Seja M M M dim m ⊂ R n \dim m \subset \R^n dim m ⊂ R n f f f
M = { x ∈ R n : F ( x ) = 0 } F : R n → R n − m f : R n → R \begin{array}{lll}
M = \{x \in \R^n: F(x) = 0\} & F: \R^n \to \R^{n-m} & f: \R^n \to \R
\end{array} M = { x ∈ R n : F ( x ) = 0 } F : R n → R n − m f : R n → R Como calcular máximo (ou mínimo) de f f f M M M
Seja x 0 x_0 x 0 M M M ∇ f ( x 0 ) \nabla f(x_0) ∇ f ( x 0 ) espaço normal de M M M
F = ( F 1 , F 2 , … , F n − m ) linhas de D F ( x 0 ) = ∇ F 1 ( x 0 ) , ∇ F 2 ( x 0 ) , … , ∇ F n − m ( x 0 ) \begin{darray}{c}
F=(F_1, F_2, \dots, F_{n-m})\\
\text{linhas de } DF(x_0) = \nabla F_1(x_0), \nabla F_2(x_0), \dots, \nabla F_{n-m}(x_0)
\end{darray} F = ( F 1 , F 2 , … , F n − m ) linhas de D F ( x 0 ) = ∇ F 1 ( x 0 ) , ∇ F 2 ( x 0 ) , … , ∇ F n − m ( x 0 ) Existem λ 1 , … , λ n − m ∈ R \lambda_1, \dots, \lambda_{n-m} \in \R λ 1 , … , λ n − m ∈ R
∇ ( f − λ 1 F 1 − λ 2 F 2 − ⋯ − λ n − m F n − m ) ( x 0 ) = 0 \nabla (f-\lambda_1 F_1 - \lambda_2 F_2 - \dots - \lambda_{n-m} F_{n-m})(x_0) = 0 ∇ ( f − λ 1 F 1 − λ 2 F 2 − ⋯ − λ n − m F n − m ) ( x 0 ) = 0 isto é, tais que x 0 x_0 x 0 f − λ 1 F 1 − ⋯ − λ n − m F n − m f - \lambda_1 F_1 - \dots- \lambda_{n-m} F_{n-m} f − λ 1 F 1 − ⋯ − λ n − m F n − m
DEFINIÇÃO
Seja M M M dim m \dim m dim m f : M → R f: M \to \R f : M → R C 1 C^1 C 1
M = { x ∈ R n : F 1 ( x ) = 0 , … , F n − m ( x ) = 0 } M = \{x \in \R^n: F_1(x) = 0, \dots, F_{n-m}(x) = 0 \} M = { x ∈ R n : F 1 ( x ) = 0 , … , F n − m ( x ) = 0 } a a a
{ ∇ ( f − λ 1 F 1 − ⋯ − λ n − m F n − m ) ( a ) = 0 F 1 ( a ) = 0 ⋮ F n − m ( a ) = 0 \begin{cases}
\nabla(f - \lambda_1 F_1 - \dots - \lambda_{n-m} F_{n-m}) (a) = 0\\
F_1(a) = 0\\
\quad \vdots\\
F_{n-m}(a) = 0
\end{cases} ⎩ ⎨ ⎧ ∇ ( f − λ 1 F 1 − ⋯ − λ n − m F n − m ) ( a ) = 0 F 1 ( a ) = 0 ⋮ F n − m ( a ) = 0 tal que λ j \lambda_j λ j Multiplicadores de Lagrange
Demonstração Seja γ \gamma γ M M M x 0 x_0 x 0 M M M γ ( 0 ) = x 0 \gamma(0) =x_0 γ ( 0 ) = x 0 f ( γ ( t ) ) f(\gamma(t)) f ( γ ( t )) t = 0 t=0 t = 0
d d t f ( γ ( t ) ) ∣ t = 0 = 0 \frac{\d}{\d t} f(\gamma(t)) \big|_{t=0} = 0 d t d f ( γ ( t )) t = 0 = 0 ∇ f ( γ ( 0 ) ) ⋅ γ ′ ( 0 ) = 0 ∇ f ( x 0 ) ⋅ γ ′ ( 0 ) = 0 ∀ γ caminho com γ ( 0 ) = 0 \begin{array}{ll}
