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 1 Z 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: 2 2 By Euler’s Theorem κn = κ1 cos θ + κ2 sin θ where θ is the the oriented angle to t1, we know 2 2 κn(θ) = κ1 cos (θ + θ0) + κ2 sin (θ + θ0). Hence Z π Z π Z π 1 1 2 1 2 κn(θ)dθ = κ1 cos (θ + θ0) + κ2 sin (θ + θ0) = κ1 + κ2 = H. π 0 π 0 π 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 2 2 2 2 k sin θ = λ1 + λ2 − 2λ1λ2 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 2 2 k||(n · N1)N2 − (n · N2)N1|| = ||λ1N2 − λ2N1|| = λ1 + λ2 − 2λ1λ2 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

2 fxxfyy − fxy K = 2 2 2 . (1 + fx + fy )

Solution σ(x, y) = (x, y, f(x, y)). 2 2 2 2 First fundamental form: (1 + fx )dx + 2fxfydxdy + (1 + fy )dy . √ fxx 2 √ 2fxy √ fyy 2 Second fundamental form: 2 2 dx + 2 2 dxdy + 2 2 dxdy . 1+fx +fy 1+fx +fy 1+fx +fy

2 2 LN − M fxxfyy − fxy K = 2 = 2 2 2 . EG − F (1 + fx + fy ) 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 ∂2φ ∂2φ where ∆φ denotes the Laplacian ∂u2 + ∂v2 of the function φ. b. Calculate the Gaussian curvature of the surface (upper half-plane model) with first funda- mental form dv2 + du2 . u2

Solution: a.) When F = 0,

1 ∂ Gu ∂ Ev 1 λu λv 1 K = − √ { (√ ) + (√ )} = − (( )u + ( )v) = − ((ln λ)uu + (ln λ)vv) 2 EG ∂u EG ∂v EG 2λ λ λ 2λ

1 u2 1 2 2 1 b.) λ = u2 , so K = − 2 ∆(ln u2 ) = u (ln u)uu = u · (− u2 ) = −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.

LN−M 2 Solution: We have several ways to show it. For example, from one side K = EG−F 2 = −1. But from the other hand, K = − √1 { ∂ ( √Gu ) + ∂ ( √Ev )} = 0. Contradiction! 2 EG ∂u EG ∂v EG 7. Find the Gaussian curvature of each surface: a. Paraboloid x2 + y2 = 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)dv2. So 1 ∂ G ∂ E p2 p2 K = − √ { (√ u ) + (√ v )} = = . 2 EG ∂u EG ∂v EG (p2 + 2pu2)2 (p2 + x2 + y2)2

Or you use 2 1 1 2 fxxfyy − f · p K = xy = p p = (1 + f 2 + f 2)2 x2+y2 2 (p2 + x2 + y2)2 x y (1 + p2 ) b.) By HW3, the first fundamental form is b2du2 + (a + b cos u)2dv2. So 1 ∂ G cos u K = − √ (√ u ) = 2 EG ∂u EG b(a + b cos u)

8. When E = G = 1 and F = cos θ, show that θ K = − uv . sin θ

1 2 1 2 1 2 Solution:Γ11 = θu cot θ, Γ11 = −θu csc θ, Γ12 = Γ12 = 0, Γ22 = −θv csc θ, Γ22 = θv cot θ. Plug 2 2 2 in first Gauss equation, K = (Γ11)v + Γ11Γ22 = −θuv csc θ. 9. Suppose that the first and second fundamental forms of a surface patch are Edu2 + Gdv2 and 2 2 L N Ldu + Ndv . Show that the principal curvatures κ1 = E and κ2 = G satisfy the equations E G (κ ) = v (κ − κ ), (κ ) = u (κ − κ ). 1 v 2E 2 1 2 u 2G 1 2

Solution: E E E G G G Γ1 = u , Γ2 = − v , Γ1 = v , Γ2 = u , Γ1 = − u , Γ2 = v . 11 2E 11 2G 12 2E 12 2G 22 2E 22 2G So the Codazzi-Mainardi equations become 1 L N 1 1 L N 1 L = E ( + ) = E (κ + κ ),N = G ( + ) = G (κ + κ ). v 2 v E G 2 v 1 2 u 2 u E G 2 u 1 2

L Since κ1 = E , Lv = (κ1)vE + κ1Ev. So 1 E (κ ) = (L − κ E ) = v (κ − κ ) 1 v E v 1 v 2E 2 1

Similarly for (κ2)u.

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