DIFFERENTIAL GEOMETRY OF SURFACES 179

Therefore, κn

d

ds

d

ds

dd

dd

= – = – with ds^2 d d

rn rn

rr

⋅ rr

⋅

⋅

≈⋅ (6.28)

We can simplify Eq. (6.28) by decomposing dr and dn along parametric lengths du and dv, that is

dr =rudu + rvdv, dn = nudu + nvdv

⇒ dr·dn =ru·nu(du)^2 + (ru·nv + rv·nu)dudv + rv·nv(dv)^2

dr·dr =ru·ru(du)^2 + (ru·rv + rv·ru)dudv + rv·rv(dv)^2

=G 11 du^2 + 2G 12 dudv + G 22 dv^2 (6.29)

Sinceru and rv are both perpendicular to n, using Eq. (6.18), we get

ru·n= 0 ⇒ ruu·n + ru·nu = 0 ⇒ru·nu = –ruu·n = –L (6.30a)

rv·n= 0 ⇒rvv·n + rv·nv = 0 ⇒rv·nv = –rvv·n = –N

ru·n= 0 ⇒ruv·n + ru·nv = 0 ⇒ru·nv = –ruv·n = –M (6.30b)

rv·n= 0 ⇒rvu·n + rv·nu = 0 ⇒rv·nu = –rvu·n = –M

Using Eqs. (6.28), (6.29) with (6.30), the expression for the normal curvature

κ

μμ

μμ

n

Ldu Mdud N d

Gdu G dud G d

LMN

GGG

=

+ 2 +

+ 2 +

=

+ 2 +

+ 2 +

22

11

2

12 22

2

2

11 12 22

2

vv

vv

(6.31)

whereμ =.
d
du

v

Equation (6.31) can be rewritten as

(G 11 + 2G 12 μ + G 22 μ^2 )κn = L + 2Mμ + Nμ^2 (6.32)

For an optimum value of the normal curvature
d
d

κn

μ

= 0. Differentiating Eq. (6.32) yields

(GGG 11 + 2 12 + 22 2 ) + 2( 12 + 22 ) = 2( + )

d

d

μμn GG n MN

κ

μ

μκ μ (6.33a)

⇒ (G 12 + G 22 μ)κn = (M + Nμ) (6.33b)

Equating Eqs. (6.31) and (6.33(b)), we get

κ

μ

μ

μμ

μμ

μμ μ

n μμ μ

MN

GG

LMN

GG G

LM MN

GG GG

=

+

+

=

+ 2 +

+ 2 +

=

( + ) + ( + )

12 22 ( + ) + ( + )

2

11 12 22

(^211121222)
which can be simplified as
κ
μ
μ
μ
n μ
MN
GG
LM
GG

• 
  


• 
  

  =

  
  


• 
  

  12 22 11 + 12
  ⇒ (M–G 12 κn) + (N–G 22 κn)μ = 0