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Unit 22: Bertrand Curves

and by (18.1)’, (18.2)’ and (20), Notes

dn (s) =  k (s)( k (s) k (s)) k (s)( k (s) k (s) (k (s)) 2 . n (s),





1
2
1
2
1
3
2
1
ds s  (s)  '(s) ( k (s) k (s))  (k (s)) 2 . t(s)   '(s) ( k (s) k (s))  (k (s)) 2 2
2
2




3
3
2
1
2
1
for all s  L. By the above equalities, we obtain
dn (s)  k ( (s)) . t( (s))  P(s) . t(s)  Q(s) . n (s),
1


ds s  (s) 1 R(s) R(s) 2
where
P(s) = –[{(k (s)) – (k (s)) – (k (s)) } + ( – 1)k (s)k (s)]
2
2
2
2
1
3
2
2
1
Q(s) = [{(k (s)) – (k (s)) – (k (s)) } + ( – 1)k (s)k (s)]
2
2
2
2
3
1
2
1
2
2
2
2

R(s) = ( + 1) ’(s) ( k (s) k (s))  (k (s))  0

2
3
1
for all s  L. We notice that, by (c), P(s)  0 for all s  L. Thus we obtain
k ( (s))

2
dn (s)
= 1  k ( (s) . t( (s))


ds s  (s) 1
2
2
2

 {(k1(s))  (k (s))  (k (s))  (g2 1)k (s)k (s)
= 2 3 1 2 > 0
2
2

 '(s)   1  k (s) k (s))  (k (s)) 2
3
1
2
for all s  L. Thus, we can define a unit vector field n (s) along C by
2
n (s) = n ( (s))
2
2
1  dn (s) 
= .  1  k ( (s)) . t( (s)) , 


1
k ( (s))   ds s  (s)  

2
that is,
1
n ( (s)) = .( t(s)   . n (s)) (21)


2
2
2
   1
for all s  L. Next we can define a unit vector field n along C by
3
n (s) = n (j(s))
3
3
1
=
2
2
 ( k (s) k (s))  (k (s)) . { k (s) . n (s) ( k (s) k (s)) . n (s)}, 3 1   1  2 3


3
2
1
that is,
n ( (s)) = –(sin  ) . n (s) + (cos  ) . n (s) (22)

3
3
0
1
0
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