Question.1.[Peng.Sec3.10] 设
F λ ( x , y , z ) = x 2 a − λ + y 2 b − λ + z 2 c − λ , a > b > c > 0 F_\lambda(x,y,z)=\dfrac {x^2}{a-\lambda}+\dfrac {y^2}{b-\lambda}+\dfrac {z^2}{c-\lambda},\quad a>b>c>0 F λ ( x , y , z ) = a − λ x 2 + b − λ y 2 + c − λ z 2 , a > b > c > 0
(1) 当 λ ∈ ( − ∞ , c ) \lambda \in (-\infty,c) λ ∈ ( − ∞ , c ) F λ = 1 F_\lambda=1 F λ = 1
(2) 当 λ ∈ ( c , b ) \lambda \in(c,b) λ ∈ ( c , b ) F λ = 1 F_\lambda=1 F λ = 1
(3) 当 λ ∈ ( b , a ) \lambda \in(b,a) λ ∈ ( b , a ) F λ = 1 F_\lambda=1 F λ = 1
证明：对 E 3 \mathbb E^3 E 3 P = ( x , y , z ) ( x y z ≠ 0 ) P=(x,y,z)(xyz\neq 0) P = ( x , y , z ) ( x y z = 0 ) P P P P P P
设 G ( λ ) = F λ ( x , y , z ) − 1 G(\lambda)=F_\lambda(x,y,z)-1 G ( λ ) = F λ ( x , y , z ) − 1 λ \lambda λ ( − ∞ , c ) (-\infty,c) ( − ∞ , c )
lim λ → − ∞ G ( λ ) = − 1 , lim λ → c − G ( λ ) = + ∞ \lim_{\lambda\to -\infty}G(\lambda)=-1,\quad \lim_{\lambda\to c^-}G(\lambda)=+\infty λ → − ∞ lim G ( λ ) = − 1 , λ → c − lim G ( λ ) = + ∞
而 ( − ∞ , c ) (-\infty,c) ( − ∞ , c ) G ( λ ) G(\lambda) G ( λ ) λ 1 ∈ ( − ∞ , c ) \lambda_1\in(-\infty,c) λ 1 ∈ ( − ∞ , c ) G ( λ 1 ) = 0 G(\lambda_1)=0 G ( λ 1 ) = 0
λ 1 < c < λ 2 < b < λ 3 < a \lambda _1<c<\lambda _2<b<\lambda_3<a λ 1 < c < λ 2 < b < λ 3 < a
记分别对应的曲面为 S 1 , S 2 , S 3 S_1,S_2,S_3 S 1 , S 2 , S 3
n ⃗ i = ∇ F λ i = ( 2 x a − λ i , 2 y b − λ i , 2 z c − λ i ) , i = 1 , 2 , 3 \vec n_i=\nabla F_{\lambda_i}=\left(\dfrac {2x}{a-\lambda_i},\dfrac {2y}{b-\lambda_i},\dfrac {2z}{c-\lambda_i}\right),\quad i=1,2,3 n i = ∇ F λ i = ( a − λ i 2 x , b − λ i 2 y , c − λ i 2 z ) , i = 1 , 2 , 3
仅需要说明 n ⃗ i ⊥ n ⃗ j \vec n_i\perp \vec n_j n i ⊥ n j
x 2 ( a − λ i ) ( a − λ j ) + y 2 ( b − λ i ) ( b − λ j ) + z 2 ( c − λ i ) ( c − λ j ) = 0 \dfrac {x^2}{(a-\lambda_i)(a-\lambda_j)}+\dfrac {y^2}{(b-\lambda_i)(b-\lambda_j)}+\dfrac {z^2}{(c-\lambda_i)(c-\lambda_j)}=0 ( a − λ i ) ( a − λ j ) x 2 + ( b − λ i ) ( b − λ j ) y 2 + ( c − λ i ) ( c − λ j ) z 2 = 0
注意到
x 2 ( a − λ i ) ( a − λ j ) = x 2 λ i − λ j ( 1 a − λ j − 1 a − λ i ) \dfrac {x^2}{(a-\lambda_i)(a-\lambda_j)}=\dfrac {x^2}{\lambda_i-\lambda_j}\left(\dfrac 1{a-\lambda_j}-\dfrac 1{a-\lambda_i}\right) ( a − λ i ) ( a − λ j ) x 2 = λ i − λ j x 2 ( a − λ j 1 − a − λ i 1 )
再由
x 2 a − λ i + y 2 b − λ i + z 2 c − λ i = 1 \dfrac {x^2}{a-\lambda_i}+\dfrac {y^2}{b-\lambda_i}+\dfrac {z^2}{c-\lambda_i}=1 a − λ i x 2 + b − λ i y 2 + c − λ i z 2 = 1
即可说明.
