Math 150A Prof. Rabin Fall 2011 Homework Assignments
Problems identified by numbers, e.g. 2.4.7, come from the course textbook Differential Geometry
and Its Applications, 2nd edition, by John Oprea. Note that they are scattered throughout the text
Math 150A Prof. Rabin Fall 2011 Homework #0
These problems will be discussed, but not turned in or graded, at the first discussion section meeting.
They are intended as a review of some ideas from Math 20E and 20F.
1. Find an equation for the plane contai
Surface of revolution. Let r : (a, b) R+ and z : (a, b) R. Dene
x : (a, b) (, ) M R3 by
x (t, ) = (r (t) cos , r (t) sin , z (t) .
= (r (t) cos , r (t) sin , z (t) ,
= (r (t) sin , r (t) cos , 0)
(gij ) =
MATH150A Final, Fall 2012 Paul Bryan
1. Arc-length parametrisations of S 1 .
Show that the only arc-length parametrisations of the unit circle S 1 R2 are of the form
() = (cos( + 0 ), sin( + 0 ) for some xed 0 [0, 2 ).
Hint : You have two equations (one d
Let x : U x (U ) = M be a simple surface in R3 . Recall
k xk ,
xij = Lij n +
where Lij = xij , n are the coecients of the second fundamental form and
are the Christoel symbols. The second fundamental
=1 xij , x g
form is the symmetric
Reminder: everything is assumed to be C .
A version of Theorem 5.9. Let M = x (U ) be a simple surface. If a
unit speed curve : [a, b] M is the shortest curve between (a) = P and
(b) = Q, then is a geodesic.
Proof. Set-up: We have x : U x (U ) = M R3 , w
Let : (a, b) x (U ) be a unit speed curve on a simple surface. The normal
curvature of is
n (s) = (s) , n ( (s) .
We may rewrite this as
(s) , n ( (s)
(s) , n ( (s) (s) , n ( (s)
= (s) , n ( (s)
n ( s) =
since (s) , n ( (s) = 0.
Review. Let x : U x (U ) = M be a simple surface in R3 . Let P M and
let TP M be the tangent plane to M at p. The rst fundamental form assigns to
X, Y TP M their Euclidean inner product X , Y . We have
spans each tangent plane. The u
October 6, 2015
Q1: In order to nd a normal vector of the plane, we cross product v1 =
(1, 2, 3) (0, 0, 0) and v2 = (1, 1, 1) (0, 0, 0) to get (1, 2, 1) . Then, the plane
1 (x 0) 2 (y 0) + 1 (z 0) = 0
and it becomes
x 2y + z = 0.
Q2. Let (