Tarea_I_especiales

Parcial 3
Cristian Camilo Perez Puentes
1000137144
crperezp@unal.edu.co
1. Encontrar factores de escala, vectores unitarios, probar su ortogonalidad, unitariedad y
regla de la mano derecha, para la coordenadas esfericas.
Dada la transformacion de coordenadas esfericas:
x = ρSin(θ)Cos(φ)
y = ρSin(θ)Sin(φ)
x = ρCos(θ)
entonces las coordenadas cartesianas quedan:
r = ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)ˆk
por lo que los factores de escala son:
h ρ = | ∂ (
)
ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)ˆk |
∂ρ
= |Sin(θ)Cos(φ)î + Sin(θ)Sin(φ)ĵ + Cos(θ)ˆk|
= √ Sin 2 (θ)Cos 2 (φ) + Sin 2 (θ)Sin 2 (φ) + Cos 2 (θ)
= √ Sin 2 (θ) + Cos 2 (θ)
= 1
∂
h φ = | (ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)) |
∂φ
= | − ρSin(θ)Sin(φ)î + ρSin(θ)Cos(φ)ĵ|
= √ a 2 Sin 2 (θ)Cos 2 (φ) + a 2 Sin 2 (θ)Sin 2 (φ)
= ρSin(θ) √ Sin 2 (φ) + Cos 2 (φ)
= ρSin(θ)
h θ = | ∂ (
)
ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)ˆk |
∂θ
= |ρCos(θ)Cos(φ)î + ρCos(θ)Sin(φ)ĵ − ρSin(θ)ˆk|
= √ a 2 Cos 2 (θ)Cos 2 (φ) + a 2 Cos 2 (θ)Sin 2 (φ) + a 2 Sin 2 (θ)
= ρ √ Sin 2 (θ) + Cos 2 (θ)
= ρ
Luego los vectores unitario quedas como
ê ρ = 1 ( ) ∂r
1 ∂ρ
= 1 ( )
∂
1 ∂ρ (ρSin(θ)Cos(φ) + ρSin(θ)Sin(φ) + ρCos(θ))
= Sin(θ)Cos(φ)î + Sin(θ)Sin(φ)ĵ + Cos(θ)ˆk
1

ê θ = 1 ρ
( ) ∂r
∂θ
= 1 ( ∂
(ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)ˆk)
ρ ∂θ
= Cos(θ)Cos(φ)î + Cos(θ)Sin(φ)ĵ − Sen(θ)ˆk
( )
1 ∂r
ê φ =
ρien(θ) ∂φ
( )
1 ∂
=
(ρSin(θ)Cos(φ)î + ρSin(θ)Sin(φ)ĵ + ρCos(θ)ˆk)
ρSin(θ) ∂φ
1
= (−Sin(θ)Sin(φ)î + Sin(θ)Cos(φ)ĵ + 0ˆk)
Sin(θ)
= −Sin(φ)î + Cos(φ)ĵ + 0ˆk
Vectores de los cuales se puede ver que:
|ê| = |Sin(θ)Sin(φ)î + Sin(θ)Cos(φ)ĵ + Cos(θ)ˆk|
= √ Sin 2 (θ)Sin 2 (φ) + Sin 2 (θ)Cos 2 (φ) + Cos 2 (θ)
= √ Sin 2 (θ) + Cos 2 (θ)
= 1
|ê| = |Cos(θ)Sin(φ) + Cos(θ)Cos(φ)î − Sen(θ)ˆk|
= √ Cos 2 (θ)Sin 2 (φ) + Cos 2 (θ)Sin 2 (φ) + Sin 2 (θ)
= √ Cos 2 (θ) + Sin 2 (θ)
= 1
|ê| = | − Sin(φ)î + Cos(φ)ĵ + 0ˆk|
= √ Sin 2 (φ) + Cos 2 (φ)
= 1
Por lo tanto se verifica que la base es unitaria. Ademas:
ê · ê = (Sin(θ)Cos(φ)î + Sin(θ)Sin(φ)ĵ + Cos(θ)ˆk) · (Cos(θ)Cos(φ)î + Cos(θ)Sin(φ)ĵ − Sin(θ)ˆk)
= Sin(θ)Cos(θ)Cos 2 (φ) + Sin(θ)Cos(θ)Sin 2 (φ) − Sin(θ)Cos(θ)
= Sin(θ)Cos(θ) − Sin(θ)Cos(θ)
= 0
ê · ê = (Sin(θ)Cos(φ)î + Sin(θ)Sin(φ)ĵ + Cos(θ)ˆk) · (−Sin(φ)î + Cos(φ)ĵ + 0ˆk)
= −Sin(θ)Cos(φ)Sin(φ) + Sin(θ)Sin(φ)Cos(φ)
= 0
ê · ê = (Cos(θ)Cos(φ)î + Cos(θ)Sin(φ)ĵ − Sin(θ)ˆk) · (−Sin(φ)î + Cos(φ)ĵ + 0ˆk)
= −Cos(θ)Cos(φ)Sin(φ) + Cos(θ)Sin(φ)Cos(φ)
= 0
Por lo que la base es ortogonal. Y tambien:
)
2

