ℝ2 → ℝ
ℝ2 → ℝ2 x
fxx (x, y) =
∂f ∂f ∂2 f ( ) = 2 (x, y) ∂x ∂x ∂x
y ∂f ∂f ∂2 f fyy (x, y) = ( ) = 2 (x, y) ∂y ∂y ∂y
fxy (x, y) =
∂f ∂f ∂2 f ( )= (x, y) ∂x ∂y ∂x ∂y
f
f(x, y)
∂2 f ∂x2 ∂3 f ∂x3
(x, y),
⋮
∂2 f ∂y2
∂3 f ∂y3
(x, y)
(x, y)
⋮ ∂n f ∂xn
f: A ⊆ ℝ2 → ℝ
(x, y)
(x, y),
∂n f ∂yn
(x, y)
𝐶2
(x0 , y0 ) ∈ A ⊆ ℝ2 ∂2 f ∂2 f = fxy (x0 , y0 ) = fyx (x0 , y0 ) = ∂x ∂y ∂y ∂x
f = y𝑒 𝑥 + 𝑥𝑦 2
∂f = y𝑒 𝑥 + 𝑦 2 , ∂x
∂2 f = 𝑒 𝑥 + 2𝑦 ∂x ∂y
∂f = 𝑒 𝑥 + 2𝑥𝑦, ∂y
∂2 f = 𝑒 𝑥 + 2𝑦 ∂y ∂x
𝛾(𝑡): [𝑎, 𝑏] → ℝ2
𝛾(𝑡): [𝑎, 𝑏] → ℝ3 γ(n) (t)
n≥2
dn γ dtn
𝐹
𝐶2
div rot F = ∇ ∙ (∇ × F) = 0 div rot F = 0
F(x, y, z) = xy + yz +
xz ∇×F 𝐢 𝐣 𝐤 ∂ ∂ ∂ ∇×F =| | = −𝑦𝐢 − z𝐣 − x𝐤 ∂x ∂y ∂z xy yz xz ∂
∂
∂
𝐹 = ∇ ∙ (∇ × F) = − (∂x , ∂y , ∂z) ∙ (𝑦, 𝑧, 𝑥) = 0 div rot F = 0
f: ℝ3 → ℝ
∂2 f ∂2 f ∂2 f ∇ f = ∇ ∙ (∇f) = 2 + 2 + 2 ∂x ∂y ∂z 2
f(x, y, z) = 𝑥 3 𝑦 2 𝑧 4 + 5𝑥𝑦𝑧
∂f = 3𝑥 2 𝑦 2 𝑧 4 + 5𝑦𝑧, ∂x
∂2 f = 6𝑥𝑦 2 𝑧 4 ∂x 2
∂f = 2𝑥 3 𝑦𝑧 4 + 5𝑥𝑧, ∂y
∂2 f = 2𝑥 3 𝑧 4 ∂y 2
∂f = 4𝑥 3 𝑦 2 𝑧 3 + 5𝑥𝑦, ∂z
∂2 f = 12𝑥 3 𝑧 2 𝑧 2 ∂z 2
∇2 f = 6𝑥𝑦 2 𝑧 4 + 2𝑥 3 𝑧 4 + 12𝑥 3 𝑧 2 𝑧 2
f: A ⊆ ℝ2 → ℝ
(𝑥0 , 𝑦0 ) ∈ 𝐴
(V(x0 ,y0 ) )
(x, y) ∈ V f(x, y) ≥ f(x0 , y0 )
f: A ⊆ ℝ2 → ℝ
(𝑥0 , 𝑦0 ) ∈ 𝐴
(V(x0 ,y0 ) )
(x, y) ∈ V f(x, y) ≤ f(x0 , y0 )
(𝑥0 , 𝑦0 ) f Df(x0 , y0 ) = 0
(𝑥0 , 𝑦0 )
V(0,0)
(x0 , y0 )
f: A ⊆ ℝ2 → ℝ Df(x0 , y0 ) = 0
V(x0,y0)
(x0 , y0 ) ∈ A (x0 , y0 )
f
f(x, y) = 2𝑥 2 + 3𝑦 2 f(x, y)
∂f ∂x
= 4x = 0
∂f ∂y
= 9y 2 = 0
f(x, y) = 2𝑥 2 + 3𝑦 2
2𝑥 2 + 3𝑦 2 ≥ 0
f: A ⊆ ℝ2 → ℝ
𝐶2
𝑉(x0 ,y0 ) ∈ A
f
f
(x0 , y0 )
∂2 f ∂2 f 2 2 2 2 2 ∂xy ∂ f ∂ f ∂ f ∂x | H𝑓(x0 , y0 ) = || 2 ( ) − ( ) 2 |= ∂x 2 ∂y 2 ∂xy ∂ f ∂ f ∂yx ∂y 2
f: A ⊆ ℝ2 → ℝ
𝐶2
(x0 , y0 ) ∈ A
(x0 , y0 )
∂2 f
Hf(x0 , y0 ) > 0
∂x2
>0
f(x0 , y0 )
∂2 f
