Math 4350, Differential Geometry Key to Mock Final Exam. Fall

Math 4350, Differential Geometry Key to Mock Final Exam.
Fall, 2010, Dr. Min Ru, University of Houston
1 . Let E be the ellipse
y
x
( )2 + ( )2 = 1,
a
b
where a, b > 0 are constants.
(a) Find a parametrization of E,
(b) Compute the curvature of E at the points (a, 0) and (0, b).
Solution. α(t) = (a sin t, b cos t).
2. Show that the curve α(t) = (sin t, t, − cos t) has constant speed. Then
find a re-parameterization of this curve by arc-length.
√
Rt
0
0
Solution.
α0 (t) = (cos
√
√ t, 1, sin t). Hence kα (t)k = 2. Let s(t) = 0 kα (u)kdu =
2t, we get t = s/ 2. Hence
√
√
√
β(s) = (sin(s/ 2), s/ 2, − cos(s/ 2)).
3. Let
α(t) = (t,
1 + t 1 − t2
,
), t > 0.
t
t
(a) Prove that the curvature is
κ(t) =
q
3/2
(t2
1
.
+ 1 + t−2 )3/2
(b) Show that α(t) is a plane curve by calculating its torsion.
noindentSolution.
α(t) = (t, 1 + t−1 , t−1 − t)
α0 (t) = (1, −t−2 , −t−2 − 1)
kα0 (t)k2 = 1 + t−4 + (1 + t−2 )2 = 2(1 + t−2 + t−4 )
α00 (t) = (0, 2t−3 , 2t−3 ) = 2t−3 (0, 1, 1)
α000 (t) = −6t−4 (0, 1, 1)
=⇒ α00 is parallel to α000 .
1
α0 × α00 = 2t−3 (1, −1, 1).
Hence,
s
kα0 × α00 k
3
1
κ(t) =
=
.
0
3
2
kα k
2 (t + 1 + t−2 )3/2
(α0 × α00 ) · α000
τ=
=0
kα0 × α00 k2
since α00 is parallel to α000 (t). Therefore α(t) is a plane curve by calculating
its torsion.
4. (a) Describe what a surface of revolution is and how to parameterize it.
(b) Observing a torus is a surface of revolution, find a profile curve of it
and parameterize the torus (with two circles of radius a and b on it). Sketch
the graph.
noindentSolution. (a) A surface of revolution is the surface obtained by
rotating a plane curve, called the profile curve, along a straight line in the
plane. Let us take the axis of rotation to be the z-axis and the plane (where
the profile curve stays) to be the xz-plane. If α(u) = (g(u), h(u), 0) is a
plane curve in xy-plane. Revolve α about the x-axis
one obtains a surface of revolution
σ(u, v) = (g(u), h(u) cos v, h(u) sin v).
(b) Consider the torus (with two circles of radius r and R on it). :
2
σ(u, v) = ((R + r cos u) cos v, (R + r cos u) sin v, r sin u),
is a parameterization of the torus.
Remark: You should know that how to parameterize other types of surfaces
as well, for example ruled surface, .....
3
5. Consider the surface M parametrized by
σ(u, v) = (sin u cos v, sin u sin v, 3 cos u), 0 < u < π, 0 < v < 2π.
(i) Sketch the surface.
(ii) Show that this surface is regular.
√
√
(iii) Let P = ( 2/2, 2/2, 0) ∈ M . Find the matrix of SP with respect
to the basis {σ u (P ), σ v (P )} in TP (M ).
(iv) Find the principal curvatures, principal directions,
√ the Gaussian cur√
vature, and the mean curvature at the point P ( 2/2, 2/2, 0) ∈ M .
Solution. (i) The graph is ellipsoid with a = 1, b = 1, and c = 3.
