DIFFERENTIAL GEOMETRY OF SURFACES 173The two curves are orthogonal to each other if
∂
∂∂
∂⎛
⎝
⎜⎞
⎠
⎟⋅∂
∂∂
∂⎛
⎝
⎜⎞
⎠
⎟rr rr(, )
+( , ) ( , )
+(, )
= 0
11 22u
udu
dtud
dtu
udu
dtud
dtvv
vvvv
vvor G
du
dtdu
dt
G
du
dtd
dtdu
dtd
dt
G
d
dtd(^11121212212212) dt
= 0
⎛ vvvv
⎝
⎞
⎠ (6.13)
If the two curves coincide, respectively, with u and viso-parametric curves of the surface, then we
may regard u≡t 1 and v≡t 2 for which the dot product in Eq. (6.12) becomes
tt
rr
rr
rr
rr
rr rr
12
11 22
12
11 22
= cos =
( , ) ( , )
= =
| || |
⋅
∂
∂
⋅
∂
∂ ⋅⋅
⋅⋅
θ
u
u
G
u
G
G
GG
u
u
u
uu
vv
v v
v
v
vv
(6.14)
Thus, the iso-parametric curves are orthogonal if G 12 = 0. To compute the area of a surface patch
r (u,v), let a small patch on the surface be formed by the curves between u = u 0 ,u = u 0 + du,v = v 0
andv = v 0 + dv. The four corners of this patch are r(u 0 ,v 0 ),r(u 0 + du,v 0 ),r(u 0 ,v 0 + dv) and
r(u 0 + du,v 0 + dv) as shown in Fig. 6.8. The infinitesimal area dA is approximated by
d A = | rruudu ××vvdvvvv | = | | rrdu d = G G 11 22 – G 122 dud = | |Gdu d
Therefore, the area of the patch is given by
Adud = | |
∫Domain
G v (6.15)
r(u 0 ,v 0 + dv)
r(u 0 ,v 0 )
ru dv
dA = | ru×rv | dudv
r(u 0 + du,v 0 )
ru dv
Figure 6.8 Infinitesimal area on the surface
6.3 Deviation of the Surface from the Tangent Plane: Second Fundamental Matrix
In Figure 6.9, let R(r(u 0 + du,v 0 + dv)) be a point on the surface a small distance away from
P(r(u 0 ,v 0 )). The deviation d of R from P along the normal n at P may be written as
d = [ r(u 0 + du,v 0 + dv)–r(u 0 ,v 0 )] ·n