Download Lecture Notes on Multivariable Calculus

Example 3.2.2.
a) (Nonlinear system of equations)
Consider the system of equations
x3 + y13 + y23 − 7
0
f (x, y1 , y2 ) =
=
.
xy1 + y1 y2 + y2 x + 2
0
The function f is zero at the point (2, −1, 0) and
3x2
3y12
3y22
Df (x, y1 , y2 ) =
,
y1 + y2 x + y2 x + y1
hence
Dy f (2, −1, 0) =
3 0
2 1
with
detDy f (2, −1, 0) = 3 6= 0.
The Implicit Function Theorem implies that there exist open neighbourhoods I of
2 and V ⊆ R2 of (−1, 0) and a continuously differentiable function g : I → V , with
g(2) = (−1, 0), such that
f (x, y1 , y2 ) = 0
⇔
y = (y1 , y2 ) = g(x) = (g1 (x), g2 (x))
for all x ∈ I, y ∈ V . Furthermore, the derivative of g in x0 = 2 is given by
−1 1
12
−4
3 0
12
1 0
=−
.
=
Dg(2) = −
−1
9
2 1
−1
3 −2 3
b) The function f : R4 → R2 is given by
f (x, y, u, v) =
x2 + uy + ev
2x + u2 − uv
.
Consider the point (2, 5, −1, 0) such that f (2, 5, −1, 0) = (0, 5)T . The Jacobian
matrix of f is
2x u
y
ev
.
Df (x, y, u, v) =
2 0 2u − v −u
Hence
D(u,v) f (x, y, u, v) =
y
ev
2u − v −u
and
D(u,v) f (2, 5, −1, 0) =
5 1
−2 1
.
Since det Df (2, 5, −1, 0) = 7 6= 0, the Implicit Function Theorem implies that
there exist open neighbourhoods U ⊂ R2 of (2, 5) and V ⊂ R2 of (−1, 0) and a
function g ∈ C 1 (U, V ) with g(2, 5) = (−1, 0) and f (x, y, g(x, y)) = (0, 5)T for all
(x, y) ∈ U . We can also compute that
−1 1
5 1
4 −1
2 −1
Dg(2, 5) = −
=−
.
−2 1
2 0
7 18 −2
20