(ii) σ u = (cos u cos v, cos u sin v, −3 sin u) and σ v = (− sin u sin v, sin u cos v, 0)
σ u × σ v = (3 sin2 u cos v, 3 sin2 u sin v, sin u cos u).
kσ u × σ v k =
=
q
9 sin4 u cos2 v + 9 sin4 u sin2 v + sin2 u cos2 u
q
9 sin4 u + sin2 u cos2 u = |sin u| 8 sin2 u + 1.
q
√
Since 8 sin2 u + 1 > 0 and sin u 6= 0 where 0 < u < π. Hence, σ is a
regular surface.
iii) To find the matrix SP , we need to find the first and second fundamental forms√of σ. √
√
√
At P = ( 2/2, 2/2, 0), we have sin u cos v = 2/2, sin u sin v = 2/2
and 3 cos u = 0. Solve the three equations, we get
u = π/2 and v0 = π/4.
√ √0
− 2
Hence σ u (P ) = (0, 0, −3), and σ v (P ) = ( 2 , 22 , 0).
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The first fundemental form: E = 9, F = 0, G = 1
σ uu = (− sin u cos v, − sin u sin v, −3 sin u),
σ vu = (− cos u sin v, − cos u cos v, 0),
σ vv = (− sin u cos v, − sin u sin v, 0).
Hence
√
√
√
√
− 2 − 2
− 2 − 2
,
, −3), σ uv (P ) = (0, 0, 0), σ vv (P ) = (
,
, −3).
σ uu (P ) = (
2
2
2
2
√
√
3 2 3 2
(σ u × σ v )P = (
,
, 0)
2
2
and kσu × σv kP = 3. The unit normal at P is
√ √
σu × σv
2 2
n=
=(
,
, 0).
kσu × σv k
2 2
The second fundamental form: e = −1, f = 0, g = −1
f F − eG
−1
eF − f E
=
, b=−
=0
2
EG − F
9
EG − F 2
gF − f G
f F − gE
c=−
=
0,
d
=
−
= −1.
EG − F 2
EG − F 2
The matrix SP with respect to the basis {σu (P ), σv (P )} is
a=−
−1/9 0
0
−1
!
.
iv) The principal curvatures : k1 = −1/9, k2 = −1
Gaussian curvature: K = k1 k2 = det(A) = ad − bc = 1/9
2
Mean curvature: H = k1 +k
= tr(A) = a+d
= −5/9
2
2
The principal vectors:
for k1 = −1/9,
0 0
0 −1
!
·
ξ1
η1
!
=
0
0
!
.
we get, ξ1 = 1, η1 = 0. This gives v1 = σu (P ) = (0, 0, −3).
5
for k1 = −1,
−1/9 0
0
0
!
ξ1
η1
·
!
=
0
0
!
,
we get, ξ1 = 0, η1 = 1,
√ √
this gives, v2 = σv (P ) = ( −2 2 , 22 , 0)
6. Suppose γ : (a, b) → R3 be a smooth unit-speed curve. The tangent
developable M is the union of tangent lines to γ which can be parametrized
by
x(u, v) = γ(u) + vγ 0 (u), u ∈ (a, b), v ∈ R.
(a) Assume that the surface M is regular, compute the principal directions
and principal curvatures, mean curvature and Gauss curvature of S. Express
your answers in terms of the curvature and torsion of γ(s).
(b) Specialize your formulas to the helix
α(t) = (cos t, sin t, t).
(Warning: This curve is not unit-speed in current form, so you need to
reparameterize it to γ : (a, b) → R3 , a unit-speed curve.)
Solution: We write
x(t, v) = γ(t) + vγ 0 (t).
We first calculate its first fundamental form.
xt = γ 0 (t) + vγ 00 (t).
By Frenet formula, γ 00 = T0 = κN, where κ is the curvature of γ and T, N
are the unit-tangent and principal normal vectors of γ resepctively. Hence
xt = γ 0 (t) + vγ 00 (t) = T(t) + κvN(t).
xv = γ 0 (t) = T(t).