Question.2.[Peng.Sec3.12] 使 F = ⟨ r u , r v ⟩ = 0 F=\langle \boldsymbol{r}_u,\boldsymbol{r}_v\rangle=0 F = ⟨ r u , r v ⟩ = 0 ( u , v ) (u,v) ( u , v ) 正交参数系 。给定一个曲面 S S S r = r ( u , v ) \boldsymbol r=\boldsymbol{r}(u,v) r = r ( u , v ) S S S P 0 = P ( u 0 , v 0 ) P_0=P(u_0,v_0) P 0 = P ( u 0 , v 0 ) P 0 P_0 P 0 D D D D D D ( s , t ) (s,t) ( s , t ) ( s , t ) (s,t) ( s , t ) S S S
不妨 ⟨ r u , r v ⟩ ≠ 0 \langle \boldsymbol{r}_u,\boldsymbol{r}_v\rangle\neq 0 ⟨ r u , r v ⟩ = 0
r u , ( r u ∧ r v ) ∧ r u = E r v − F r u \boldsymbol r_u,\quad (\boldsymbol r_u\wedge \boldsymbol r_v)\wedge \boldsymbol r_u=E\boldsymbol r_{v}-F\boldsymbol{r}_u r u , ( r u ∧ r v ) ∧ r u = E r v − F r u
我们构造参数 ( s , t ) (s,t) ( s , t )
r s = r u , r t = E r v − F r u \boldsymbol r_s=\boldsymbol r_u,\quad \boldsymbol r_t=E\boldsymbol r_{v}-F\boldsymbol{r}_u r s = r u , r t = E r v − F r u
这同时给出了 ∂ ( u , v ) ∂ ( s , t ) \frac {\partial (u,v)}{\partial (s,t)} ∂ ( s , t ) ∂ ( u , v ) D D D
Question.3.[Peng.Sec3.13] 在曲面 S : r = r ( u , v ) S:\boldsymbol r=\boldsymbol r(u,v) S : r = r ( u , v ) P d u d u + 2 Q d u d v + R d v d v = 0 P\mathrm du\mathrm du+2Q\mathrm du\mathrm dv+R\mathrm dv\mathrm dv=0 P d u d u + 2 Q d u d v + R d v d v = 0 E R − 2 F Q + G P = 0 ER-2FQ+GP=0 E R − 2 F Q + G P = 0 E , F , G E,F,G E , F , G
由方程确定的两个切向量满足
d u d v = − Q ± Q 2 − P R P = α + or α − \dfrac {\mathrm du}{\mathrm dv}=\dfrac {-Q\pm \sqrt{Q^2-PR}}P=\alpha^+\text{ or }\alpha^- d v d u = P − Q ± Q 2 − P R = α + or α −
所以这两个切向量可表示为
v + = r u α + + r v , v − = r u α − + r v \boldsymbol v^+=\boldsymbol r_u \alpha^+ +\boldsymbol r_v,\quad \boldsymbol v^-=\boldsymbol r_u \alpha^- +\boldsymbol r_v v + = r u α + + r v , v − = r u α − + r v
那么它们相互正交，等价于
0 = ⟨ v + , v − ⟩ = E α + α − + F ( α + + α − ) + G = E R − 2 F Q + G P P 0=\langle \boldsymbol v^+,\boldsymbol v^-\rangle=E\alpha^+\alpha^-+F(\alpha^++\alpha^-)+G=\dfrac {ER-2FQ+GP}{P} 0 = ⟨ v + , v − ⟩ = E α + α − + F ( α + + α − ) + G = P E R − 2 F Q + G P
Question.4.[Peng.Sec3.14] 求曲面的第二基本形式
(1) r ( u , v ) = ( u cos v , u sin v , b v ) \boldsymbol r(u,v)=(u\cos v,u\sin v,bv) r ( u , v ) = ( u cos v , u sin v , b v )