\nabla f(\gamma (0)) \cdot \gamma'(0) = 0 & \nabla f(x_0) \cdot \gamma'(0) = 0 \forall \gamma \text{ caminho com } \gamma(0) = 0
\end{array} ∇ f ( γ ( 0 )) ⋅ γ ′ ( 0 ) = 0 ∇ f ( x 0 ) ⋅ γ ′ ( 0 ) = 0∀ γ caminho com γ ( 0 ) = 0 ⟹ ∇ f ( x 0 ) ∈ ( T x 0 M ) ⊥ = L { linhas de D F ( x 0 ) } \implies \nabla f(x_0) \in (T_{x_0}M)^{\perp} = \mathcal{L}\{\text{linhas de } DF(x_0)\} ⟹ ∇ f ( x 0 ) ∈ ( T x 0 M ) ⊥ = L { linhas de D F ( x 0 )} F = ( F 1 , F 2 , … , F n − m ) linhas de D F ( x 0 ) = ∇ F 1 ( x 0 ) , ∇ F 2 ( x 0 ) , … , ∇ F n − m ( x 0 ) \begin{array}{l}
F=(F_1, F_2, \dots, F_{n-m})\\
\text{linhas de } DF(x_0) = \nabla F_1(x_0), \nabla F_2(x_0), \dots, \nabla F_{n-m}(x_0)
\end{array} F = ( F 1 , F 2 , … , F n − m ) linhas de D F ( x 0 ) = ∇ F 1 ( x 0 ) , ∇ F 2 ( x 0 ) , … , ∇ F n − m ( x 0 ) ⟹ \implies ⟹ λ 1 , … , λ n − m ∈ R \lambda_1, \dots, \lambda_{n-m} \in \R λ 1 , … , λ n − m ∈ R
∇ f ( x 0 ) = λ 1 ∇ F 1 ( x 0 ) + ⋯ + λ n − m ∇ F n − m ( x 0 ) \nabla f(x_0) = \lambda_1 \nabla F_1(x_0) + \dots + \lambda_{n-m} \nabla F_{n-m}(x_0) ∇ f ( x 0 ) = λ 1 ∇ F 1 ( x 0 ) + ⋯ + λ n − m ∇ F n − m ( x 0 ) ∇ ( f − λ 1 F 1 − λ 2 F 2 − ⋯ − λ n − m F n − m ) ( x 0 ) = 0 \nabla (f-\lambda_1 F_1 - \lambda_2 F_2 - \dots - \lambda_{n-m} F_{n-m})(x_0) = 0 ∇ ( f − λ 1 F 1 − λ 2 F 2 − ⋯ − λ n − m F n − m ) ( x 0 ) = 0 Conclusão: existem λ 1 , … , λ n − m ∈ R \lambda_1, \dots, \lambda_{n-m} \in \R λ 1 , … , λ n − m ∈ R x 0 x_0 x 0 f − λ 1 F 1 − ⋯ − λ n − m F n − m f - \lambda_1 F_1 - \dots- \lambda_{n-m} F_{n-m} f − λ 1 F 1 − ⋯ − λ n − m F n − m
Exemplos Considerando M M M f f f
M = { x 2 + y 2 + z 2 = 1 } f ( x , y , z ) = z \begin{array}{ll}
M= \{x^2+y^2+z^2=1\} & f(x,y,z) = z
\end{array} M = { x 2 + y 2 + z 2 = 1 } f ( x , y , z ) = z Mostre que f f f ( 0 , 0 , 1 ) (0,0,1) ( 0 , 0 , 1 ) ( 0 , 0 , − 1 ) (0,0,-1) ( 0 , 0 , − 1 ) M M M
Podemos escrever M M M F ( x , y , z ) = x 2 + y 2 + z 2 − 1 F(x,y,z) = x^2+y^2+z^2-1 F ( x , y , z ) = x 2 + y 2 + z 2 − 1
Assim, pelo método dos multiplicadores de Lagrange
{ ∇ ( f − λ F ) = 0 F = 0 ⇔ { ∇ ( z − λ ( x 2 + y 2 + z 2 − 1 ) ) = 0 x 2 + y 2 + z 2 = 1 ⇔ { − 2 λ x = 0 − 2 λ y = 0 1 − 2 λ z = 0 x 2 + y 2 + z 2 = 1 \begin{array}{l}
\begin{cases}
\nabla(f - \lambda F) = 0 \\
F=0
\end{cases}\\
\Leftrightarrow
\begin{cases}