Hence, E = xt · xt = 1 + v 2 κ2 , F = xt · xv = 1, G = xv · xv = kTk2 = 1, To
find the second fundamental form, we fist compute the unit-normal n of the
surface. Note that
xt × xv = (T(t) + κvN(t)) × T(t) = −κvB(t)
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where B(t) is the binormal vectors of γ. Hence
n=
xt × x v
= −B(t).
kxt × xv k
Also, by the Frenet formula again,
xtt = T0 (t) + κvN0 (t) = κN(t) + κv(−κT(t) + τ B(t)),
xvt = T0 (t) = κN,
xvv = 0.
We get
e = xtt · n = xtt · (−B(t)) = −κτ v,
f = xvt · n = κN · (−B(t)) = 0, g = 0, where κ and τ are the curvature and
the torsion of the curve γ. Hence
K=0
and
H=
eG − 2f F + gE
−vκτ
−τ
=
=
.
2(EG − F 2 )
v 2 κ2
2vκ
The matrix of the shape operator is
FI−1 FII
τ
= 2 2
v κ
1 0
−1 0
!
.
So the eignevalues are 0 and τ /tκ, and the principal directions are e1 =
w1 /kw1 k where w1 = xv (with ξ1 = 0, η1 = 1) and e2 = w2 /kw2 k where
w2 = xu − xv (with ξ2 = 1, η2 = −1).
(b) Take the helix
γ(t) = (cos t, sin t, t).
You can compute its curvature and torsion to get κ = 1/2, τ = 1/2. So
H = −1/2v, K = 0.
Remarks: (1) This problem is a perfect example how the materials of the
first part of the semester (the theory of curves, Frenet formula) the materials
of the second part are combined. This type of problem will be in the final.
7
(2) Please compare this problem with Problem 5. The method is the
same. This one is just more abstract.
7. Let S be the unit shpere x2 + y 2 + z 2 = 1, and let γ ⊂ S be the
intersection of S and the plane z = c (c is constant). Find the curvature,
normal curvature, and geodesic curvature of γ at every point of the curve.
Remark: See the key to HW12 (last homework) to see how to compute the
normal curvature, and geodesic curvature.
Solution: The unit shpere x2 + y 2 + z 2 = 1 has the parametrization
σ(u, v) = (sin u cos v, sin u sin v, cos u),
0 < u < π, 0 < v < 2π.
The curve γ ⊂ S which is the intersection of S and the plane z = c (c is
constant) is x = sin u cos v, y = sin u sin v and x = cos u = c. So
γ(t) = (sin θ0 cos t, sin θ0 sin t, cos θ0 ),
where cos θ0 = c.
It is not a unit-speed curve, since γ 0 (t) = (− sin θ0 sin t, sin θ0 cos t, 0) and
kγ 0 (t)k = sin θ0 . its arc-length parameterization is
γ(s) = (sin θ0 cos(s/ sin θ0 ), sin θ0 sin(s/ sin θ0 ), 0).
γ 00 (s) =
1
( − cos(s/ sin θ0 ), − sin(s/ sin θ0 ), 0).
sin θ0
Its curvature is
1
.
sin θ0
To find its normal and geodesic curvature, we nned find n(s). By the calculation,
σu × σv
= ±(sin u cos v, sin u sin v, cos u).
n=
kσ u × σ v k
κ(s) = kγ 00 (s)k =
Write γ(s) = σ(u(s), v(s)) we find
u(s) = θ0 ,
v(s) =
s
.
sin θ0
Hence, the restriction of n to γ(s) is
n(s) = ±(sin θ0 cos(s/ sin θ0 ), sin θ0 sin(s/ sin θ0 ), cos θ0 ).
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By the formula, the normal curvature of γ is
κn (s) = n(s) · γ 00 (s) = ± cos2 (s/ sin θ0 ) + sin2 (s/ sin θ0 ) = ±1.
