that atifie thi equation the acceleration i perpendicular to the velocity, o the peed will be contant, a it i for our helix The force on a particle with electric charge q moving with velocity v in a magnetic field B i F = qv B

2 The acceleration due to thi force i given by F = mf (t where m i the ma of the particle So the motion of the particle atifie mf (t = qf (t B Our helix will atify thi equation if B = ωm k The frequency ω = q B i independent of the peed (provided the peed i low enough that the ma can be regarded q π π m a contant Thi fact played a role in the deign of the cyclotron particle accelerator by Lawrence and Livington in 93 Curve To tudy propertie of the curve itelf rather than propertie of motion along the curve, we conider the curve parameterized by arclength If c( i a vector function of a parameter, the condition that be arclength i that the peed c ( = Then T ( = c ( i the unit tangent vector Since T ( T ( =, differentiating we find T ( T ( = 0, o T ( T ( We define the curvature of c to be k( = T ( If k( 0, define the principal normal vector by N( = k( T ( Then T ( = k(n( The vector T ( and N( are perpendicular to each other and of length one The binormal vector defined by B( = T ( N( ha length one and i perpendicular to both T and N The triple T, N, and B i an orthonormal bai of R 3 and can be obtained from the tandard orthonormal bai i, j, and k by a rotation of 3-pace Jut a with the tandard bai, an arbitrary vector can be written a a um in the form V = at + bn + cb To find the coefficient a, b, and c, ue the dot product: V T = a, V N = b, and V B = c Since N( N( =, N ( N(, and ince N( T ( = 0 we have N ( T ( = N( T ( = k( We define the torion of c to be τ( = N ( B( Then N ( = k(t ( + τ(b( Finally by differentiation we find B = T N +T N = kn N +T ( kt +τb = τn At each point c( along the curve c we have a unit vector T tangent to the curve, N perpendicular to T and pointing in the direction that T i turning, and B completing an

3 orthonormal bai The rotation of thee vector a we move along the curve i decribed by the Frenet-Serret formula for their derivative: T = kn N = kt B = τn τb If we pecify a given point p, direction u, and function, k( (nonvanihing and τ( which have third derivative, then a fundamental theorem of differential equation implie the exitence of a curve c( with c(0 = p, c (0 = u, c ( =, and with curvature and torion equal to the given function k and τ Furthermore the curve i uniquely determined by the initial condition c(0 = p and c (0 = u and the curvature and torion function Example The helix above with ω = ha contant peed a + b Replacing t by / a + b we get c( = ( a co c ( = T ( = c ( = T ( = So k( = N ( = a + b, a in (a in a + b ( co a a + b a a + b and N( = a + b (in B = T N = τ( = N B = a + b, b a + b a + b, a co a + b, in ( co a + b, co a + b (b in b a + b a + b, b a + b, 0 a + b, in a + b, 0, b co a + b a + b, 0 a + b, a Motion along a curve Suppoe a particle follow the curve c( at varying peed It ditance along the curve will be given by ome function (t It poition i given by f(t = c((t The velocity i f (t = c ((t (t = (tt ((t, the peed i f (t and the acceleration i f (t = (tt ((t + ( (t T ((t = (tt ((t + k((t( (t N((t = (t, The tangential component of acceleration i the rate of change of peed, the normal component i the curvature time the quare of the peed 3