Hf(x0 , y0 ) > 0
∂x2
<0
f(x0 , y0 )
Hf(x0 , y0 ) < 0 f(x0 , y0 ) Hf(x0 , y0 ) = 0
f(x, y) = 𝑥 4 + 𝑦 4 − 2𝑥𝑦 + 2
f ∂f = 4x 3 − 2y = 0 ∂x ∂f = 4y 3 − 2x = 0 ∂y
1
(0,0) (
,
1
)
√2 √2
(−
1 √2
,−
1
)
√2
f Hf(x0 , y0 ) ∂2 f ∂x2
= 12x 2
∂2 f ∂y2
∂2 f
= 12y 2
= −2
∂xy
2
∂2 f ∂2 f ∂2 f ) Hf(x0 , y0 = 2 ( 2 ) − ( ) = 144x 2 y 2 − 2 ∂x ∂y ∂xy
Hf(0,0) = −2 < 0 Hf (
1
,
1
) = 34 >
√2 √2
Hf (−
1 √2
,−
1
1
f(
) = 34 > 0
√2
,
1
3
)=2>0
√2 √2
f (−
1 √2
,−
1
3
)=2>0
√2
(±
1 √2
,±
1
)
√2
f: C ⊆ ℝ2 → ℝ
C f(x0 , y0 )
f(x1 , y1 )
(x0 , y0 ), (x1 , y1 ) ∈ C
C f
C
f
f: C ⊆ ℝ2 → ℝ f(x, y) = x 2 − 2𝑥 − 2𝑦 + 𝑥𝑦 + y 2 C = {(x, y)│ − 5 ≤ x ≤ 5 , −5 ≤ y ≤ 5}
x 2 − 2𝑥 − 2𝑦 + 𝑥𝑦 + y 2
C
f
f
∂f = 2𝑥 − 2 + 𝑦 = 0 ∂x ∂f = 2𝑦 + 𝑥 − 2 = 0 ∂y
2 2
f
(3 , 3)
2 2
4
f (3 , 3) = − 3
f
𝑦 = −5 7
f (2 , −5) =
C
−5 ≤ x ≤ 5 f(x, −5) = 𝑥 2 − 7𝑥 + 35
11
f(−5, −5) = 95
4
3
−5 ≤ x ≤ 5 f(x, 5) = x 2 + 3x + 15
𝑦=5
f (− 2 , 5) =
f(5,5) = 55
x=5 3
f (5, − 2) =
51
f(5, −5) = 25
4
x = −5 7
f (−5, 2) =
91 4
−5 ≤ y ≤ 5 f(5, y) = y 2 + 3y + 15
−5 ≤ y ≤ 5 f(5, y) = y 2 − 7y + 35 f(−5,5) = 25
2 2
(3 , 3)
91 4
2 2
4
(−5, −5)
f (3 , 3) = − 3
f(−5, −5) =
95
f: D ⊆ ℝ2 → ℝ
𝑔 = ℝ2 → ℝ
C1
S = {(x, y) ∈ ℝ2 │ g(x, y) = k} S ∇f(x0 , y0 ) = 𝜆∇g(x0 , y0 )
f│S
∇f(x0 , y0 ) = 𝜆∇g(x0 , y0 )
∂f ∂g −𝜆 =0 ∂x ∂x ∂f ∂g −𝜆 =0 ∂y ∂y
ℝn → ℝ
f(x, y) = 2(x 2 − y 2 )
(x0 , y0 ) ∈ 𝐷 g(x0 , y0 ) = 𝑘 ∇𝑔(x0 , y0 ) ≠ 0 𝜆∈ℝ
f│S
f(x, y) = 2(x 2 − y 2 ) ∇f(x0 , y0 )
g(x, y) = x 2 + y 2 = 1
𝜆∇g(x0 , y0 )
∂f ∂f ∇f(x0 , y0 ) = ( , ) = (4𝑥, −4𝑦) ∂x ∂y ∂g ∂g
𝜆∇g(x0 , y0 ) = 𝜆 (∂x , ∂y) = (2𝑥𝜆, 2𝑦𝜆) ≠ 0
x2 + y2 = 1
4𝑥 − 2𝑥𝜆 = 0 −4𝑦 − 2𝑦𝜆 = 0 x2 + y2 − 1 = 0
(1,0) (−1,0)
𝜆=2
(0,1) (0, −1)
𝜆 = −2
f
S = {(x, y) ∈ ℝ2 │ x 2 + y 2 = 1}
f│S