Lanko diferencijalas . Jeigu paviršius užrašytas formoje (3) arba (3a),
M
(
u
,
v
)
{\displaystyle M(u,v)}
N
(
u
+
d
u
,
v
+
d
v
)
{\displaystyle N(u+du,v+dv)}
MN ant paviršiaus apytiksliai išreiškiamas diferencialu lanko arba linijiniu elementu paviršiaus pagal formulę
d
s
2
=
E
d
u
2
+
2
F
d
u
d
v
+
G
d
v
2
,
(
1
)
{\displaystyle ds^{2}=E\;du^{2}+2F\;du\;dv+G\;dv^{2},\quad (1)}
kur
E
=
r
1
2
=
(
∂
x
∂
u
)
2
+
(
∂
y
∂
u
)
2
+
(
∂
z
∂
u
)
2
,
{\displaystyle E=\mathbf {r} _{1}^{2}=({\frac {\partial x}{\partial u}})^{2}+({\frac {\partial y}{\partial u}})^{2}+({\frac {\partial z}{\partial u}})^{2},}
F
=
r
1
r
2
=
∂
x
∂
u
∂
x
∂
v
+
∂
y
∂
u
∂
y
∂
v
+
∂
z
∂
u
∂
z
∂
v
,
{\displaystyle F=\mathbf {r} _{1}\mathbf {r} _{2}={\frac {\partial x}{\partial u}}{\frac {\partial x}{\partial v}}+{\frac {\partial y}{\partial u}}{\frac {\partial y}{\partial v}}+{\frac {\partial z}{\partial u}}{\frac {\partial z}{\partial v}},}
G
=
r
2
2
=
(
∂
x
∂
v
)
2
+
(
∂
y
∂
v
)
2
+
(
∂
z
∂
v
)
2
.
{\displaystyle G=\mathbf {r} _{2}^{2}=({\frac {\partial x}{\partial v}})^{2}+({\frac {\partial y}{\partial v}})^{2}+({\frac {\partial z}{\partial v}})^{2}.}
Dešinė dalis formulės (1) vadinasi taip pat pirma kvadratine forma paviršiaus, užrašytos formoje (2); jos koeficientai E, F, G priklauso nuo taško paviršiaus.
Pavyzdis : Sferai:
r
=
a
(
cos
(
u
)
sin
(
v
)
i
+
sin
(
u
)
sin
(
v
)
j
+
cos
(
v
)
k
)
,
{\displaystyle \mathbf {r} =a(\cos(u)\sin(v)\mathbf {i} +\sin(u)\sin(v)\mathbf {j} +\cos(v)\mathbf {k} ),}
E
=
a
2
sin
2
v
,
F
=
0
,
G
=
a
2
;
{\displaystyle E=a^{2}\sin ^{2}v,\quad F=0,\quad G=a^{2};}
pirma kvadratinė forma:
d
s
2
=
a
2
(
sin
2
(
v
)
d
u
2
+
d
v
2
)
.
{\displaystyle ds^{2}=a^{2}(\sin ^{2}(v)du^{2}+dv^{2}).}
Paviršiui, užrašytam formoje (2):
E
=
1
+
p
2
,
F
=
p
q
,
G
=
1
+
q
2
,
{\displaystyle E=1+p^{2},\quad F=pq,\quad G=1+q^{2},}
p
=
∂
z
∂
x
,
q
=
∂
z
∂
y
.
{\displaystyle p={\frac {\partial z}{\partial x}},\quad q={\frac {\partial z}{\partial y}}.}
Tada
d
s
2
=
E
d
u
2
+
2
F
d
u
d
v
+
G
d
v
2
=
{\displaystyle ds^{2}=E\;du^{2}+2F\;du\;dv+G\;dv^{2}=}
=
E
d
x
2
+
2
F
d
x
d
y
+
G
d
y
2
=
{\displaystyle =E\;dx^{2}+2F\;dx\;dy+G\;dy^{2}=}
=
(
1
+
p
2
)
d
x
2
+
2
p
q
d
x
d
y
+
(
1
+
q
2
)
d
y
2
=
{\displaystyle =(1+p^{2})\;dx^{2}+2pq\;dx\;dy+(1+q^{2})\;dy^{2}=}
=
(
1
+
(
∂
z
∂
x
)
2
)
d
x
2
+
2
∂
z
∂
x
∂
z
∂
y
d
x
d
y
+
(
1
+
(
∂
z
∂
y
)
2
)
d
y
2
.
{\displaystyle =(1+({\frac {\partial z}{\partial x}})^{2})\;dx^{2}+2{\frac {\partial z}{\partial x}}{\frac {\partial z}{\partial y}}\;dx\;dy+(1+({\frac {\partial z}{\partial y}})^{2})\;dy^{2}.}
Nežinau ar taip tada išeina, kad
d
s
2
=
E
d
u
2
+
2
F
d
u
d
v
+
G
d
v
2
=
(
1
+
(
∂
z
∂
x
)
2
)
d
x
2
+
2
∂
z
∂
x
∂
z
∂
y
d
x
d
y
+
(
1
+
(
∂
z
∂
y
)
2
)
d
y
2
.
{\displaystyle ds^{2}=E\;du^{2}+2F\;du\;dv+G\;dv^{2}=(1+({\frac {\partial z}{\partial x}})^{2})\;dx^{2}+2{\frac {\partial z}{\partial x}}{\frac {\partial z}{\partial y}}\;dx\;dy+(1+({\frac {\partial z}{\partial y}})^{2})\;dy^{2}.}
