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(MATH321)[2005](f)final~ma_yxf^_10504.pdf
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MATH321 Final/2005
~
1. (40 Marks) Consider the surface S given by the parametrization X(u, v)= (u, v, uv2).
(a)
Find the .rst fundamental form under this parametrization.

(b)
Find the second fundamental form under the parametrization.

(c)
Compute the Gauss curvature K of the surface.

(d)
Find the principal directions and principal curvatures at the ori-gin.

(e)
Find asymptotic curves.

(f)
Find normal curvature of the curve (t)=(t, t, t3) on the surface at t = 1.

(g)
Find elliptic points, hyperbolic points, parabolic points and plan-ner points.


2. (15 Marks) Let p be a point on a regular surface S. Assume p is not an umbilical point. Suppose that, for any two tangent directions ~v and

w~of S at p with angle , where 0 , the sum of the normal
2

curvatures along ~v and w~is a constant, prove that =.
2
3.
(15 Marks) If a sphere intersects a surface S at a constant angle, prove that the curve of intersection is a line of curvature on S.

4.
(15 Marks) Let (s) be a regular curve parametrized by arc length ~


s with non-vanishing curvature. Consider the surface S: X(s, v)=
(s)+ v~n(s), where ~n(s) is the unit normal vector to . Prove that the ~
tangent planes of S along any coordinate curve X(s0,v) for a .xed s0 are the same if and only if is a plane curve.
5. (15 Marks) Let (s) be a regular curve on a regular surface S where s
is an arc length parameter. Suppose that the curvature of is nonzero ~
everywhere. Let N(s) be the unit normal vector of the surface along
~
the curve . Let w~(s)= N(s) 0(s). Prove that the vector .eld w~(s) is parallel along if and only if is a geodesic.
2),
4
Xu Xv Xu Xu
Xu Xv Xv Xv
Xuu Xuv
~
Xu Xv
~~
~
~
eg . f2 .4v2
~~
(c) K == .
EG . F 2 (v4 +4u2v2 + 1)2
~
(d) Since at the origin e = f = g = 0, the second fundamental form is a zero form. Therefore principal curvatures are zeros and all
~
directions are principal directions.
~~~
1. (a)
= (1, 0,v

= (0, 1, 2uv). E =

=1+ v

,

3 22
F

=2uv

G =

=1+4u



=

v

,

.

Xvv = (0, 0, 2u). (.v2 , .2uv, 1)
N(u, v)=
~~
(b)
= (0, 0, 0),

= (0, 0, 2v),

2
=(.v

, .2uv, 1).



.

v4 +4u2v2 +1
e = Xuu N =0,
~
2v
~


f

= Xuv N =
~~
,

v4 +4u2v2 +12u

~N

g = Xvv
~
=

.

v4 +4u2v2 +1
(e) Let
~
X(u(t),v(t)) be an asymptotic curve, then we have the equa-tion eu 0(t)2 +2fu0(t)v 0(t)+ gv 0(t)2 =0,
4vu 0 v 0 +2uv 0 v 0 =0.
Thus either v0 = 0 or (u2v)0 = 0. Therefore the asymptotic curves are v = const, u = 0, or u2v = c where c = 0. 6

0(1) = (1, 1, 3). Thus |0(1)| =
1, v0(t) = 1. Thus we can

~
X(t, t). u(t)= t, v(t)= t. We have u
(f) Write (t) 11. Let
=

0(t)
=

write
0(1) 1

(1, 1).



~
~
XuXv
Therefore the normal curvature is

0 000
uuvv6
e(1, 1)( )2 +2f(1, 1) + g(1