Mathematical Physics Vol 1

3.2 Vector analysis

75

Given that n · b = 0 it follows that d n d s · b + n · d b d s

= 0 ⇒

d n d s ·

d b d s

b = − n ·

= − n ( − k 2 n )= k 2 .

By derivation of n , we obtain

d d s ⇒

n = b × t

d n d s

d b d s ×

d t d s

t + b ×

=

=

1 ρ

1 ρ

− k 2 n × t + b ×

n = k 2 b +

( − t ) .

(3.11)

Relations (3.7), (3.10) and (3.11) represent the so called Frenet-Serret formulas d t d s = 1 ρ n ( k 1 n ) , d n d s = − k 1 t + k 2 b = − 1 ρ t + 1 T b , d b d s = − k 2 n = − 1 T n . (3.12)

Problem32 Prove that the curvature radius of a curve defined by parametric equations x = x ( s ) , y = y ( s ) , z = z ( s ) , is defined by ρ = " d 2 x d s 2 2 + d 2 y d s 2 2 + d 2 z d s 2 2 # − 1 2 . (3.13)

Proof Given that the position vector of an arbitrary point on the curve is r = x ( s ) i + y ( s ) j + z ( s ) k ,

it follows that

d r d s

d x ( s ) d s

d y ( s ) d s

d z ( s ) d s

t =

i +

j +

k

=

and

d 2 x d s 2

d 2 y d s 2

d 2 z d s 2

d t d s

i +

j +

k .

=