item [2.] Encontrar factores de escala, vectores unitarios, probar su ortogonalidad, unitariedad
y regla de la mano derecha, para la coordenadas esferoidales oblatas.
x = aCosh(ξ)Cos(η)Cos(φ)
y = aCosh(ξ)Cos(η)Sin(φ)
x = aSinh(ξ)Sin(η)
entonces las coordenadas cartesianas quedan:
r = aCosh(ξ)Cos(η)Cos(φ)î + aCosh(ξ)Cos(η)Sin(φ)ĵ + aSinh(ξ)Sin(η)ˆk
por lo que los factores de escala son:
h ξ = | ∂ (
)
aCosh(ξ)Cos(η)Cos(φ)î + aCosh(ξ)Cos(η)Sin(φ)ĵ + aSinh(ξ)Sin(η)ˆk |
∂ξ
= |aSin(ξ)Cos(η)Cos(φ)îaSin(ξ)Cos(η)Sin(φ)ĵ + aCos(ξ)Sin(η)ˆk|
= a √ Sinh 2 (ξ)Cos 2 (η)Cos 2 (φ) + Sinh 2 (ξ)Cos 2 (η)Sin 2 (φ) + Cosh 2 (ξ)Sin 2 (η)
= a √ Sinh 2 (ξ)Cos 2 (η) + Cosh 2 (ξ)Sin 2 (η)
= a √ Sinh 2 (ξ)Cos 2 (η) + (1 + Sinh 2 (ξ))Sin 2 (η)
= a √ Sinh 2 (ξ)Cos 2 (η) + Sin 2 (η) + Sinh 2 (ξ)Sin 2 (η)
= a √ Sinh 2 (ξ) + Sin 2 (η)
∂
h φ = | (aCosh(ξ)Cos(η)Cos(φ)î + aCosh(ξ)Cos(η)Sin(φ)ĵ + aSinh(ξ)Sin(η)) |
∂φ
= | − aCosh(ξ)Cos(η)Sin(φ)î + aCosh(ξ)Cos(η)Cos(φ)ĵ|
= √ a 2 Cosh 2 (ξ)Cos 2 (η)Cos 2 (φ) + a 2 Cosh 2 (ξ)Cos 2 (η)Sin 2 (φ)
= aCosh(ξ)Cos(η) √ Sin 2 (φ) + Cos 2 (φ)
= aCosh(ξ)Cos(η)
h η = | ∂ (
)
aCosh(ξ)Cos(η)Cos(φ)î + aCosh(ξ)Cos(η)Sin(φ)ĵ + aSinh(ξ)Sin(η)ˆk |
∂η
= |aCosh(ξ)Sin(η)Cos(φ)î + aCosh(ξ)Sin(η)Sin(φ)ĵ − aSinh(ξ)Cos(η)ˆk|
= √ a 2 Cosh 2 (ξ)Sin 2 (η)Cos 2 (φ) + a 2 Cosh 2 (ξ)Sin 2 (η)Sin 2 (φ) + a 2 Sinh 2 (ξ)Cos 2 (η)
= a √ Cosh 2 (ξ)Sin 2 (η) + Sinh 2 (ξ)Cos 2 (η)
= a √ Sinh 2 (ξ)Cos 2 (η) + (1 + Sinh 2 (ξ))Sin 2 (η)
= a √ Sinh 2 (ξ)Cos 2 (η) + Sin 2 (η) + Sinh 2 (ξ)Sin 2 (η)
= a √ Sinh 2 (ξ) + Sin 2 (η)
Luego los vectores unitario quedas como
ê ξ = 1 ( ) ∂r
h ξ ∂ξ
= 1 ( )
∂
h ξ ∂ξ (aCosh(ξ)Cos(η)Cos(φ) + aCosh(ξ)Cos(η)Sin(φ) + aSinh(ξ)Sin(η))
= a )
(Sinh(ξ)Cos(η)Cos(φ)î + Sinh(ξ)Cos(η)Sin(φ)ĵ + Cosh(ξ)Sin(η)ˆk
h ξ
3

