x0

?0?0,D(f1,···,f(?0?D(?1,?m?1(?0?非奇異Bλ(?0RBμ(?0Rm?1

D(f1,···,f(?0?

??(?)Bμ(0), ?,?(?))X3,···,

?(?)

?(?0Bλ(?0(?0[Bλ(?0(?0

(?01 11:(?0.(?0Bμ(?

?(?)

?(?0

((?)(0)?(?)

?(?)

?可確定曲線Γ?) ?)??)(?)(?) x?→0 ?

(?,?())

D?(?) RmΣ={x∈mf(x)=0∈R}基于隱映照定理針對上述約束有以下結論x0

x0?0Df(x?,?)?f

,?)

(?? ?(?0,(?0 ?=??)Bμ(?0 ??(?))2:隱映照定理論的幾何刻畫,如圖2所示.Bλ?0Bμ(?0Rm中,Σ

?(?)

?(?

?R.現(xiàn)為 中的曲 Σ(?):Rm?1?

((?(?

)?(?)

?DΣ(?)=:=D?(x?)1

·· (?∈

TxΣ

g1,···,

(?)?相應地,n(x)Rm, (D?)T(?)

n(x)=0∈1(R3中曲線（隱式表示形式）).R3{ f(x,y,z)=0∈Γ ∈Rg(x,y,z)=0∈z{ f(x,y,z)= ?(f,

g(x0,y0,z0)=

?(xz(x,y,z?0(F

[x])

f(x,y,

∈ g(x,y,(F

f(x0,y0, =0∈g(x0,y0,

(x,y,z)R2×2

Bλ(y0R,μ

R2?yB(y),ξ(y

B (

[[=[

f(x(y),y,

=0∈ g(x(y),y,亦即 Γ，表所以曲線示可以用向

Γ(y):Bλ(y0)?y7→Γ(y)

∈Γ

ddΓ(y)=DΓ(y) ∈ λ y, =0∈R2,?y∈Bλ

+D[x

])

)=0∈)

z

z

(x(y),y,z(y))

(x(y),y,

x = =

(x(y),y,

(x(y),y, ?(f,= ?(y,

(x(y),y,?(f,g)(x(y),y,?(x,

?(f,?(x,?(f,—?(y,z)(x(y),y,?(f, ?(x,

?(f,?(y,?(f,dy(y)

?(z, (x(y),y, ?y∈?(f, ?(f,—?x,—?x, (x(y),y,

?(x,?(f,?(x,

?(f,?(y, + ?(f, (x(y),y,z(y))∈?(z,

?(f,?(x,?(f,

x?

=?(f,

y?

=?(f,

z??(y,

(x(y0),y0,

(x(y0),y0, ?z,

(x(y0),y0, ?x, 2(R4中曲線（隱式表示形式）).R4f(x,y,z,θ)=0∈x∈Γ g(x,y,z,θ)=0∈h(x,y,z,θ)=0∈θ f(x,y,z,θ)=

)=

?(f,g,h)(x,y,z,θ?0 h(x0,y0,z0,θ0)=

?(x,y,

f(x,y,z, g(x,y,z,

∈ h(x,y,z, f(x0,y0,z0,

g(x0,y0,z0, =0∈h(x0,y0,z0, D[D[x z0 yθ0

(xy,z,θR3×3 ?Bλ(z0)?R,??z∈Bλ(z0),?ξ(z)∈

g(x(z),y(z),z,h(x(z),y(z),z,

=0∈

Γ(z):B(z)?z7→Γ(z) ∈ Γ

(z)=DΓ(z)

∈

=0∈ ?z∈ +D[x =0∈

yθ

(x(z),y(z),z,θ(z))

(x(z),y(z),z, =

=

(x(z),y(z),z,

(x(z),y(z),z,

?(f,g,?(z,y,=? ?(f,g,?(x,y,θ)(x(z),y(z),z,

?(f,g, (x(z),y(z),z,?(x,z,?(f,g,?(x,y,?(f,g,?(z,y,—?(f,g,h)(x(z),y(z),z,?(x,y,?(f,g,

?(f,g,—?(z,y, ?(x,z,

—?(f,g,?Γ(z)?