(2) r ( u , v ) = ( a ( u + v ) , b ( u − v ) , u 2 + v 2 ) \boldsymbol r(u,v)=(a(u+v),b(u-v),u^2+v^2) r ( u , v ) = ( a ( u + v ) , b ( u − v ) , u 2 + v 2 )
直接计算
(1) r u = ( cos v , sin v , 0 ) , r v = ( − u sin v , u cos v , b ) \boldsymbol r_u=(\cos v,\sin v,0),\boldsymbol r_v=(-u\sin v,u\cos v,b) r u = ( cos v , sin v , 0 ) , r v = ( − u sin v , u cos v , b )
n = 1 b 2 + u 2 ( b sin v , − b cos v , u ) \boldsymbol n=\dfrac {1}{\sqrt {b^2+u^2}}(b\sin v,-b\cos v,u) n = b 2 + u 2 1 ( b sin v , − b cos v , u )
进而
L = ⟨ r u u , n ⟩ = 0 , M = ⟨ r u v , n ⟩ = − b b 2 + u 2 , N = ⟨ r v v , n ⟩ = 0 L=\langle \boldsymbol r_{uu},\boldsymbol n\rangle=0,\quad M=\langle \boldsymbol r_{uv},\boldsymbol n\rangle=\dfrac {-b}{\sqrt {b^2+u^2}},\quad N=\langle \boldsymbol r_{vv},\boldsymbol n\rangle=0 L = ⟨ r u u , n ⟩ = 0 , M = ⟨ r u v , n ⟩ = b 2 + u 2 − b , N = ⟨ r v v , n ⟩ = 0
所以
I I = − 2 b b 2 + u 2 d u d v II=-\dfrac {2b}{\sqrt {b^2+u^2}}\mathrm du\mathrm dv I I = − b 2 + u 2 2 b d u d v
(2) r u = ( a , b , 2 u ) , r v = ( a , − b , 2 v ) \boldsymbol r_u=(a,b,2u),\boldsymbol r_v=(a,-b,2v) r u = ( a , b , 2 u ) , r v = ( a , − b , 2 v ) Δ = a 2 b 2 + a 2 ( u − v ) 2 + b 2 ( u + v ) 2 \Delta=a^2b^2+a^2( u-v)^2+b^2(u+v)^2 Δ = a 2 b 2 + a 2 ( u − v ) 2 + b 2 ( u + v ) 2
n = 1 Δ ( b ( u + v ) , a ( u − v ) , − a b ) \boldsymbol n=\dfrac 1{\Delta}\left(b(u+v),a(u-v),-ab\right) n = Δ 1 ( b ( u + v ) , a ( u − v ) , − a b )
进而
L = ⟨ r u u , n ⟩ = − 2 a b Δ , M = ⟨ r u v , n ⟩ = 0 , N = ⟨ r v v , n ⟩ = − 2 a b Δ L=\langle \boldsymbol r_{uu},\boldsymbol n\rangle=\dfrac {-2ab}{\Delta},\quad M=\langle \boldsymbol r_{uv},\boldsymbol n\rangle=0,\quad N=\langle \boldsymbol r_{vv},\boldsymbol n\rangle=\dfrac {-2ab}{\Delta} L = ⟨ r u u , n ⟩ = Δ − 2 a b , M = ⟨ r u v , n ⟩ = 0 , N = ⟨ r v v , n ⟩ = Δ − 2 a b
所以
I I = − 2 a b Δ ( d u 2 + d v 2 ) II=\dfrac {-2ab}{\Delta}(\mathrm du^2+\mathrm dv^2) I I = Δ − 2 a b ( d u 2 + d v 2 )
Question.5.[Peng.Sec3.15] 求曲面 z = f ( x , y ) z=f(x,y) z = f ( x , y )
直接计算，r x = ( 1 , 0 , f x ) , r y = ( 0 , 1 , f y ) \boldsymbol r_x=(1,0,f_x),\boldsymbol r_y=(0,1,f_y) r x = ( 1 , 0 , f x ) , r y = ( 0 , 1 , f y )
