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  7. Cálculo (102)

Summary

Sumario Cálculo en varias variables.
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  • Course
  • Cálculo (102)
  • Institution
  • Universidad De Málaga

- Dominios y curvas de nivel. - Cálculo de límites. - Derivadas parciales. Plano tangente. - Función diferenciable en dos variables. - Derivada direccional. Gradiente. - Teorema de Schwartz. - Regla de la cadena. - Formula de Taylor. - Lagrangiano

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Document information

  • September 14, 2021
  • 6
  • 2020/2021
  • Summary

Subjects

  • ingeniería
  • cálculo
  • varias variables
  • matemáticas
  • ingeniería industrial
  • ingeniería electrónica industrial

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  • Universidad de Málaga
  • Ingeniería Electrónica Industrial
  • Cálculo (102)
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Tema 2. Cálculo en varias variables.
Dominios y curvas de nivel.
2 2
Dada una función 𝑓: 𝑅 → 𝑅 se define el dominio de la función como 𝑓 = {(𝑥, 𝑦) ∈ 𝑅 | 𝑓(𝑥, 𝑦) 𝑒𝑥𝑖𝑠𝑡𝑒}
Ejemplos:
𝑥𝑦 2
𝑓(𝑥, 𝑦) = 2 2 ; 𝐷𝑜𝑚𝑓 = 𝑅 − {(0, 0)}
𝑥 +𝑦
2 2
2 2 2
𝑓(𝑥, 𝑦) =
𝑥 +𝑦 −9
𝑥 {
; 𝐷𝑜𝑚𝑓 = (𝑥, 𝑦) ∈ 𝑅 |𝑥 + 𝑦 − 9 ≥ 0 ∧ 𝑥 ≠ 0 }
2 2 2 2 2
con la igualdad: 𝑥 + 𝑦 − 9 = 0; 𝑥 + 𝑦 = 3
2
𝑓(𝑥, 𝑦) = 𝑎𝑟𝑐𝑠𝑒𝑛(𝑥 + 𝑦); 𝐷𝑜𝑚𝑓 = (𝑥, 𝑦) ∈ 𝑅 | − 1 ≤ 𝑥 + 𝑦 ≤ 1 { }
−1≤𝑥+𝑦 ∧ 𝑥+𝑦≤1
𝑦 =− 𝑥 − 1 ∧ 𝑦=1−𝑥

2
Dada 𝑓: 𝑅 → 𝑅 tal que a cada valor de x se le hace corresponder el valor de
𝑓(𝑥, 𝑦), se llama curva de nivel k, a la curva 𝑓(𝑥, 𝑦) = 𝑘.

Ejemplos:
2 2
𝑓(𝑥, 𝑦) = 25 − 𝑥 − 𝑦
2 2 2 2 2
NIVEL 0: 25 − 𝑥 − 𝑦 = 0; 𝑥 + 𝑦 = 5
2 2 2 2 2
NIVEL 9: 25 − 𝑥 − 𝑦 = 9; 𝑥 + 𝑦 = 4
2 2 2 2 2
NIVEL 16: 25 − 𝑥 − 𝑦 = 16; 𝑥 + 𝑦 = 3
2 2 2 2 2
NIVEL 21: 25 − 𝑥 − 𝑦 = 21; 𝑥 + 𝑦 = 2
2 2 2 2 2
NIVEL 24: 25 − 𝑥 − 𝑦 = 24; 𝑥 + 𝑦 = 1
2 2 2 2 2
NIVEL 25: 25 − 𝑥 − 𝑦 = 25; 𝑥 + 𝑦 = 0

𝑓(𝑥, 𝑦) = |𝑥𝑦|
NIVEL 0: |𝑥𝑦| = 0
NIVEL 1: |𝑥𝑦| = 1
NIVEL 4: |𝑥𝑦| = 4

Cálculo de límites.
Se dice que lim 𝑓(𝑥) = 𝑙 si ∀ε > 0, ∃δ > 0, tal que si 𝑥|0 < |𝑥 − 𝑎| < δ entonces
𝑥→𝑎
|𝑓(𝑥) − 𝑙| < ε.

Ejemplo:
1 2
lim 2 𝑥 + 1 = 3
𝑥→2
∀ε > 0, ∃δ, 𝑥|0 < |𝑥 − 2| < δentonces
| 1 𝑥2 + 1 − 3| < δ
|2 |
| 𝑥 − 2| = | 1 𝑥2 − 4 | = | 1 (𝑥 + 2)(𝑥 − 2)| =
1 2
( ) 1
|𝑥 + 2||𝑥 − 2| < ε
|2 | |2 | |2 | 2
Suponemos que: 𝑓: [1. 5, 2. 5] → 𝑅
1 2ε 2ε
2
· 4. 5 · |𝑥 − 2| < ε; |𝑥 − 2| < 4.5
⇒δ = 4.5