—(f,g,h)(x(z),y(z),z,

?(x,z, (x(z),y(z),z, ?(x,y,1

?(f,g,?(x,y,?(f,g,?(f,g,?(f,g,?(x,y,

x(z),y(z),z,

?(x,y,?(f,g,?(y,z,?(f,g,=

(x(z),y(z),z, ?z∈?(f,g,?(x,y,?(f,g,—?(x,y,3(Rp中曲線（隱式表示形式）).Rp Γ X∈Rp|f(X)=0∈f ) ∈設有X0∈Γ亦即f(X0)

(X0)

.

?(X1,···,Xα,···,

=f(X)∈X.XX1X0

. =f0

)=0∈0.p0X1X0.. D(f1,···,fX0D0.?

X 000.

D(X1,···,Xα,···,

(X∈R(p?1)×(p?1)X.

pX1X0.

.? ，對?X∈B(X),?ξ(X) p

(Xα)?R,?

(Xα) p1?.?0 0..XpX0

F(Xα,ξ(Xα))=0∈

..

.0Γ(Xα):Bλ(Xα)?Xα7→Γ(Xα) ∈0.Γ

dX1(X.α(Xα)=DΓ(Xα) α.

∈.?0F(Xα,ξ(Xα))= =0∈ ?Xα∈Bλ0.1X X DXα + X.

. (

(Xα)=0∈ . ?f ?f ?f

?f ?f ·· ·· ?.α

(Xα)=

.?f

.. ?f

.?f

.. ?f

?f.

··

··

?(?1)α?2?(f1,···,??(X1,···,. ?(f1,···, ?(X1,···,Xα?1,···,?0= ?(f1,···, ?(f1,···,?0 ?(X1,···,Xα,···, ?(X1,···,Xα+1,···,.?(?1)p?α?1?(f1,···,??(X1,···,

?(?1)α?2?(f1,···,??(X1,···,.?(f1,···,

(?1) ?(X1,···,Xα?1,···, αdΓ(Xα) α

?(f1,···,α α ?(X1,···,Xα,···, ?(f1,···,??(X1,···,Xα+1,···,.(?1)p?α+1?(f1,···,

?(X1,··?(?1)0?(f1,···,??(X1,···,

,

?(f1,···, ? ?(X1,X2,···,.

∈.曲

p?1?(f1,···, ??(X1,···, 4(R32維曲面).R32 Σ f(x,y,z)=03z

Σf(x0y0z00

?f(x0,y0,z0)?=0([x ,z

=f(x,y,z)∈ x0,y0

=f(x0,y0,z0)=y

,

=?f(x0,y0,z0)?= x

[x

x則按隱映照定理，?

?R2,?B(y)?R，對 ∈

,?!ξ(x,z)Bμ(y0)?R

([xFz

,ξ(x,

=f(x,ξ(x,z),z)=([x ([x [x Σz

:

7→

ξ(x, ∈z

[?Σ?Σ

∈

?x,?z(x,z)= ?x(x, ?z(x,

[

x × (x,z)= (x, ∈

?ξ(x,

?ξ(x,再確定隱函數(shù)的偏導數(shù)?ξ(x,z)和?ξ(x,z) F

([xz

),ξ(x,z)= ([ ([ D[x x,ξ(x,z)+D x,ξ(x,z)D[x]ξ(x,z)=0∈y ?f,

(x,ξ(x,z),z)+?f(x,ξ(x,z),

?ξ,

(x,z)=0∈?x ?x ?x(x,z)=?(x,ξ(x,z), (x,z)=??z(x,ξ(x,z), n

×

(x,z)= (x,ξ(x,z),z)