n = 1 1 + f x 2 + f y 2 ( − f x , − f y , 1 ) \boldsymbol n=\dfrac 1{\sqrt {1+f_x^2+f_y^2}}(-f_x,-f_y,1) n = 1 + f x 2 + f y 2 1 ( − f x , − f y , 1 )
记 1 + f x 2 + f y 2 = Δ \sqrt{1+f_x^2+f_y^2}=\Delta 1 + f x 2 + f y 2 = Δ
L = ⟨ r x x , n ⟩ = f x x Δ , M = ⟨ r x y , n ⟩ = f x y Δ , N = ⟨ r y y , n ⟩ = f y y Δ L=\langle \boldsymbol r_{xx},\boldsymbol n\rangle=\dfrac {f_{xx}}{\Delta},\quad M=\langle \boldsymbol r_{xy},\boldsymbol n\rangle=\dfrac {f_{xy}}{\Delta},\quad N=\langle \boldsymbol r_{yy},\boldsymbol n\rangle=\dfrac {f_{yy}}{\Delta} L = ⟨ r x x , n ⟩ = Δ f x x , M = ⟨ r x y , n ⟩ = Δ f x y , N = ⟨ r y y , n ⟩ = Δ f y y
所以
I I = f x x Δ d x 2 + 2 f x y Δ d x d y + f y y Δ d y 2 II=\dfrac {f_{xx}}{\Delta}\mathrm dx^2+\dfrac {2f_{xy}}{\Delta}\mathrm dx\mathrm dy+\dfrac {f_{yy}}{\Delta}\mathrm dy^2 I I = Δ f x x d x 2 + Δ 2 f x y d x d y + Δ f y y d y 2
Question.6.[Peng.Sec3.19] 定义 I I I = ⟨ d n , d n ⟩ III=\langle \mathrm d\boldsymbol n,\mathrm d\boldsymbol n\rangle I I I = ⟨ d n , d n ⟩ K I − 2 H I I + I I I = 0 KI-2HII+III=0 K I − 2 H I I + I I I = 0
取 Weingarten 算子 W \mathcal W W
W ( d r ) = − d n \mathcal W(\mathrm d\boldsymbol r)= -\mathrm d\boldsymbol n W ( d r ) = − d n
这说明
I I I = ⟨ d n , d n ⟩ = ⟨ W ( d r ) , W ( d r ) ⟩ = ⟨ W 2 ( d r ) , d r ⟩ III=\langle \mathrm d\boldsymbol n,\mathrm d\boldsymbol n\rangle=\langle \mathcal W(\mathrm d\boldsymbol r),\mathcal W(\mathrm d\boldsymbol r)\rangle=\langle \mathcal W^2(\mathrm d\boldsymbol r),\mathrm d\boldsymbol r\rangle I I I = ⟨ d n , d n ⟩ = ⟨ W ( d r ) , W ( d r ) ⟩ = ⟨ W 2 ( d r ) , d r ⟩
K I − 2 H I I = K ⟨ d r , d r ⟩ − 2 H ⟨ W ( d r ) , d r ⟩ = ⟨ ( K i d − 2 H W ) ( d r ) , d r ⟩ KI-2HII=K\langle \mathrm d\boldsymbol r,\mathrm d\boldsymbol r\rangle -2H\langle \mathcal W(\mathrm d\boldsymbol r),\mathrm d\boldsymbol r\rangle=\langle (K {\mathrm {id}}-2H\mathcal W)(\mathrm d\boldsymbol r),\mathrm d\boldsymbol r\rangle K I − 2 H I I = K ⟨ d r , d r ⟩ − 2 H ⟨ W ( d r ) , d r ⟩ = ⟨ ( K i d − 2 H W ) ( d r ) , d r ⟩
而注意到 W \mathcal W W k 1 , k 2 k_1,k_2 k 1 , k 2 k 1 + k 2 = 2 H , K = k 1 k 2 k_1+k_2=2H,K=k_1k_2 k 1 + k 2 = 2 H , K = k 1 k 2
