Solutions of Tenth Homework
Solution of 9.3 Ex. 8: Define the function H(s, t) = (st, s + t),
(s, t) ∈ R2 . Then, we have G(s, t) = (F ◦ H)(s, t) for (s, t) ∈ R2 . By
the chain rule, we have
#
"
∂f1
∂f1
(st,
s
+
t)
(st,
s
+
t)
t s
∂x
∂y
dG(s, t) = dF (H(s, t))◦dH(s, t) = ∂f2
2
(st, s + t) ∂f
(st, s + t) 1 1
∂x
∂y
"
#
∂f1
∂f1
∂f1
∂f1
(st,
s
+
t)t
+
(st,
s
+
t)
(st,
s
+
t)s
+
(st,
s
+
t)
∂x
∂y
∂x
∂y
= ∂f
.
∂f2
∂f2
∂f2
2
(st, s + t)t + ∂y (st, s + t) ∂x (st, s + t)s + ∂y (st, s + t)
∂x
Solution of 9.3 Ex. 14: Since matrix multiplication is a continuous
function and the functions h 7−→ A(h) and h 7−→ h are also continuous,
we have
lim A(h)h = A(0)0 = 0.
h→0
Solution of 9.5 Ex. 1: The first partial derivatives of f are
∂f
∂f
(x, y) = 2x + y,
(x, y) = x.
∂x
∂y
The second partial derivatives are
∂ 2f
∂ 2f
∂ 2f
(x,
y)
=
2,
(x,
y)
=
1,
(x, y) = 0.
∂x2
∂x∂y
∂y 2
Therefore, all third partial derivatives vanish.
We have f (1, 2) = 3 and df (1, 2) = (4, 1). It follows that Taylor
formula is
f (x, y) = 3 + 4(x − 1) + (y − 2) + (x − 1)2 + (x − 1)(y − 2).
Solution of 9.5 Ex. 8: The first partial derivatives of f are:
∂f
∂f
(x, y) = −4x − 2y,
(x, y) = −2x − 2y.
∂x
∂y
Since they are continuous functions, f is differentiable and the differential is given by
df (x, y) = −4x − 2y −2x − 2y .
The coordinates of stationary points of f satisfy the equations
−4x − 2y = 0
− 2x − 2y = 0.
1
2
From the second equation we see that x = −y. From the first we
conclude that 0 = −4x − 2y = −4x + 2x = −2x. Hence x = 0 and
y = 0, i.e., the only stationary point is (0, 0).
The matrix of second derivatives is
−4 −2
.
−2 −2
Its determinant is 4. Hence, the quadratic form is negative definite and
the point (0, 0) is a local maximum.
Solution of 9.5 Ex. 9: The first partial derivatives of f are:
∂f
∂f
(x, y) = 2x − 2y,
(x, y) = 3y 2 + 2y − 2x − 3.
∂x
∂y
Since they are continuous functions, f is differentiable and the differential is given by
df (x, y) = 2x − 2y 3y 2 + 2y − 2x − 3 .
The coordinates of stationary points of f satisfy the equations
2x − 2y = 0
3y 2 + 2y − 2x − 3 = 0.
From the first equation we see that x = y. From the second we conclude
that 3y 2 − 3 = 0. Hence y 2 = 1 and y = ±1. Hence, the stationary
points are (1, 1) and (−1, −1).
The matrix of second derivatives is
2
−2
.
−2 6y + 2
Hence at the stationary point (1, 1) it is equal to
2 −2
.
−2 8
Its determinant is 12. Hence, the quadratic form is positive definite and
the point (1, 1) is a local minimum. At the stationary point (−1, −1)
it is equal to
2 −2
.
−2 −4
Its determinant is −12. Hence, the quadratic form is indefinite and the
point (1, 1) is a saddle point.