κg = γ 00 (s) · (n(s) × γ 0 (s)) = − cot θ0 .
Remark. An alternative way to calculate the normal curvature is through
the second fundamental form: By a simple calculation, we have
e = −1,
f = 0, g = − sin2 u.
On the other hand, as we did above, after re-parameterized by arc-length
parameter, we get
γ(s) = (sin θ0 cos(s/ sin θ0 ), sin θ0 sin(s/ sin θ0 ), 0).
Write γ(s) = σ(u(s), v(s)) we find
u(s) = θ0 ,
v(s) =
s
.
sin θ0
Thus, u0 (s) = 0, v 0 (s) = 1/ sin θ0 . Hence
κn (s) = II(γ 0 (s), γ 0 (s)) = e(u0 (s))2 + 2f u0 (s)v 0 (s) + g(v 0 (s))2
2
1
2
= −1.
= − sin θ0
sin θ0
Since the principal normal vector can have two directions, we kave κn (s) =
±1.
κ2g = κ2 − κ2n ,
so
κg = − cot θ0 .
8. Let α(t) be a unit-speed curve in R3 with principal normal N(t) and
binormal B(t). We define a surface (called the tube of radius a > 0 around
the curve α) by
σ(t, θ) = α(t) + a(N(t) cos θ + B(t) sin θ),
where a > 0 is a constant. Show that, for any fixed t0 , the curve γ(θ) =
σ(t0 , θ) is geodesic (i.e. its geodesic curvature is identically zero) on σ.
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Solution:
γ(θ) = σ(t0 , θ) = α(t0 ) + a(N(t0 ) cos θ + B(t0 ) sin θ).
=⇒
γ 0 (θ) = a(−N(t0 ) sin θ + B(t0 ) cos θ)
=⇒ kγ 0 (θ)k = a. It is not a unit-speed curve, we reparametrize it to be
(unit-speed)
γ(θ) = σ(t0 , θ) = α(t0 ) + a(N(t0 ) cos(θ/a) + B(t0 ) sin(θ/a).
Now
γ 0 (θ) = −N(t0 ) sin(θ/a) + B(t0 ) cos(θ/a)
1
γ 00 (θ) = (−N(t0 ) cos(θ/a) − B(t0 ) sin(θ/a)).
a
To find n the unit-normal to the surafce, write
σ(u, v) = α(u) + a(N(u) cos v + B(u) sin v),
we calculate, by the Frenet formula
σ u = α0 (u) + a(N0 (u) cos v + B0 (u) sin v)
= T(u) + a(−κT(u) + τ B(u)) cos v − τ N(u) sin v)
= (1 − aκ cos v)T(u) + aτ cos vB(u) − aτ sin θN(u).
σ v = a(−N(u) sin v + B(u) cos v).
So
σ u × σ v = −(1 − aκ cos v)a sin vB(u) − (1 − aκ cos v)a cos vN(u).
Hence
n=
σu × σv
= − sin vB(u) − cos vN(u).
kσ u × σ v k
Now write (note here γ(θ) is fater re-paramaetrization) γ(θ) = σ(u(θ), v(θ)),
we get u(theta) = t0 , b(theta) = θ/a. Hence
n(θ) = − sin(θ/a)B(t0 ) − cos(θ/a)N(t0 ).
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Thus
n(θ) × γ 0 (θ) = (− sin(θ/a)B(t0 ) − cos(θ/a)N(t0 )) × (−N(t0 ) sin(θ/a) + B(t0 ) cos(θ/a)
= − sin2 (θ/a)T(t0 ) − cos2 (θ/a)T(t0 ) = −T(t0 ).
So the geodeisc curvature is
κg = γ 00 (θ) · (n(θ) × γ 0 (θ))
1
(−N(t0 ) cos(θ/a) − B(t0 ) sin(θ/a)) · (−T(t0 )) = 0.