4 Thee formula for velocity and acceleration can be ued to give a formula for curvature when the parameter i not necearily arclength ( f (t f (t = T + k N T Hence k(t = f (t f (t f (t 3 = k 3 N T = k 3 B Taylor erie We expand c( in a Taylor erie around = 0 The derivative are given by: The expanion of c( i: c ( = T ( c ( = T ( = k(n( c ( = k (N( + k(n ( = k T + k N + kτb c( = c(0 + T (0 + k(0n( ( k(0 T (0 + k (0N(0 + k(0τ(0b(0 + The leading term behavior for each component i c( c(0 + T (0 + k(0n( k(0τ(0b(0 The projection of thi approximation onto the plane panned by T and N (the oculating plane i (, k(0/ The projection onto the plane panned by T and B (the rectifying plane i (, 3 k(0τ(0/6 The projection onto the plane panned by N and B (the normal plane i (k(0/(, 3 τ(0/3 N T Oculation plane View of a curve B T Rectifying plane B Normal plane An activity Find a clear cylindrical drinking gla and ome narrow colored tape If you place the tape on the gla tarting down at an angle from the top it will follow a helix Be careful not to bend the tape to the ide which will caue it to pucker and deviate from a helix Viualize the vector T, N, and B at ome point on the helix Then view the curve 4 N

5 by looking traight down firt B, then N, and then T What you ee hould look like the view of a curve above Geometric conequence of differential aumption If τ = 0, c i a plane curve Proof Since B ( = τ(n(, we have B ( = 0, o B( i a contant vector, B We will how that the curve lie in a plane perpendicular to B, that i that c( c(0 B The derivative of (c( c(0 B i c ( B = T B = 0, o thi dot product i a contant Taking = 0 we ee that c( c(0 B Exercie: If k = 0, c lie on a traight line If τ = 0 and k( i contant, then c( lie on a circle Proof The curvature of a circle of radiu a i /a The center i at c( + an( Hence with our hypothee, A( = c( + k( N( hould be a contant vector Differentiate A( and ue the Frenet-Serret formula to conclude that A( i contant, ay equal to A Then c( A = /k i contant and (c( A B = 0 Therefore c( lie on the circle of radiu /k about c(0 + (/kn(0 in the plane perpendicular to B through that center point Exercie: If k and τ are contant, c i a helix Curve on a phere Suppoe c( lie on the phere of radiu r We know that a traight line i not curved enough to lie on a phere What i the leat curved curve that lie on the phere? The condition that c( lie on the phere i c( c( = r Differentiating we find c c = 0 and c c + c c = 0 Since we aume c i parameterized by arclength, c c = So c c = c c = and therefore kc N = If θ i the angle between c and N, c N = c N co θ Therefore c N c N = r, o k /r Thi how /r i the leat poible curvature at any point along a curve on the phere of radiu r If k = /r, then c N = r, θ = π, and N = (/rc Then N = (/rc = (/rt By the Frenet-Serret formula thi implie that τ = 0 Therefore B i contant and c B = rn B = 0, o c lie in the plane through the origin perpendicular to B which meet the phere in a great circle The leat curved curve which lie on a phere i a great circle Sweeping area Conider the orbit f(t of a particle and two nearby point f(t and f(t + h The area wept out by the line from the origin to the orbit a time goe from t to t + h i approximately half the area of the parallelogram panned by f(t and f(t + h Area(h f(t f(t + h 5

6 Rate of weep of area = lim h 0 + Area(h h = lim f(t f(t + h h 0 + h = lim (f(t + h f(t f(t h 0 h = f(t f (t A central force i a force which, at each point, i directed along the line toward (or away from the origin, F ( v i a calar function time v It follow that v F( v = 0 For example the gravitational force i given by F ( v = GmM v v v If a particle of ma m i moving in an orbit f(t according to Newton law mf (t = F (f(t, then f(t f (t = 0 Differentiating (f f = f f + f f = 0 Hence f f = A i a contant vector Therefore the rate of weep of area i contant, f(t weep out equal area in equal time A i perpendicular to f and f and alo to f becaue it i a multiple of f by the central force aumption Hence A T and A N, o A i a multiple of B and B i contant Therefore the motion lie in a plane Thi argument derive Kepler econd law from Newton dynamic Exercie: Let x(t i + y(t j, a t b, be a curve in the plane which encloe a region R Ue the rate of weep of area formula to how that the area of R i given by b a x(ty (t x (ty(t dt 6

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