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 Tt rrtt   Curvature:   r s   3 ds rt rt 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 rrs   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 t 0, 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( )  02sr rr Q

curvature at PQ> curvature at P

r '(t ) Unit tangent vector T   r '(s ) |r '(t ) |

dT  Ts   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 Tt a 1 a    curvature   ds rt 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   rt r  t r  t ab22  2 rrtt      http://en.wikipedia.org/wiki/Frenet_frame Decompose the acceleration vector a r ''(t )

use v  r ' and a r '' aaaTN 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 Nt    dt2 dt T r

d2 s ds ds aT  hence T T  N  r  N  N dt2 dt dt d2 s ds ds a2 T N dt dt dt 2 2 2 ds ds d2 s ds a   aT  2 N  a2 T  N dt dt dt dt d 2 aT   r  a   r dt N aaaTN 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 1011 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 rC 0

 the planet moves in a plane orthogonal to C !