(x,ξ(x,z),

ξ(x0,

(x,ξ(x,z), =

[xz

∈B

(x?x0)?x(x0,ξ(x0,z0),z0)+(y?ξ(x0,z0))?y(x0,ξ(x0,z0),z0)+(z?z0)?z(x0,ξ(x0,z0),z0)=

Σ (x?x0)?x(x0,y0,z0)+(y?y0)?y(x0,y0,z0)+(z?z0)?z(x0,y0,z0)=(DΣ)T(x,z)n=0∈ ??x(x,ξ(x,z), ??z(x,ξ(x,z),

=0∈

n10??x(x,ξ(x,z), n ??z(x,ξ(x,z),

=0∈

?x(x,ξ(x,z),n3=?z(x,ξ(x,z),?x 1 1n ∥

(x,ξ(x,z),z)

(x,ξ(x,z),n ?f n?z 5(R43維曲面).R43xΣ ∈R4f(x,y,z,θ)=0∈zθ

Σf(xy,z,θ0

?f(x,y,z,θ?=0 x , =f(x,y,z,θ)∈θ ,

=f(x0,y0,z0,θ0)=

z , =?y(x0,y0,z0,θ0)?=x?x?zθ∈,?!ξ(x,z,θ)

θxxxΣzθ:?zθx7→ θ

ξ(x,z,=zθ

∈

?ξ

?Σ?Σ

(x,z, (x,z, (x,z, θ

(x,z,θ)

∈x =θzθ

z + ,ξ(x,z, D[x]ξ(x,z,θ)=0∈zθ ?f,?f,?f(x,ξ(x,z,θ),z,θ)+?f(x,ξ(x,z,θ),z,θ)?ξ,?ξ,?ξ(x,z,θ)=0∈?x?z

?x??x(x,z,θ)=??f(x,ξ(x,z,θ),z, (x,z,θ)=??z(x,ξ(x,z,θ),z,?ξ(x,z,θ)=??θ(x,ξ(x,z,θ),z,

(DΣ)T(x,z,θ)n=0∈1??x(x,ξ(x,z,θ),z,θ)00??z(x,ξ(x,z,θ),z,θ)1

4

0∈

0??x(x,ξ(x,z,θ),z,θ)0 100??x(x,ξ(x,z,θ),z,010??z(x,ξ(x,z,θ),z,

2

0∈001??x(x,ξ(x,z,θ),z, 1 n=?f(x,ξ(x,z,θ),z,n3=?z(x,ξ(x,z,θ),z,n4=?θ(x,ξ(x,z,θ),z, nn

(x,ξ(x,z,θ),z, ξ(x0,z0, ∈Σ點的切平面方程 y ξ(x0,z0,

(x,ξ(x,z,θ),z, = 0

∈ (x?

(x0,ξ(x0,z0,θ0),z0,θ0)+(y?ξ(x0,

(x0,ξ(x0,z0,θ0),z0, +(z? (x0,ξ(x0,z0,θ0),z0,θ0)+(θ?

(x0,ξ(x0,z0,θ0),z0,θ0)= (x?

(x0,y0,z0,θ0)+(y? +(θ?

(x0,y0,z0,θ0)+(z?(x0,y0,z0,θ0)=

(x0,y0,z0,6(Rp+1p維曲面（隱式表示形式）).Rp+1p Σ X∈Rp+1f(X)=0∈

)=0Σ，亦即有f(X ?Xα(X0)?=0，其中α為1,···,p+1 )=0.? ,

=f(X1,···,

,···,

)∈.?X1X0.. ,

=f(X1,···,Xα,···,Xp+1)=0X 0.XX0. ?

, (X1,···,Xα,···,Xp+1)?= X 0.XX

??

X0X.?..XX

.?..

∈

X0X.? ..XX00ξ(X1···,Xα,···,Xp+1)∈Bμ(Xα)?R00.? .

,

?1,···,Xα,···,

=

1

∈

ξ(X,···,Xα,···, .XXX XXX0