Page 66 - Vector Analysis

62 CHAPTER 2. Differentiation of Functions of Several Variables

2. If (x0, y0) = (0, ´1), then Fy(x0, y0) = ´2 ‰ 0; thus the implicit function theorem

implies that locally y can be expressed as a function of x.

3. If ( 1 , ? ) then Fx(x0, y0) = ´1 ‰ 0 and ? thus the
(x0, y0) = ´ 3 , Fy(x0, y0) = 3 ‰ 0;

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implicit function theorem implies that locally x can be expressed as a function of y

and locally y can be expressed as a function of x.

Example 2.70. Suppose that (x, y, u, v) satisfies the equation

" xu + yv2 = 0
xv3 + y2u6 = 0

and (x0, y0, u0, v0) = (1, ´1, 1, ´1). Let F (x, y, u, v) = (xu + yv2, xv3 + y2u6). Then

F (x0, y0, u0, v0) = 0.

 B F1 B F1  []

1. Since (Dx,yF )(x0, y0, u0, v0) =  Bx By  (x0, y0, u0 , v0 ) = 1 1 is invertible,
B F2 B F2 ´1 ´2

Bx By

locally (x, y) can be expressed in terms of u, v; that is, locally x = x(u, v) and y =

y(u, v).

 B F1 B F1  [ 1] is invertible,
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2. Since (Dy,uF )(x0, y0, u0, v0) =  By Bu  (x0, y0, u0, v0) = 1
B F2 B F2 ´2

By Bu

locally (y, u) can be expressed in terms of x, v.

Example 2.71. Let f : R3 Ñ R2 be given by

f (x, y, z) = (xey + yez, xez + zey) .

Then f is of class C 1, f (´1, 1, 1) = (0, 0) and

[ ] [ xey + ez yez ]
(Df )(x, z) ey zey xez + ey
y, = ez .

[0 e ]
e 0
Since (Dy,zf )(´1, 1, 1) = is invertible, the implicit function theorem implies that the

system

" xey + yez = 0
xez + zey = 0