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Unit 22: Bertrand Curves

for all s  L. By theorem A, this fact is a contradiction. This we must consider only the case of sin Notes
T  0. Then (6.2) implies
0
k (s) – k (s)  0 (s  L),
3
2
that is, we obtain the relation (a).
The fact sin T  0 and (7) simply
0
k (s) + {(cos T ) (sin T ) } (k (s) – k (s)) = 1.
–1
0
2
3
1
0
From this, we obtain
k (s) + (k (s) – k (s) = 1
2
3
1
for all s  L, where  = (cos T ) (sin T ) is a constant number. Thus we obtain the relation (b).
–1
0
0
Differentiating (5)’ with respect to s and using the Frenet equations, we obtain
 '(s)k ( (s)) . n ( (s)) (k (s)cosT  k (s)sin T ).n (s) k (s)sin T .n (s)




0
1
0
1
1
3
1
0
2
3
for all s  L. From the above equality, (6.1), (6.2) and (b), we obtain
{ '(s)k ( (s))} 2


1
2
= {k (s)cosT  k (s)sinT }  {k (s)sin T } 2
0
3
0
0
2
1
= (k (s) – k (s)) [(k (s) – k (s)) + (k (s)) ] (’(s)) .
2
2
2
–2
3
2
2
1
3
for all s  L. From (4) and (b), it holds
(’(s)) = ( + 1) (k (s) – k (s)) . 2
2
2
2
3
Thus we obtain
2
2
{ '(s)k ( (s))} = 1 {( k (s) k (s) k (s)) } (8)





1
2
  1 1 2 3
By (6.1), (6.2) and (b), we can set

n ( (s)) = (cos (s)).n (s) (sin (s).n (s), 1   3 (9)
1
where


( k (s)   k (s))( k (s) k (s))

cos (s) = 2 3 1 2 (10.1)
k ( (s))( '(s)) 2


1

3
sin (s) = ( k (s)  k (s))k (s) (10.2)
3
2
k ( (s))( '(s)) 2


1
for all s  L. Here, h is a C -function on L.

Differentiating (9) with respect to s and using the Frenet equations, we obtain
 j'(s)k ( (s)).t(j(s))  j'(s)k (j(s)).n ( (s))


2
1
2
dcos (s) dsin (s)


= .n (s)  .n (s) k (s)(cos (s)) . t(s) (k (s)(cos (s)) k (s)(sin (s)).n (s) 1   2   3  2
ds ds
3
1
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