=
a
Show that, for any fixed t0 , the curve γ(θ) = σ(t0 , θ) is geodesic (i.e. its
geodesic curvature is identically zero) on σ.
Remark: (1) This problem is a perfect example how the materials of the
first part of the semester (the theory of curves, Frenet formula) the materials
of the second part are combined. This type of problem will be in the final.
(2) Please compare this problem with Problem 7. The method is the
same. This one is just more abstract.
9 State and prove Euler’s theorem. (as well as other theorems covere) Theorem 4.1.3(Euler’s Formula). Let e1 , e2 be unit vectors in the principal directions at P corresponding principal curvatures κ1 , κ2 . Suppose that
v = cos θe1 + sin θe2 for some θ ∈ [0, 2π). Then
IIP (v, v) = κ1 cos2 θ + κ2 sin2 θ,
i.e. normal curvature κn,v can be exxpressed by κ1 , κ2 as
κn,v = κ1 cos2 θ + κ2 sin2 θ.
Proof. This is a straightforward computation: Since SP (e1 ) = κ1 e1 , SP (e2 ) =
κ2 e2 ,
IIP (v, v) =
=
=
=
SP (v) · v
(SP (cosθe1 + sin θe2 )) · (cosθe1 + sin θe2 )
(cosθκ1 e1 + sin θκ2 e2 ) · (cosθe1 + sin θe2 )
κ1 cos2 θ + κ2 sin2 θ,
11
note that here we use the important property that e1 · e2 = 0. Here
is the proof: Since Sp is symmetric, (Sp (e1 )) · e2 = e1 · (Sp (e2 )). This
combining with the fact that Sp (e1 ) = κ1 e1 , Sp (e2 ) = κ2 e2 , we get
κ1 e1 · e2 = κ2 e1 · e2 , =⇒ (κ1 − κ2 )e1 · e2 = 0 =⇒ e1 · e2 = 0 . Q.E.D.
10. Let x(u, v) = (f (v) cos u, f (v) sin u, v), v ∈ I, 0 < u < 2π, be a parameterization of a surface of revolution S. Show that S is minimal (i.e. its Gauss
curvature is identically zero) then S is part of a catenoid (i..e you need to
find out what f is).
Solution: xu = (−f (v) sin u, f (v) cos u, 0), xv = (f 0 (v) cos u, f 0 (v) sin u, 1), so
the ffirst fundamental form is E = f (v), F = 0 and G = 1 + (f 0 (v))2 . Also,
we have xu × xv = (f (v) cos u, f (v) sin u, −f (v)f 0 (v)), hence
n=
(f (v) cos u, f (v) sin u, −f (v)f 0 (v))
xu × xv
=
.
kxu × xv k
f (v)(1 + f 02 (v))1/2
Notice that, xuu = (−f (v) cos u, −f (v) sin u, 0), xuv = (−f 0 (v) sin u, f 0 (v) cos u, 0),
xvv = (f 00 (v) cos u, f 00 (v) sin u, 0), we have the second fundamental form
e = n · xuu =
−f (v)
.
f (v)(1 + f 02 (v))1/2
f =0
and
g = n · xvv =
f (v)f 00 (v)
.
f (v)(1 + f 02 (v))1/2
By the formula for mean curvature
H=
eG − 2f F + gE
2(EG − F 2 )
and by the definition, we have that S is minimal if only if H ≡ 0, so eG −
2f G + gE = 0, i.e. we have
−f (v)(1 + f 02 (v)) + f 2 (v)f 00 (v) = 0.
Solving this differential equation yields f (v) = cosh v. So S is part of a
catenoid.
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11. Let M be a surface in R3 whose first fundamental form is given by
E = 1, F = 0 and G = u2 .
(i) Compute its Christoffel symbols.
(ii) Calculate its Gauss curvature.
Solution: Omitted.
12. The type of problems that the first and second fundamental forms are
given, verify whether the Gauss equations and Codazzi-Mainardi equations
hold (so the surface exists).
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