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In-Depth Information
Compare this with Example 8.2.2. This parameterization is differentiable everywhere
but not regular and property (4) fails when x is ±1. Nonregularity may be a problem
if one is interested in the tangent plane at those points where the parameterization is
not regular because the standard way to find an equation for the tangent plane is to
compute the normal using Corollary 8.4.4. Therefore, if condition (4) is important,
then one needs to check that the parameterization at hand is in fact a regular para-
meterization at the points in question.
p
=
(
)
To find the tangent plane
X
to
S
2
at the point
1
2
1
2
8.4.5. Example.
,,
.
Solution.
Let us use the parameterization
(
)
2
2
(
)
=
F uv
,
uv
, ,
1
-
u
-
v
for the upper hemisphere. Then
∂
∂
F
u
uv
∂
∂
F
v
uv
Ê
Ë
ˆ
¯
Ê
Ë
ˆ
¯
=
10
,,
-
and
=
01
,,
-
.
u
v
2
2
2
2
1
--
1
--
(
)
=
p
1
2
Since
,
F
,
0
∂
∂
F
1
2
∂
∂
F
1
2
Ê
Ë
ˆ
¯
Ê
Ë
ˆ
¯
(
)
=
(
)
,
0101
=-
, ,
and
,
0
, ,
0
u
v
are a basis for
X
. A normal vector for
X
is (1,0,1) = (1,0,-1) ¥ (0,1,0). These answers
clearly agree with one's intuition of what the plane should be.
Using a parameterization and Corollary 8.4.4 to determine the tangent plane
involves a fair amount of computation. It turns out that if we are able to present our
surface as the zeroes of a function, then it is much easier to get equations for the
tangent planes. Compare the next result to Proposition 4.5.7.
8.4.6. Proposition.
Let
M
be a manifold in
R
n
and suppose that f :
R
n
Æ
R
is a dif-
ferentiable function such that
M
= V(f). Let
a
Œ
M
. The tangent plane
X
of
M
at
a
is
defined by the equation
—
()
∑-
(
)
=
f
apa
0.
(8.5)
If
a
= (a
1
,a
2
,...,a
n
) and
p
= (x
1
,x
2
,...,x
n
), then this equation can be written as
∂
∂
f
x
∂
∂
f
x
()
(
)
+
()
(
)
=
a
xa
-
...
+
a
xa
-
0
.
(8.6)
11
nn
1
n
Proof.
If g :[-c,c] Æ
M
is any curve in
M
through
a
, then
(
g
()
)
= 0
ft
