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