ê η = 1 ( ) ∂r
h η ∂η
= 1 ( )
∂
h η ∂η (aCosh(ξ)Cos(η)Cos(φ) + aCosh(ξ)Cos(η)Sin(φ) + aSinh(ξ)Sin(η))
= a )
(−Sin(η)Cosh(ξ)Cos(φ)î − Sin(η)Cosh(ξ)Sin(φ)ĵ + Cos(η)Sinh(ξ)ˆk
h η
ê φ = 1 ( ) ∂r
h φ ∂φ
( )
∂
(aCosh(ξ)Cos(η)Cos(φ)î + aCosh(ξ)Cos(η)Sin(φ)ĵ + aSinh(ξ)Sin(η)ˆk)
∂φ
1
=
aCosh(ξ)Cos(η)
1
=
(−Cosh(ξ)Cos(η)Sin(φ)î + Cosh(ξ)Cos(η)Cos(φ)ĵ + 0ˆk)
Cosh(ξ)Cos(η)
= −Sin(φ)î + Cos(φ)ĵ + 0ˆk
Vectores de los cuales se puede ver que:
|ê ξ | = a h ξ
|Sinh(ξ)Cos(η)Cos(φ)îSinh(ξ)Cos(η)Sin(φ)ĵ + Cosh(ξ)Sin(η)ˆk|
= a h ξ
√
Sinh2 (ξ)Cos 2 (η)Sin 2 (φ) + Sinh 2 (ξ)Cos 2 (η)Cos 2 (φ) + Cosh 2 (ξ)Sin 2 (η)
= a h ξ
√
Sinh2 (ξ)Cos 2 (η) + Cosh 2 (ξ)Sin 2 (η)
= a h ξ
√
Sinh2 (ξ)Cos 2 (η) + (1 + Sinh 2 (ξ))Sin 2 (η)
= a h ξ
√
Sinh2 (ξ)Cos 2 (η) + Sin 2 (η) + Sinh 2 (ξ)Sin 2 (η)
= a h ξ
√
Sinh2 (ξ) + Sin 2 (η)
a
=
a √ √
Sinh2 (ξ) + Sin 2 (η)
Sinh 2 (ξ) + Sin 2 (η)
= 1
|ê η | = a | − Sin(η)Cosh(ξ)Cos(φ)î − Sin(η)Cosh(ξ)Sin(φ)ĵ + Cos(η)Sinh(ξ)ˆk|
h η
= a √
Sin2 (η)Cosh
h 2 (ξ)Sin 2 (φ) + Sin 2 (η)Cosh 2 (ξ)Cos 2 (φ) + Cos 2 (η)Sinh 2 (ξ)
η
= a √
Sin2 (η)Cosh
h 2 (ξ) + Cos 2 (η)Sinh 2 (ξ)
η
= a √
Sinh2 (η)Cos
h 2 (ξ) + (1 + Sinh 2 (η))Sin 2 (ξ)
η
= a √
Sinh2 (η)Cos
h 2 (ξ) + Sin 2 (ξ) + Sinh 2 (η)Sin 2 (ξ)
η
= a √
Sinh2 (η) + Sin
h 2 (ξ)
η
a
=
a √ √
Sinh2 (η) + Sin 2 (ξ)
Sinh 2 (η) + Sin 2 (ξ)
= 1
4

|ê φ | = | − Sin(φ)î + Cos(φ)ĵ + 0ˆk|
= √ Sin 2 (φ) + Cos 2 (φ)
= 1
Por lo tanto se verifica que la base es unitaria. Ademas:
ê ξ · ê η = a )
(Sinh(ξ)Cos(η)Cos(φ)î + Sinh(ξ)Cos(η)Sin(φ)ĵ + Cosh(ξ)Sin(η)ˆk ·
h ξ
a
(
)
−Sin(η)Cosh(ξ)Cos(φ)î − Sin(η)Cosh(ξ)Sin(φ)ĵ + Cos(η)Sinh(ξ)ˆk
h η
= − Sinh(ξ)Cosh(ξ)Cos(η)Sin(η)Cos 2 (φ)+
− Sinh(ξ)Cosh(ξ)Cos(η)Sin(η)Sin 2 (φ)+
+ Sinh(ξ)Cosh(ξ)Cos(η)Sin(η)
= −Sinh(ξ)Cosh(ξ)Cos(η)Sin(η) + Sinh(ξ)Cosh(ξ)Cos(η)Sin(η)
= 0
ê ξ · ê φ = a )
(Sinh(ξ)Cos(η)Cos(φ)î + Sinh(ξ)Cos(η)Sin(φ)ĵ + Cosh(ξ)Sin(η)ˆk ·
h ξ
(−Sin(φ)î + Cos(φ)ĵ + 0ˆk))
= −Sinh(ξ)Cos(η)Cos(φ)Sin(φ) + Sinh(ξ)Cos(η)Sin(φ)Cos(φ)
= 0
ê η · ê φ = a )
(−Sin(η)Cosh(ξ)Cos(φ)î − Sin(η)Cosh(ξ)Sin(φ)ĵ + Cos(η)Sinh(ξ)ˆk ·
h η
= Sin(η)Cosh(ξ)Cos(φ)Sin(φ) − Sin(η)Cos(ξ)Sin(φ)Cos(φ)
= 0
Por lo que la base es ortogonal. Y tambien:
(−Sin(φ)î + Cos(φ)ĵ + 0ˆk))
5