Page 270 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 22: Bertrand Curves

Differentiating (3) and using the Frenet equations, we obtain Notes

dcos (s) dsinq(s)






k ( (s)) '(s).n ( (s))  ds . t(s) (k (s)cos (s) k (s)sin (s).n (s)  ds .n (s) k (s)sin (s).n (s).




1
1
2
1
3
3
2
1
Since n ( (s)) = ±n (s) for all s  L, we obtain

1
1
k (s) sin q(s)  0. (5)
3
By k (s)  0 (  s  L) and (5), we obtain that sin (s)  0. Thus, by k2(s) > 0(  s  L) and (4.2), we
3
obtain that  = 0. Therefore, (1)’ implies that C coincides with C. This is a contradiction. This
completes the proof of Theorem A.
22.3 (1, 3)-Bertrand Curves in E 4
By the results in the previous section, the notion of Bertrand curve stands only on E and E . Thus,
2
3
we will try to get the notion of generalization of Bertrand curve in E (n  4).
n
Let C and C be C -special Frenet curves in E and  : L  L a regular C -map such that each
4



point c(s) of C corresponds to the point c(s)  c( (s)) of C for s  L. Here s and s arc-length
parameters of C and C respectively. If the Frenet (1, 3)-normal plane at each point c(s) of C
coincides with the Frenet (1, 3)-normal plane at each point c(s) of C coincides with the Frenet
(1, 3)-normal plane at corresponding point c(s)  c( (s)) of C for all s  L, then C is called a

(1, 3)-Bertrand curve in E and C is called a (1, 3)-Bertrand mate of C. We obtain a characterization
4
of (1, 3)-Bertrand curve, that is, we obtain Theorem B.
Proof of Theorem B. (i) We assume that C is a (1, 3)-Bertrand curve parametrized by arc-length
s. The (1, 3)-Bertrand mate C is given by
c(s)  c( (s)) = c(s) + (s) . n (s) + (s) . n (s) (1)

1
3
for all s  L. Here  and  are C -functions on L, and s is the arc-length parameter of C .

Differentiating (1) with respect to s, and using the Frenet equations, we obtain
’(s) . t( (s)) = (1 – (s)k (s)) . t(s) + ’(s) . n (s) + ((s)k (s) – (s)k (s)) . n (s) + ’(s) . n (s)

1
2
3
2
1
3
for all s  L.
Since the plane spanned by n (s) and n (s) coincides with the plane spanned by n ( (s)) and
1
1
3
n ( (s)), we can put

3
n ( (s)) = (cos (s)) . n (s) + (sin (s)) . n (s) (2.1)

1
1
3
n ( (s)) = (– sin (s)) . n (s) + (cos (s)) . n (s) (2.2)

3
1
3
and we notice that sin (s)  0 for all s  L. By the following facts
0 =  '(s) . t( (s)),n ( (s))   's .(cos (s)  '(s) .(sin (s))




1

0   '(s) . t( (s)),n ( (s))   's .(sin (s)   '(s) .(cos (s)),



3
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