\nabla(z-\lambda(x^2+y^2+z^2-1)) = 0\\
x^2+y^2+z^2=1
\end{cases}\\
\Leftrightarrow
\begin{cases}
-2\lambda x = 0\\
-2 \lambda y = 0\\
1-2\lambda z = 0\\
x^2+y^2+z^2=1
\end{cases}
\end{array} { ∇ ( f − λ F ) = 0 F = 0 ⇔ { ∇ ( z − λ ( x 2 + y 2 + z 2 − 1 )) = 0 x 2 + y 2 + z 2 = 1 ⇔ ⎩ ⎨ ⎧ − 2 λ x = 0 − 2 λ y = 0 1 − 2 λ z = 0 x 2 + y 2 + z 2 = 1 Se λ = 0 \lambda = 0 λ = 0 1 − 2 λ z = 0 ⇔ 1 = 0 1-2\lambda z = 0 \Leftrightarrow 1=0 1 − 2 λ z = 0 ⇔ 1 = 0 λ ≠ 0 \lambda\ne 0 λ = 0
{ y = 0 e x = 0 2 λ z = 1 ⟹ λ = ± 1 2 z 2 = 1 ⟹ z = ± 1 \begin{cases}
y = 0 \text{ e } x = 0\\
2\lambda z = 1 \implies \lambda = \pm \frac{1}{2}\\
z^2=1 \implies z=\pm 1
\end{cases} ⎩ ⎨ ⎧ y = 0 e x = 0 2 λ z = 1 ⟹ λ = ± 2 1 z 2 = 1 ⟹ z = ± 1 Soluções do sistema:
( x , y , z ) = ( 0 , 0 , 1 ) λ = 1 2 ( x , y , z ) = ( 0 , 0 , − 1 ) λ = − 1 2 \begin{array}{ll}
(x,y,z) = (0,0,1) & \lambda = \frac{1}{2}\\
(x,y,z) = (0,0,-1) & \lambda = -\frac{1}{2}
\end{array} ( x , y , z ) = ( 0 , 0 , 1 ) ( x , y , z ) = ( 0 , 0 , − 1 ) λ = 2 1 λ = − 2 1 O critério indica os candidatos a máximo ou mínimo.
Como nos encontramos numa esfera, ( 0 , 0 , 1 ) (0,0,1) ( 0 , 0 , 1 ) ( 0 , 0 , − 1 ) (0,0,-1) ( 0 , 0 , − 1 )
O método dos multiplicadores de Lagrange
Podemos assim recorrer ao Teorema de Weierstrass
f : M → R f: M \to \R f : M → R M M M f f f
Então f f f M M M
Ficamos assim com dois passos que temos de fazer quando queremos
encontrar extremos condicionados:
Justificar a existência de máximo (ou mínimo) - Teorema de Weierstrass
Utilizar um critério para encontrar os candidatos a máximo ou mínimo - método dos multiplicadores de Lagrange
Exemplos Queremos encontrar o paralelipípedo com maior volume de entre aqueles que satisfazem x + y + z = 1 x+y+z = 1 x + y + z = 1 x , y , z ≥ 0 x,y,z \geq 0 x , y , z ≥ 0
Para isso, temos de maximizar o volume V = x y z V = xyz V = x y z F = x + y + z − 1 = 0 F=x+y+z-1=0 F = x + y + z − 1 = 0
Será que existe máximo?V = x y z V = xyz V = x y z
{ x + y + z = 1 x , y , z ≥ 0 \begin{cases}
x+y+z=1\\
x,y,z\geq 0
\end{cases} { x + y + z = 1 x , y , z ≥ 0 M M M
Então, pelo Teorema de Weierstrass, V V V M M M
{ ∇ ( V − λ F ) = 0 F = 0 ⇔ x + y + z = 1 ⟹ { ∇ ( x y z − λ ( x + y + z − 1 ) ) = 0 x + y + z = 1 ⟹ { y z − λ = 0 x z − λ = 0 x y − λ = 0 x + y + z = 1 ⟹ { y z = λ x z = y z = x y x + y + z = 1 \begin{array}{l}