( W ) 2 − 2 H W + K i d = ( W − k 1 i d ) ( W − k 2 i d ) = 0 (\mathcal W)^2 -2H\mathcal W +K{\mathrm {id}}=(\mathcal W-k_1{\mathrm {id}})(\mathcal W-k_2{\mathrm {id}})=0 ( W ) 2 − 2 H W + K i d = ( W − k 1 i d ) ( W − k 2 i d ) = 0
因此 K I − 2 H I I + I I I = 0 KI-2HII+III=0 K I − 2 H I I + I I I = 0
Question.7.[Peng.Sec3.20] 设曲面 S 1 S_1 S 1 S 2 S_2 S 2 C C C κ \kappa κ C C C S i S_i S i k i , i = 1 , 2 k_i,i=1,2 k i , i = 1 , 2 C C C S 1 S_1 S 1 S 2 S_2 S 2 θ \theta θ
κ 2 sin 2 θ = k 1 2 + k 2 2 − 2 k 1 k 2 cos θ \kappa^2\sin^2\theta=k_1^2+k_2^2-2k_1k_2\cos\theta κ 2 sin 2 θ = k 1 2 + k 2 2 − 2 k 1 k 2 cos θ
考虑标架 { t , n 1 , n 2 } \{\boldsymbol t,\boldsymbol n_1,\boldsymbol n_2\} { t , n 1 , n 2 } t \boldsymbol t t C C C n i \boldsymbol n_i n i S i S_i S i r s s ⊥ t \boldsymbol r_{ss}\perp \boldsymbol t r s s ⊥ t α , β ∈ R \alpha,\beta\in\mathbb R α , β ∈ R
r s s = α n 1 + β n 2 \boldsymbol{r}_{ss}=\alpha \boldsymbol n_1+\beta \boldsymbol n_2 r s s = α n 1 + β n 2
考虑到 r s s ⋅ n i = k i \boldsymbol r_{ss}\cdot \boldsymbol n_i=k_i r s s ⋅ n i = k i n 1 ⋅ n 2 = cos θ \boldsymbol n_1\cdot \boldsymbol n_2=\cos\theta n 1 ⋅ n 2 = cos θ
α = k 1 − k 2 cos θ sin 2 θ , β = k 2 − k 1 cos θ sin 2 θ \alpha=\dfrac {k_1-k_2\cos\theta}{\sin^2\theta},\quad \beta=\dfrac {k_2-k_1\cos\theta}{\sin^2\theta} α = sin 2 θ k 1 − k 2 cos θ , β = sin 2 θ k 2 − k 1 cos θ
所以
κ 2 = ⟨ r s s , r s s ⟩ = α 2 + β 2 + 2 α β cos θ = k 1 2 + k 2 2 − 2 k 1 k 2 cos θ sin 2 θ \kappa^2=\langle \boldsymbol r_{ss},\boldsymbol r_{ss}\rangle=\alpha^2+\beta^2+2\alpha\beta\cos\theta=\dfrac {k_1^2+k_2^2-2k_1k_2\cos\theta}{\sin^2\theta} κ 2 = ⟨ r s s , r s s ⟩ = α 2 + β 2 + 2 α β cos θ = sin 2 θ k 1 2 + k 2 2 − 2 k 1 k 2 cos θ
Question.8.[Peng.Sec3.21] 求下列曲面的 Gauss 曲率和平均曲率：
(1) 单叶双曲面
x 2 a 2 + y 2 b 2 − z 2 c 2 = 1 \frac {x^2}{a^2}+\frac {y^2}{b^2}-\frac {z^2}{c^2}=1 a 2 x 2 + b 2 y 2 − c 2 z 2 = 1
(2) 环面
( x , y , z ) = ( R + r cos u ) cos v , ( R + r cos u ) sin v , r sin u ) (x,y,z)=(R+r\cos u)\cos v,(R+r\cos u)\sin v,r\sin u) ( x , y , z ) = ( R + r cos u ) cos v , ( R + r cos u ) sin v , r sin u )
直接计算
(1) 局部参数表示为
