Vector Function Continuity of Vector Function Differentiability of a ...

Acceleration 6• Example Let a be a positive constant. Find the curvature of the curver(t) = (a sin t, a cos t, t).soWe haveConsequentlyT =drdtdTdt= (a cos t, −a sin t, 1),∥ ds ∥∥∥∥dt = dr ∥ ∥∥dt ∥ = √ a 2 + 1,drdt∥ dr∥dt=(a cos t, −a sin t, 1)√ ,a2 + 1(−a sin t, −a cos t, 0)= √ .a2 + 1∥ ∥ dT∥∥ dTκ =∥ ds ∥ = ∥dtdsdt= aa 2 + 1 .NormalThe tangent vector T satisfies T · T = 1. When this identity is differentiated withrespect to s, we obtaindTds · T + T · dTds = 0.Therefore T · dTds = 0. At a point where κ= ∥ ∥ ∥dTdsis given byThusdTdsN = ∥ ∥ dT= 1 dT∥ κ ds .dsdTds = κN.∥ ∥∥ is nonzero, the principal unit normalAccelerationSince the velocity v satisfiesv = drdt = drdsdsdt = T dsdt , March 17, 2001

Curvature for a Plane Curve 7the acceleration a satisfiesa = dvdt = dT dsdt dt + T d2 sdt = dT2 ds= κ( ) 2dsN + d2 sdt dt T. 2( ) 2ds+ T d2 sdt dt 2In this way, the quantity d2 sis expressed as the tangential component of accelerationdt 2while the quantity κ ( ) 2dsdt is expressed as the normal component of acceleration. Since⎛v · a = T ds ( ) ⎞2dsdt · ⎝κ N + d2 sdt dt T 2the tangential component of acceleration isThe computationv × a = dsdt T × ⎛⎝κd 2 sdt = v · a2 dsdtand the fact ‖T × N‖ = 1 show that= r′ (t) · r ′′ (t)‖r ′ (t)‖ .( ) ⎞2dsN + d2 sdt dt T ⎠ = κ2⎠ = dsdtd 2 sdt 2 ,( ) 3dsT × Ndtκ =‖v × a‖( dsdt) 3= ‖r′ (t) × r ′′ (t)‖‖r ′ (t)‖ 3 .Therefore, the normal component of acceleration may be expressed as‖r ′ (t) × r ′′ (t)‖.‖r ′ (t)‖Curvature for a Plane CurveSuppose that the plane curve is given by (x(t), y(t)). Regard it as a space curve r(t) =(x(t), y(t), 0), its curvature equalsκ = ‖r′ (t) × r ′′ (t)‖‖r ′ (t)‖ 3= ‖(x′ , y ′ , 0) × (x ′′ , y ′′ , 0)‖‖(x ′ , y ′ , 0)‖ 3= |x′ y ′′ − y ′ x ′′ |[(x ′ ) 2 + (y ′ ) 2 ] 3/2 . March 17, 2001

Radius,Center,Circle of Curvature 8Suppose that the plane curve is the graph y = f(x) of the function f(x). It can beregarded as the parametric curve(x, f(x), 0).Hence its curvature equalsκ = |1 · f ′′ (x) − f ′ (x) · 0|[1 2 + (f ′ (x)) 2 ] 3/2 =|f ′′ (x)|[1 + (f ′ (x)) 2 ] 3/2 .Geometric Interpretation∣ ∣κ = ∣∣ ∣∣∣dφds ∣φ =the angle that the tangent makes with the x-axisSince tan φ = y ′ , it follows that φ = tan −1 y ′ soIt follows from the chain rule thatdφdx = y ′′1 + (y ′ ) 2 .dφdx = dφ dsds dx = dφ √1 + (yds′ ) 2 .Hence∣ dφ∣ ∣∣ =∣ ds∣ dφ∣dx√1 + (y ′ ) = |y ′′ |= κ.2 (1 + (y ′ ) 2 3/2)Radius,Center,Circle of CurvatureThe reciprocalρ = 1 κof the curvature is called the radius of curvature. A point at a distance ρ from thecurve in the direction of principal normal is called the center of curvature. Thecircle with center at center of curvature and radius ρ is called the osculating circle(also called the circle of curvature.) Thus the position vector of the center of curvatureis r + N κ = r + ρN. March 17, 2001

Binormal 10for some scalar function τ which is known as the torsion of the curve. We now showhow this function may be computed as function of t :Differentiating both sides ofwith respect to t we obtainSinceand sinceit follows thatd 2 ( ) 2r dsdt = a = κ N + d2 s2 dt dt T 2⎡d 3 rdt = d ( ) ⎤ 2⎣ d2 s ds3 dt dt T + κN ⎦2 dt( )= d3 sdt T + d2 s dT ds d 2 ( ) 2 ( ) 2 3 dt 2 dt + 2 s dsdt dt κN + dκ ds2 dt dt N + κ dN dt dt .dTdt = κds dt NdNdt = dN dsds dt = dsdt (dB ds × T + B × dTds )= ds (−τN × T + B × (κN))dt= ds (τB − κT ),dtd 3 [r d 3 ] [dt = s3 dt − (ds 3 dt )3 κ 2 T + 3( ds sdt )(d2 )κ + (dsdt2 dt )2 ( dκdt ) ]N + ( dsdt )3 κτB.Also we haveThereforeConsequently[drdt × d2 rd 2 ]dt = (ds 2 dt T ) × sdt T + (ds2 dt )2 κNτ == ( dsdt )3 κT × N = ( dsdt )3 κB.d 3 [ ]r drdt · 3 dt × d2 r= ( dsdt 2 dt )6 κ 2 τ.dr · d2 r× d3 rdt dt 2 dt[ 3( dsdt) 3κ] 2=dr · d2 rdt dt 2× d3 rdt 3∥ ∥∥ drdt × d2 rdt 2 ∥ ∥∥ 2 . March 17, 2001

Formula of Frenet 11Formula of FrenetDifferentiating N · N = 1 with respect to s we haveTherefore dNdsdNds · N = 0.can be expressed asdNds = α 1T + α 2 Bfor some scalars α 1 , α 2 . To determine these scalars we haveα 1 = α 1 T · T = (α 1 T + α 2 B) · T = dNds · T = −dT ds · N= −κN · N = −κandThereforeSummery:α 2 = α 2 B · B = (α 1 T + α 2 B) · B = dNds · B = −dB ds · NdT= dsdNdsdNds= τN · N = τ.= −κT + τB.κN= −κT +τBdBds = −τB March 17, 2001