\begin{cases}
\nabla(V-\lambda F) = 0\\
F=0 \Leftrightarrow x+y+z=1
\end{cases}
\implies
\begin{cases}
\nabla(xyz-\lambda(x+y+z-1))=0\\
x+y+z=1
\end{cases}\\
\implies
\begin{cases}
yz-\lambda = 0\\
xz-\lambda = 0\\
xy-\lambda = 0\\
x+y+z=1
\end{cases}
\implies
\begin{cases}
yz=\lambda\\
xz=yz=xy\\
x+y+z=1
\end{cases}
\end{array} { ∇ ( V − λ F ) = 0 F = 0 ⇔ x + y + z = 1 ⟹ { ∇ ( x y z − λ ( x + y + z − 1 )) = 0 x + y + z = 1 ⟹ ⎩ ⎨ ⎧ y z − λ = 0 x z − λ = 0 x y − λ = 0 x + y + z = 1 ⟹ ⎩ ⎨ ⎧ y z = λ x z = y z = x y x + y + z = 1 Será que interessa x = 0 ∧ y = 0 ∧ z = 0 x=0 \land y=0 \land z=0 x = 0 ∧ y = 0 ∧ z = 0 V = 0 V=0 V = 0
{ x z = y z ⟹ x = y y z = x y ⟹ z = x x + y + z = 1 ⟹ { x = y z = x 3 x = 1 ⟹ { x = 1 3 y = 1 3 z = 1 3 \begin{cases}
xz=yz\implies x=y\\
yz=xy\implies z=x\\
x+y+z=1
\end{cases}
\implies
\begin{cases}
x=y\\
z=x\\
3x=1
\end{cases}
\implies
\begin{cases}
x=\frac{1}{3}\\
y=\frac{1}{3}\\
z=\frac{1}{3}
\end{cases} ⎩ ⎨ ⎧ x z = y z ⟹ x = y y z = x y ⟹ z = x x + y + z = 1 ⟹ ⎩ ⎨ ⎧ x = y z = x 3 x = 1 ⟹ ⎩ ⎨ ⎧ x = 3 1 y = 3 1 z = 3 1 Só há um candidato a máximo. Como sabemos que ele existe, o máximo tem de ser atingido em ( 1 3 , 1 3 , 1 3 ) (\frac{1}{3},\frac{1}{3},\frac{1}{3}) ( 3 1 , 3 1 , 3 1 )
Quais os pontos que estão na interseção do plano x + z = 1 x+z=1 x + z = 1 z = x 2 + y 2 z = x^2+y^2 z = x 2 + y 2
M = { ( x , y , z ) ∈ R 3 : x + z − 1 = 0 , x 2 + y 2 − z = 0 } M = \{(x,y,z) \in \R^3 : x+z-1 = 0, x^2+y^2-z = 0 \} M = {( x , y , z ) ∈ R 3 : x + z − 1 = 0 , x 2 + y 2 − z = 0 } Como minimizar a distância é a mesma coisa que minimizar distância ao quadrado, podemos encontrar
o mínimo da seguinte função na variedade M M M
f ( x , y , z ) = ( dist a ˆ ncia a ˋ origem ) 2 = ( x 2 + y 2 + z 2 ) 2 = x 2 + y 2 + z 2 f(x,y,z) = (\text{distância à origem})^2 = (\sqrt{x^2+y^2+z^2})^2 = x^2+y^2+z^2 f ( x , y , z ) = ( dist a ˆ ncia a ˋ origem ) 2 = ( x 2 + y 2 + z 2 ) 2 = x 2 + y 2 + z 2 Pelo método dos multiplicadores de Lagrange
{ ∇ ( f − λ 1 F 1 − λ 2 F 2 ) = 0 F 1 = 0 F 2 = 0 ⇒ { ∇ ( x 2 + y 2 + z 2 − λ 1 ( x + y − 1 ) − λ 2 ( x 2 + y 2 − z ) ) = 0 x + z = 1 z = x 2 + y 2 ⇒ { 2 x − λ 1 − λ 2 ⋅ 2 x = 0 2 y − 2 λ 2 y = 0 2 z − λ 1 + λ 2 = 0 x + z = 1 z = x 2 + y 2 \begin{array}{l}