r ( x , y ) = ( x , y , ± c x 2 a 2 + y 2 b 2 − 1 ) \boldsymbol r(x,y)=(x,y,\pm c\sqrt {\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}-1}) r ( x , y ) = ( x , y , ± c a 2 x 2 + b 2 y 2 − 1 )
当 z ≠ 0 z\neq 0 z = 0
z x = c 2 x a 2 z , z y = c 2 y b 2 z z_x=\dfrac {c^2x}{a^2z},\quad z_y=\dfrac {c^2y}{b^2z} z x = a 2 z c 2 x , z y = b 2 z c 2 y
z x x = c 2 a 2 z ( 1 − c 2 x 2 a 2 z 2 ) , z x y = − c 4 x y a 2 b 2 z 3 , z y y = c 2 b 2 z ( 1 − c 2 y 2 b 2 z 2 ) z_{xx}=\dfrac {c^2}{a^2z}\left(1-\dfrac {c^2x^2}{a^2z^2}\right),\quad z_{xy}=-\dfrac {c^4xy}{a^2b^2z^3},\quad z_{yy}=\dfrac {c^2}{b^2z}\left(1-\dfrac {c^2y^2}{b^2z^2}\right) z x x = a 2 z c 2 ( 1 − a 2 z 2 c 2 x 2 ) , z x y = − a 2 b 2 z 3 c 4 x y , z y y = b 2 z c 2 ( 1 − b 2 z 2 c 2 y 2 )
利用 [Peng.Sec3.22] 的结果，得到
K = − 1 a 2 b 2 c 2 ( x 2 a 4 + y 2 b 4 + z 2 c 4 ) 2 K=-\dfrac {1}{a^2b^2c^2(\frac {x^2}{a^4}+\frac {y^2}{b^4}+\frac {z^2}{c^4})^2} K = − a 2 b 2 c 2 ( a 4 x 2 + b 4 y 2 + c 4 z 2 ) 2 1
H = x 2 + y 2 + z 2 − a 2 − b 2 + c 2 2 a 2 b 2 c 2 ( x 2 a 4 + y 2 b 4 + z 2 c 4 ) 3 2 H=\dfrac{x^2+y^2+z^2-a^2-b^2+c^2}{2a^2b^2c^2(\frac {x^2}{a^4}+\frac {y^2}{b^4}+\frac {z^2}{c^4})^{\frac 32}} H = 2 a 2 b 2 c 2 ( a 4 x 2 + b 4 y 2 + c 4 z 2 ) 2 3 x 2 + y 2 + z 2 − a 2 − b 2 + c 2
(2) 直接计算
r u = ( − r sin u cos v , − r sin u sin v , r cos u ) \boldsymbol r_u=(-r\sin u\cos v,-r\sin u\sin v,r\cos u) r u = ( − r sin u cos v , − r sin u sin v , r cos u )
r v = ( − ( R + r cos u ) sin v , ( R + r cos u ) cos v , 0 ) \boldsymbol r_v=(-(R+r\cos u)\sin v,(R+r\cos u)\cos v,0) r v = ( − ( R + r cos u ) sin v , ( R + r cos u ) cos v , 0 )
则
n = ( − cos u cos v , − cos u sin v , − sin u ) \boldsymbol n=(-\cos u\cos v,-\cos u\sin v,-\sin u) n = ( − cos u cos v , − cos u sin v , − sin u )
r u u = ( − r cos u cos v , − r cos u sin v , − r sin u ) \boldsymbol r_{uu}=(-r\cos u\cos v,-r\cos u\sin v,-r\sin u) r u u = ( − r cos u cos v , − r cos u sin v , − r sin u )
r u v = ( r sin u sin v , − r sin u cos v , 0 ) \boldsymbol r_{uv}=(r\sin u\sin v,-r\sin u\cos v,0) r u v = ( r sin u sin v , − r sin u cos v , 0 )
r v v = ( − ( R + r cos u ) cos v , − ( R + r cos u ) sin v , 0 ) \boldsymbol r_{vv}=(-(R+r\cos u)\cos v,-(R+r\cos u)\sin v,0) r v v = ( − ( R + r cos u ) cos v , − ( R + r cos u ) sin v , 0 )
