MA3D9 Example Sheet 4
Without otherwise mentioned, all curves are smooth and regular.
1. Complete all the exercises mentioned in the class and in the lecture notes.
2. Show that the mean curvature H at p ∈ S is given by
Z
1 π
H=
κn (θ)dθ,
π 0
where κn (θ) is the normal curvature at p along a direction making an angle θ with a fixed
direction.
Solution:
By Euler’s Theorem κn = κ1 cos2 θ + κ2 sin2 θ where θ is the the oriented angle to t1 , we know
κn (θ) = κ1 cos2 (θ + θ0 ) + κ2 sin2 (θ + θ0 ). Hence
Z
Z π
Z π
1 π
1
1
2
κn (θ)dθ = κ1
cos (θ + θ0 ) + κ2
sin2 (θ + θ0 ) = κ1 + κ2 = H.
π 0
π
π
0
0
3. If the surface S1 and S2 intersect along a regular curve C, then the curvature k of C at p is
given by
k 2 sin2 θ = λ21 + λ22 − 2λ1 λ2 cos θ,
where λ1 and λ2 are the normal curvatures at p, along the tangent line to C, of S1 and S2 ,
respectively, and θ is the angle made up by the normal vectors of S1 and S2 at p.
Solution: Let N1 and N2 be unit normal vectors of S1 and S2 respectively, and n be the
principal normal of C.
So λ1 = kN1 · n, λ2 = kN2 · n. Hence
q
k||(n · N1 )N2 − (n · N2 )N1 || = ||λ1 N2 − λ2 N1 || = λ21 + λ22 − 2λ1 λ2 cos θ
because N1 · N2 = cos θ.
On the other hand,
||(n · N1 )N2 − (n · N2 )N1 || = ||n × (N1 × N2 )|| = ||N1 × N2 || = | sin θ|.
This completes the equality.
4. Show that the Gaussian curvature of the surface z = f (x, y), where f is a smooth function, is
K=
2
fxx fyy − fxy
.
(1 + fx2 + fy2 )2
Solution
σ(x, y) = (x, y, f (x, y)).
First fundamental form: (1 + fx2 )dx2 + 2fx fy dxdy + (1 + fy2 )dy 2 .
fxx
dx2
1+fx2 +fy2
Second fundamental form: √
K=
+√
2fxy
1+fx2 +fy2
dxdy + √
fyy
1+fx2 +fy2
dxdy 2 .
2
fxx fyy − fxy
LN − M 2
=
.
EG − F 2
(1 + fx2 + fy2 )2
5. a. Show that if σ is an isothermal parametrization, that is, E = G = λ(u, v) and F = 0, then
the Gaussian curvature
1
K = − ∆(ln λ),
2λ
where ∆φ denotes the Laplacian
∂2φ
∂u2
+
∂2φ
∂v 2
of the function φ.
b. Calculate the Gaussian curvature of the surface (upper half-plane model) with first fundamental form
dv 2 + du2
.
u2
Solution:
a.) When F = 0,
Ev
λv
1
∂ Gu
∂
1 λu
1
(√
{ (√
)+
)} = − (( )u + ( )v ) = − ((ln λ)uu + (ln λ)vv )
K=− √
∂v EG
2λ λ
λ
2λ
2 EG ∂u EG
b.) λ =
1
,
u2
2
so K = − u2 ∆(ln u12 ) = u2 (ln u)uu = u2 · (− u12 ) = −1.
6. Show that there exists no surface σ(u, v) such that E = G = 1, F = 0 and L = 1, M = 0 and
N = −1.
2
−M
Solution: We have several ways to show it. For example, from one side K = LN
= −1.
EG−F 2
1
∂ √Gu
∂ √Ev
√
But from the other hand, K = − 2 EG { ∂u ( EG ) + ∂v ( EG )} = 0. Contradiction!
7. Find the Gaussian curvature of each surface:
a. Paraboloid x2 + y 2 = 2pz.
b. Torus σ(u, v) = ((a + b cos u) cos v, (a + b cos u) sin v, b sin u); 0 < b < a, 0 ≤ u, v ≤ 2π.
2
Solution:
√
√
a.) σ(u, v) = ( 2pu cos v, 2pu sin v, u2 ). By HW3, the first fundamental form is (2p +
4u2 )du2 + (2pu2 )dv 2 . So
∂ Gu
∂
Ev
p2
1
p2
{ (√
)+
(√
)} = 2
K=− √
=
.
∂v EG
(p + 2pu2 )2
(p2 + x2 + y 2 )2
2 EG ∂u EG
Or you use
1 1
2
·
fxx fyy − fxy
p2
p p
=
K=
=
2
2
(1 + fx2 + fy2 )2
(p2 + x2 + y 2 )2
(1 + x p+y
)2
2
b.) By HW3, the first fundamental form is b2 du2 + (a + b cos u)2 dv 2 . So
1
∂ Gu
cos u
K=− √
(√
)=
b(a + b cos u)
2 EG ∂u EG
8. When E = G = 1 and F = cos θ, show that
K=−
θuv
.
sin θ
Solution: Γ111 = θu cot θ, Γ211 = −θu csc θ, Γ112 = Γ212 = 0, Γ122 = −θv csc θ, Γ222 = θv cot θ. Plug
in first Gauss equation, K = (Γ211 )v + Γ211 Γ222 = −θuv csc θ.
9. Suppose that the first and second fundamental forms of a surface patch are Edu2 + Gdv 2 and
Ldu2 + N dv 2 . Show that the principal curvatures κ1 = EL and κ2 = N
satisfy the equations
G
(κ1 )v =
Ev
Gu
(κ2 − κ1 ), (κ2 )u =
(κ1 − κ2 ).
2E
2G
Solution:
Γ111 =
Eu 2
Ev
Ev 2
Gu 1
Gu
Gv
, Γ11 = − , Γ112 =
, Γ12 =
, Γ22 = − , Γ222 =
.
2E
2G
2E
2G
2E
2G
So the Codazzi-Mainardi equations become
L N
1
1
L N
1
1
Lv = Ev ( + ) = Ev (κ1 + κ2 ), Nu = Gu ( + ) = Gu (κ1 + κ2 ).
2
E
G
2
2
E
G
2
Since κ1 =
L
,
E
Lv = (κ1 )v E + κ1 Ev . So
(κ1 )v =
1
Ev
(Lv − κ1 Ev ) =
(κ2 − κ1 )
E
2E
Similarly for (κ2 )u .
3