Page 274 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 22: Bertrand Curves

for all s  L. Thus, by the relation (b), we obtain Notes

dc(s) k (s)) . ( . t(s) n (s))
ds = ( k (s) 2   3   2 (12)
for all s  L. Since the relation (a) holds, the curve C is a regular curve. Then there exists a
regular map  : L  L defined by

s dc(t)


s   (s)   dt ( s L),
0 dt
where s denotes the arc-length parameter of C, and we obtain

2


’(s) =    1 ( k (s)  k (s)) 0, (13)
3
2
where  = 1 if k (s) – k (s) > 0, and  = –1 if k (s) – k (s) < 0, for all s  L. Thus the curve C is
3
2
2
3
rewritten as
c(s) = c( (s))

= c(s) +  . n (s) +  . n (s)
3
1
for all s  L. Differentiating the above equality with respect to s, we obtain
dc(s)
 '(s) . = (k (s) – k (s)) . { . t(s) + n (s)}. (14)
ds s  (s) 2 3 2
We can define a unit vector field t along C by t(s) = dc(s)/ds for all s  L. By (13) and (14),
we obtain
t( (s)) = ( + 1) –1/2 . { . t(s) + n (s)} (15)
2

2
for all s  L. Differentiating (15) with respect to s and using the Frenet equations, we obtain
dt(s)
2



 '(s) .  (    1)  1/2 . {( k (s) k (s)) . n (s) k (s) . n (s)}
2
1
3
3
ds
1
s  (s)
and
2
dt(s)  ( k (s) k (s))  (k (s)) 2 .


2
3
1
ds s  (s)  '(s)   1
2
By the fact that k (s) > 0 for all s  L, we obtain
3
dt(s)
k ( (s)) =  0 (16)

1
ds
s  (s)
for all s  L. Then we can define a unit vector field n along C by
1
n (s) = n ( (s))

1
1
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