DIFFERENTIAL GEOMETRY OF CURVES 81

The unit bi-normal vector Bmay be obtained from Eq. (3.30b) as

B ̇r ̇ ̇r

̇r ̇ ̇r

=

| |

×

×

̇ ̇rijk = – cos – sin + 0atat

r ̇ ̇ ̇r

ijk

= – sin cos ijk

• cos – sin 0

× = sin – cos +^2

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

at a t b

at at

ab t ab t a

| | = ( ̇r× ̇ ̇r ab sin ) + (–t^222222 ab cos ) + (t a ) = a a + b

Therefore, B
ijk

sin – cos +

(^22) +
btbta
ab
The normal vector N is given by
N = B×T
Therefore
N
ijk
=^1 ijk

• sin – cos

• sin cos

22 = – (cos + sin + 0 )

ab

bt b ta

at a t b

tt

The curvature is given by Eq. 3.30(c).

κ = | |

| |

=

+

( + )

=

( + )

3.

22

223/2 22

r ̇ ̇ ̇r

̇r

× aa b

ab

a

ab

Therefore, the radius of curvature ρ =
( + ab^22 )
a
.
From Eq. (3.33), the torsion τ is given as

τ =

d

ds

B

d

ds

d

dt

ds

dt

d

dt

d

dt

BB B r

= ⎛ =

⎝

⎞

⎠

d

dt

btbt

ab

Bij

=

cos + sin

(^22) +
d
dt
ab
r
=^22 +
⇒ τ =
cos + sin
(+ )

(^22) (+ ) 22
btbt
ab
b
ab
ij