\begin{cases}
\nabla(f-\lambda_1 F_1 - \lambda_2 F_2) = 0\\
F_1=0\\
F_2=0
\end{cases}\\
\Rightarrow
\begin{cases}
\nabla(x^2+y^2+z^2-\lambda_1 (x+y-1) - \lambda_2 (x^2+y^2-z)) = 0\\
x+z=1\\
z=x^2+y^2
\end{cases}\\
\Rightarrow
\begin{cases}
2x-\lambda_1-\lambda_2\cdot 2x=0\\
2y-2\lambda_2 y = 0\\
2z-\lambda_1 + \lambda_2 = 0\\
x+z=1\\
z=x^2+y^2
\end{cases}
\end{array} ⎩ ⎨ ⎧ ∇ ( f − λ 1 F 1 − λ 2 F 2 ) = 0 F 1 = 0 F 2 = 0 ⇒ ⎩ ⎨ ⎧ ∇ ( x 2 + y 2 + z 2 − λ 1 ( x + y − 1 ) − λ 2 ( x 2 + y 2 − z )) = 0 x + z = 1 z = x 2 + y 2 ⇒ ⎩ ⎨ ⎧ 2 x − λ 1 − λ 2 ⋅ 2 x = 0 2 y − 2 λ 2 y = 0 2 z − λ 1 + λ 2 = 0 x + z = 1 z = x 2 + y 2 Se y = 0 y=0 y = 0
{ 2 x − λ 1 − λ 2 ⋅ 2 x = 0 2 z − λ 1 + λ 2 = 0 x + z = 1 z = x 2 + y 2 \begin{cases}
2x-\lambda_1-\lambda_2\cdot 2x=0\\
2z-\lambda_1 + \lambda_2 = 0\\
x+z=1\\
z=x^2+y^2
\end{cases} ⎩ ⎨ ⎧ 2 x − λ 1 − λ 2 ⋅ 2 x = 0 2 z − λ 1 + λ 2 = 0 x + z = 1 z = x 2 + y 2 x 2 = 1 − x ⟹ x 2 + x − 1 = 0 ⟹ x = − 1 ± 5 2 x^2=1-x\implies x^2+x-1=0 \implies x= \frac{-1 \pm \sqrt{5}}{2} x 2 = 1 − x ⟹ x 2 + x − 1 = 0 ⟹ x = 2 − 1 ± 5 z = 1 − x = 1 − − 1 ± 5 2 = 3 2 ∓ 5 2 z=1-x=1-\frac{-1\pm \sqrt{5}}{2} = \frac{3}{2} \mp \frac{\sqrt{5}}{2} z = 1 − x = 1 − 2 − 1 ± 5 = 2 3 ∓ 2 5
Pontos com y = 0 y=0 y = 0 ( − 1 ± 5 2 , 0 , 3 ∓ 5 2 ) (\frac{-1 \pm \sqrt{5}}{2}, 0, \frac{3 \mp \sqrt{5}}{2}) ( 2 − 1 ± 5 , 0 , 2 3 ∓ 5 )
Se λ 2 = 1 \lambda_2 = 1 λ 2 = 1
{ 2 x − λ 1 − 2 x = 0 2 z − λ 1 + 1 = 0 x + z = 1 z = x 2 + y 2 ⇒ { λ 1 = 0 2 z + 1 = 0 x + z = 1 z = x 2 + y 2 ⇒ { λ 1 = 0 z = − 1 2 x + z = 1 x 2 + y 2 = − 1 2 \begin{cases}
2x-\lambda_1-2x=0\\
2z-\lambda_1 + 1 = 0\\
x+z=1\\
z=x^2+y^2
\end{cases}
\Rightarrow
\begin{cases}
\lambda_1 = 0\\
2z+1=0\\
x+z=1\\
z=x^2+y^2
\end{cases}
\Rightarrow
\begin{cases}
\lambda_1 = 0\\
z=-\frac{1}{2}\\
x+z=1\\
\smartcolor{orange}{x^2+y^2=-\frac{1}{2}}
\end{cases} ⎩ ⎨ ⎧ 2 x − λ 1 − 2 x = 0 2 z − λ 1 + 1 = 0 x + z = 1 z = x 2 + y 2 ⇒ ⎩ ⎨ ⎧ λ 1 = 0 2 z + 1 = 0 x + z = 1 z = x 2 + y 2 ⇒ ⎩ ⎨ ⎧ λ 1 = 0 z = − 2 1 x + z = 1 x 2 + y 2 = − 2 1 É impossível (a soma do quadrado de dois números nunca pode ser negativa), logo os únicos candidatos são:
( − 1 − 5 2 , 0 , 3 + 5 2 ) (\frac{-1 - \sqrt{5}}{2}, 0, \frac{3 + \sqrt{5}}{2}) ( 2 − 1 − 5 , 0 , 2 3 + 5 ) ( − 1 + 5 2 , 0 , 3 − 5 2 ) (\frac{-1 + \sqrt{5}}{2}, 0, \frac{3 - \sqrt{5}}{2}) ( 2 − 1 + 5 , 0 , 2 3 − 5 )
