Graphics Reference
In-Depth Information
by hypothesis. Differentiating this equation using the chain rule yields
—
()
∑
()
=
f
a
g
0 .
In other words, the gradient of f is orthogonal to every tangent vector. It follows that
—f gives us a normal vector of the tangent plane that we are trying to find.
8.4.7. Example.
To redo Example 8.4.5 using Propositions 8.4.6.
Solution.
Define
2
2
2
fxyz
(
,,
)
=++-
x
y
z
1
.
The sphere is the zero set of f and —f = (2x,2y,2z). According to Proposition 8.4.6,
1
2
1
2
Ê
Ë
ˆ
¯
=
(
)
—
f
,,
0
20 2
,,
will be a normal vector to the plane. This agrees with the result in Example 8.4.5.
8.4.8. Example.
To show that the lines normal to the tangent planes of an arbitrary
sphere pass through its center.
Solution.
Consider the sphere
S
of radius r with center
p
0
. If
2
2
,
()
=-
f
ppp
-
r
0
then
S
= V(f). It is easy to check that
—
()
=-
(
)
f
ppp
2
0
.
Since the point-normal form of the line
L
through
p
and
p
0
is
p
+ t(
p
-
p
0
), the normal
—f(
p
) is a direction vector and we are done. See Figure 8.14.
Now we look at how differentiable maps induce natural maps on tangent spaces.
Let
n
k
f
:
M
Æ N
be a differentiable map between differentiable manifolds. Let
p
be a point of
M
n
and
let
q
= f(
p
). Define
(
)
Æ
(
)
n
k
Df
:
TM
T
N
pp
q
as follows: Let
v
be any vector in the tangent space T
p
(
M
n
) and let g(t) be any curve
lying
M
n
with g(0) =
p
and g¢(0) =
v
. Then
