13.5 Torsion of a curve Tangential and Normal Components of Acceleration Recall: b Length of a curve |r '(t ) | dt a t ds Arc length function s( t ) r u du |r '(t ) | a dt Arc length parametrization rr(ss ) with | '( ) | 1 r '(t ) Unit tangent vector Tr'(s) |r '(t ) | dT Tt rrtt Curvature: r s 3 ds rt rt t Arc length function s( t ) r u du s measures distance traveled starting at t a ds a |r '(t ) | measures speed of motion rrs t s "arc length parametrization" dt dr dr drr d dt r '(t ) if s is arc length parameter, then = = dt hence |r' (s ) |dt 1 "you travel with speed 1" ds dt ds ds |rr '(tt ) | | '( ) | dt If ss is arc length parameter, then |r '( ) | 1 Assume that tt is a parameter with |r '( ) |1 : tt If your basepoint is t0, then s( t ) r u du 1 du t 00 So s t, which means t is already the arclength parameter. tt If I assume your basepoint is ta , then s( t ) r u du 1 du t a aa ta is still an arc length parameter, it just measures distance starting at in either case, distance traveled from ss to is simply Examples: a) arc length parametrization of a straight line: r(s) r0 s v with | v |= 1 ss b) arc length parametrization of a circle x2+y 2 =r 2 : r(s) r cos( ), r sin( ) 02sr rr Q curvature at PQ> curvature at P r '(t ) Unit tangent vector T r '(s ) |r '(t ) | dT Ts r s ds curvature measure how quickly we turn if we travel at speed 1 Frenet Frame: r '(t ) dT dT T is also called the curvature vector |r '(t ) | ds ds dT dT dt dTT d dt Principal unit normal : N ds since dT dT ds dt ds (N is only defined when 0 !) dt ds dt and 0 is a scalar ds since TTTTTN 1, we have ' 0 or 0 N is orthogonal to T a third vector is the binormal B T N BTN is orthogonal to and and of unit length: | B | |T | | N | sin( ) 1 2 Altogether, we have Frenet frame (or TNB frame) T,N,B They are all of unit length and orthogonal to each other (like i, j,k) they form a moving frame: http://en.wikipedia.org/wiki/Frenet_frame Torsion: dT dT ds 1 dT dT N dT or N ds ds ds ds B T N dB Claim : is parallel to N : ds dB B B 1 2 B 0 ds ddBTdB dB B T 0 0 = T B T B N = T ds ds ds ds dBBB d d Since B 0 and T 0 we see is a multiple of N ds ds ds This multiple (up to sign) is called torsion: dB dB N or N ds ds B is the normal vector to the plane spanned by T and N dB ddBB measure the "tilt'' of this plane since N we also have ds ds ds (up to sign) measures the magnitude of the tilt Example: a circle of radius r: r( t ) r cos( t ), r sin( t ),0 ss arc length parametrization: r(s ) r cos( ), r sin( ),0 rr ss dT 11 s s Tr'(s ) sin( ),cos( ),0 cos( ), sin( ),0 rr ds r r r r dT dT 1 ss N ds cos( ), sin( ),0 ds r dT rr ds i j k ss 22ss 퐵 = 푇 × 푁 sin( ) cos( ) 0 sin ( ) cos ( ) k = k rr rr ss cos( ) sin( ) 0 rr dB N 0 ds for every plane curve B T N k and torsion 0 ! Example: Compute T,N,B of the circular helix: r(t ) a cos( t ), a sin(t), bt asin( t ), a cos( t ), b r '(t ) a sin( t ), a cos( t ), b hence T ab22 dT 1 a dT acos( t ), a sin( t ),0 a2cos 2 ( t ) a 2 sin 2 ( t ) 22 22 dt ab22 dt ab ab dT Tt a 1 a curvature ds rt a2 b 2 a 2 b 2 ab22 dT principle unit normal N dt dT cos(tt ), sin( ),0 dt i j k 1 1 binormal B T N asin( t ) a cos( t ) b bsin( t )i b cos(t) j a k ab22 ab22 cos(tt ) sin( ) 0 What is the torsion of the circular helix? circular helix: r(t ) a cos( t ), a sin(t), bt asin( t ), a cos( t ), b T N cos(tt ), sin( ),0 ab22 1 a B bsin( t ), b cos( t ), a 22 ab22 ab dB N but ts is not arc length parameter ! ds we need a formula for the torsion in a general parameter t a computation shows where r(t ) x ( t ), y (t), z ( t ) that for the helix we have: and v r', a r '' b rt r t r t ab22 2 rrtt http://en.wikipedia.org/wiki/Frenet_frame Decompose the acceleration vector a r ''(t ) use v r' and a r '' aaaTN T N ds v r T T ds r '(t ) dt Recall: r T dt |r '(t ) | 2 d s ds T T v T T Nt dt2 dt T r d2 s ds ds aT hence T T N r N N dt2 dt dt d2 s ds ds a2 T N dt dt dt 2 2 2 ds ds d2 s ds a aT 2 N a2 T N dt dt dt dt d 2 aT r a r dt N aaaTN T N d2 s d tangential acceleration: a = (r ) T dt2 dt 2 ds 2 normal acceleration: aN r dt ds2 if a car travels along a curve, it feels an internal acceleration of dt 2 2 and a force of magnitude maN m r (centrifugal force) large curvature (tight curve) and large speed2 = problems ! if you travel at unit speed, then amT 0, and force other formulas: a v r' r '' v a r' r '' a aT a aN (try to show this....) T vr' N vr' 22 22 also useful: aa aaTN aaNT |a| Example: A car travels along a track of radius ra with velocity d 2 1 a (r ) 0 aa r 2 T dt N r 13.6 Acceleration in Polar Coordinates Newton’s law of gravitation (1687): r is the vector from the center of the sun to the planet GmM r M is the mass of the sun m is the mass of the planet F 2 | r | | r | G is the gravitational constant G = 6.674 1011 N m 2 kg 2 (from 1798) Inverse square law GM r F =m a a r'' = | r |2 | r | d rr' r' r ' r r '' rr'' 0 dt since r'' is parallel to r by Newton's law hence r r' is a constant vector C in particular rC0 the planet moves in a plane orthogonal to C !.