进而
L = r , M = 0 , N = ( R + r cos u ) cos u L=r,\quad M=0,\quad N=(R+r\cos u)\cos u L = r , M = 0 , N = ( R + r cos u ) cos u
E = r 2 , F = 0 , G = ( R + r cos u ) 2 E=r^2,\quad F=0,\quad G=(R+r\cos u)^2 E = r 2 , F = 0 , G = ( R + r cos u ) 2
所以
K = L N − M 2 E G − F 2 = cos u r ( R + r cos u ) K=\dfrac {LN-M^2}{EG-F^2}=\dfrac {\cos u}{r(R+r\cos u)} K = E G − F 2 L N − M 2 = r ( R + r cos u ) cos u
H = E N − 2 F M + G L 2 ( E G − F 2 ) = R + 2 r cos u 2 r ( R + r cos u ) H=\dfrac {EN-2FM+GL}{2(EG-F^2)}=\dfrac {R+2r\cos u}{2r(R+r\cos u)} H = 2 ( E G − F 2 ) E N − 2 F M + G L = 2 r ( R + r cos u ) R + 2 r cos u
Question.9.[Peng.Sec3.22] 求曲面 z = f ( x , y ) z=f(x,y) z = f ( x , y )
直接计算，r x = ( 1 , 0 , f x ) , r y = ( 0 , 1 , f y ) \boldsymbol r_x=(1,0,f_x),\boldsymbol r_y=(0,1,f_y) r x = ( 1 , 0 , f x ) , r y = ( 0 , 1 , f y )
n = 1 1 + f x 2 + f y 2 ( − f x , − f y , 1 ) \boldsymbol n=\dfrac 1{\sqrt {1+f_x^2+f_y^2}}(-f_x,-f_y,1) n = 1 + f x 2 + f y 2 1 ( − f x , − f y , 1 )
记 1 + f x 2 + f y 2 = Δ \sqrt{1+f_x^2+f_y^2}=\Delta 1 + f x 2 + f y 2 = Δ
r x x = ( 0 , 0 , f x x ) , r x y = ( 0 , 0 , f x y ) , r y y = ( 0 , 0 , f y y ) \boldsymbol r_{xx}=(0,0,f_{xx}),\quad \boldsymbol r_{xy}=(0,0,f_{xy}),\quad \boldsymbol r_{yy}=(0,0,f_{yy}) r x x = ( 0 , 0 , f x x ) , r x y = ( 0 , 0 , f x y ) , r y y = ( 0 , 0 , f y y )
所以
E = 1 + f x 2 , F = f x f y , G = 1 + f y 2 E=1+f_x^2,\quad F=f_xf_y,\quad G=1+f_y^2 E = 1 + f x 2 , F = f x f y , G = 1 + f y 2
L = f x x Δ , M = f x y Δ , N = f y y Δ L=\dfrac {f_{xx}}{\Delta},\quad M=\dfrac {f_{xy}}{\Delta},\quad N=\dfrac {f_{yy}}{\Delta} L = Δ f x x , M = Δ f x y , N = Δ f y y
得到
K = L N − M 2 E G − F 2 = f x x f y y − f x y 2 Δ 4 K=\dfrac {LN-M^2}{EG-F^2}=\dfrac {f_{xx}f_{yy}-f_{xy}^2}{\Delta^4} K = E G − F 2 L N − M 2 = Δ 4 f x x f y y − f x y 2
H = E N − 2 F M + G L 2 ( E G − F 2 ) = ( 1 + f y 2 ) f x x − 2 f x f y f x y + ( 1 + f x 2 ) f y y 2 Δ 3 H=\dfrac {EN-2FM+GL}{2(EG-F^2)}=\dfrac {(1+f_y^2)f_{xx}-2f_xf_yf_{xy}+(1+f_x^2)f_{yy}}{2\Delta^3} H = 2 ( E G − F 2 ) E N − 2 F M + G L = 2 Δ 3 ( 1 + f y 2 ) f x x − 2 f x f y f x y + ( 1 + f x 2 ) f y y
