Graphics Reference
In-Depth Information
(
(
)
)
=
(
)(
)
fxk
,
xy
,
f G
FG
FG
o
xy
,
=
(
(
)
)(
)
p
p
p
oo
oo
xy
xy
,
,
(
(
)
)(
)
=
(
)
=
xy
,
=
y
.
Letting g(
x
) = k(
x
,
0
) proves the theorem.
Note that the chain rule for differentiation could easily be used to compute the
derivative of the function g in Theorem 4.4.7.
Let us see how the implicit function theorem applies to our earlier example
dealing with the unit circle and the function f in equation (4.10). We have that
∂
∂
f
x
∂
∂
f
y
=
2
x
and
=
2 .
y
Evaluating ∂f/∂x and ∂f/∂y at the points
A
,
B
, and
C
and checking when the values are
nonzero will show that the implicit function theorem gives us the same answers as
before.
Finally, it is important to realize that the implicit function theorem says nothing
about the
existence
of solutions to equations
(
)
=
f
xy
,
0
,
but rather is typically used to assert that the solution set,
if
it exists, can be parame-
terized locally by
(
()
)
xxx
Æ
,g
using some function g.
Consider the set of points
C
in
R
3
defined by the equations
4.4.8. Example.
2
x z
xyz
++=
-++=.
10
20
2
2
2
To show that a neighborhood of the point
p
= (1,2,-1) on
C
is a curve that can be
parameterized by a function g(x).
Solution.
Define a function
2
2
f:
RR
¥Æ
R
by
(
)
(
)
=++ -++
2
2
2
2
fxyz
,,
x
yz
1
,
x
y
z
2
.