, | |
∀ε > 0, ∃δ/||(𝑥, 𝑦) − (𝑎, 𝑏)|| < δ entonces ||𝑓(𝑥, 𝑦) − lim |< ε
|
| (𝑥,𝑦) → (𝑎,𝑏) |




Ejemplo:
3
2𝑥
Probar que lim 2 2 =0
(𝑥,𝑦) → (0,0) 𝑥 +𝑦

2 2 | 2𝑥3 |
∀ε > 0 ∃δ/ 𝑥 + 𝑦 < δ entonces | 2 2 | < ε
| 𝑥 +𝑦 |
3 2 2
| 2𝑥 | | 𝑥 | 𝑥 2 2
| 2 2 | = |2𝑥|| 2 2 | = |2𝑥| 2 2 ≤ |2𝑥| · 1 = 2|𝑥| < 2 𝑥 + 𝑦
| 𝑥 +𝑦 | | 𝑥 +𝑦 | 𝑥 +𝑦


2 2 2 2 ε ε
si quiero que 2 𝑥 + 𝑦 < ε, es decir, 𝑥 +𝑦 < 2
, basta tomar δ ≤ 2


Procedimiento para calcular un límite:
1.) Sustituir
lim 2𝑥𝑦 + 4 = 8
(𝑥,𝑦) → (1,2)
2.) Manipular y simplificar
3 3 2 2
lim
𝑥 𝑦−𝑥𝑦
= lim ( ) =
𝑥𝑦 𝑥 −𝑦 1
4 4 2 2 2 2 2
(𝑥,𝑦) → (1,1) 𝑥 −𝑦 (𝑥,𝑦) → (1,1) (𝑥 +𝑦 )(𝑥 −𝑦 )
lim
𝑥+𝑦+1−1
= lim ( 𝑥+𝑦+1−1)( 𝑥+𝑦+1+1) = lim
(𝑥+𝑦)
=
1
2 2
(𝑥,𝑦) → (1,−1) 𝑥 −𝑦 (𝑥,𝑦) → (1,−1) (𝑥2−𝑦2)( 𝑥+𝑦+1+1) (𝑥,𝑦) → (1,−1)
(𝑥+𝑦)(𝑥−𝑦)( 𝑥+𝑦+1+1) 4

2 2
𝑥
−𝑙𝑛 1+ 𝑦+1 ( 𝑥
)
𝑥
( )− ( ) +...⎤⎥⎦
−⎡⎢
𝑥 1 𝑥 1 𝑥
𝑥 2
lim
(𝑥,𝑦) → (0,0)
𝑦+1

𝑦
2 = lim
(𝑥,𝑦) → (0,0)
𝑦+1
⎣ 𝑦+1

𝑦
2
2 𝑦+1
= lim
(𝑥,𝑦) → (0,0)
2 (𝑦+1)2

𝑦
2 = lim
(𝑥,𝑦) → (0,0)
1
2 ( ) 𝑦



Definición. Dada una curva 𝐺(𝑥, 𝑦) = 0 que pasa por (𝑎, 𝑏) cuyos puntos
(𝑥, 𝑦) ∈ 𝐷𝑜𝑚𝑖𝑛𝑖𝑜 𝑑𝑒 𝑓
Supongamos 𝑦 = 𝑔(𝑥)
lim 𝑓(𝑥, 𝑦): = lim 𝑓(𝑥, 𝑔(𝑥))
(𝑥,𝑦) → (𝑎,𝑏) 𝑥→𝑎


Teorema. Si existe el límite, entonces coincide con el límite según cualquier dirección.
𝑥 𝑥
lim 𝑦
= [𝑥 = 𝑦] = lim 𝑥
=1
(𝑥,𝑦) → (0,0) 𝑥→0
𝑥 𝑥
lim 𝑦
= [𝑦 =− 𝑥] = lim −𝑥
=− 1
(𝑥,𝑦) → (0,0) 𝑥→0
𝑥 𝑥 1
lim 𝑦
= [𝑦 = 𝑚𝑥] = lim 𝑚𝑥
= 𝑚
⇒𝑚 ≠ 0
(𝑥,𝑦) → (0,0) 𝑥→0


3.) Intentar probar que el límite no existe.
3 3 4 3 2
𝑥𝑦 𝑚𝑥 𝑚𝑥
lim 2 6 = [𝑦 = 𝑚𝑥] = lim 2 6 6 = lim 6 4 =0
(𝑥,𝑦) → (0,0) 𝑥 +𝑦 𝑥→0 𝑥 +𝑚 𝑥 𝑥→0 1+𝑚 𝑥
3 3 3 6
3
lim
𝑥𝑦
2
𝑥 +𝑦
6 [
= 𝑥 = 𝑦 = lim ] 𝑦𝑦
6
𝑦 +𝑦
6 = lim
𝑦
2𝑦
6 =
1
2
(𝑥,𝑦) → (0,0) 𝑦→0 𝑦→0

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