f f f
( − 1 − 5 2 ) 2 + 0 2 + ( 3 + 5 2 ) 2 = 1 + 2 5 + 5 4 + 9 + 6 5 + 5 4 \left(\frac{-1 - \sqrt{5}}{2} \right)^2 + 0^2 + \left(\frac{3 + \sqrt{5}}{2} \right)^2 = \frac{1+2\sqrt{5} + 5}{4} + \frac{9+6\sqrt{5} + 5}{4} ( 2 − 1 − 5 ) 2 + 0 2 + ( 2 3 + 5 ) 2 = 4 1 + 2 5 + 5 + 4 9 + 6 5 + 5
( − 1 + 5 2 ) 2 + 0 2 + ( 3 − 5 2 ) 2 = 1 − 2 5 + 5 4 + 9 − 6 5 + 5 4 \left(\frac{-1 + \sqrt{5}}{2} \right)^2 + 0^2 + \left(\frac{3 - \sqrt{5}}{2} \right)^2 = \frac{1-2\sqrt{5} + 5}{4} + \frac{9-6\sqrt{5} + 5}{4} ( 2 − 1 + 5 ) 2 + 0 2 + ( 2 3 − 5 ) 2 = 4 1 − 2 5 + 5 + 4 9 − 6 5 + 5
O último tem a menor imagem
Logo o ponto mais próximo da origem é ( − 1 + 5 2 , 0 , 3 − 5 2 ) \left(\frac{-1 + \sqrt{5}}{2}, 0, \frac{3 - \sqrt{5}}{2} \right) ( 2 − 1 + 5 , 0 , 2 3 − 5 )
Pelo Teorema de Weierstrass podemos também afirmar que este ponto é mesmo um mínimo, pois:
M M M O mínimo de f f f M M M R R R
Então, M ∩ B R ( 0 ) ‾ M \cap \overline{B_R(0)} M ∩ B R ( 0 )
Logo, pelo Teorema de Weierstrass:
f f f M ∩ B R ( 0 ) M \cap B_R(0) M ∩ B R ( 0 ) f f f M M M Dada uma variedade M M M f : M → R f: M \to \R f : M → R ∫ M f \int_M f ∫ M f
DEFINIÇÃO
Considerando dim M = 1 \dim M = 1 dim M = 1 M M M R 2 \R^2 R 2
∫ M f d γ = ∫ a b f ( g ( t ) ) ⋅ ∣ ∣ g ′ ( t ) ∣ ∣ d t \int_M f \d \gamma = \int^b_a f(g(t)) \cdot ||g'(t)|| \d t ∫ M f d γ = ∫ a b f ( g ( t )) ⋅ ∣∣ g ′ ( t ) ∣∣ d t A estes integrais chamamos Integral de um campo escalar numa variedade de dim 1 \dim 1 dim 1 Integral de linha de um campo escalar
Exemplos Seja a circunferência x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1
Qual o perímetro (comprimento) da circunferência?
Começamos por parametrizar esta variedade:
g ( t ) = ( cos t , sin t ) t ∈ ] 0 , 2 π [ g ′ ( t ) = ( − sin t , cos t ) ∣ ∣ g ′ ( t ) ∣ ∣ = sin 2 t + cos 2 t = 1 \begin{array}{cc}
g(t) = (\cos t, \sin t) & t \in ]0, 2\pi[\\
g'(t) = (-\sin t, \cos t) & ||g'(t)|| = \sqrt{\sin^2t + \cos^2t} = 1
\end{array} g ( t ) = ( cos t , sin t ) g ′ ( t ) = ( − sin t , cos t ) t ∈ ] 0 , 2 π [ ∣∣ g ′ ( t ) ∣∣ = sin 2 t + cos 2 t = 1 Assim, basta-nos calcular o integral do campo escalar:
∫ x 2 + y 2 = 1 1 d γ = ∫ 0 2 π 1 ⋅ ∣ ∣ g ′ ( t ) ∣ ