Lecture Notes for 120 - UCLA Department of Mathematics

B.4. SURFACES GIVEN BY AN EQUATION 186
Next we make the claim that
II (X, Y ) =
=
1
|rF | I(D XrF, Y )
1
|rF | Y · D XrF
where D X is the directional derivative. We can only evaluate II on tangent vectors,
but Y · D X rF clearly makes sense for all vectors. This has the advantage that we
can even use Cartesian coordinates in R 3 for our tangent vectors. First we show
that
L (X) =
Select a parametrization q (u, v) such that
@q
@u ⇥ @q
@v
@q
@u ⇥ @q
@v
D X N
= rF
|rF |
The Weingarten equations then tell us that
✓ ◆ @q
L = @N
@w @w =
D N @q
@w
We can now return to the second fundamental form. Let Y be another tangent
vector then, Y ·rF =0so
II (X, Y ) = I(L (X) ,Y)
= Y · D X N
✓
1
= Y · D X
|rF |
1
= Y ·
|rF | D XrF
◆
rF + Y ·
1
|rF | D XrF
Note that even when X is tangent it does not necessarily follow that D X rF is also
tangent to the surface.
In case @F
@z
6=0we get a relatively simple orthogonal basis for the tangent
space. In case @F
@x = @F
@y
=0we can simply use
X =(1, 0, 0) ,Y=(0, 1, 0)
otherwise we obtain an orthogonal basis by using
✓ @F
X =
@y , @F ◆
@x , 0
Y =
@F @F
@z @x , @F @F
@z @y ,
✓ ◆ 2 ✓ ◆ !! 2 @F @F
+
@x @y
With that basis the Weingarten map can then be calculated as
[L] = [I]
1 [II]
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2 |X| 0
=
0 |Y |
2
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II (X, X) II(X, Y )
II (X, Y ) II(Y,Y )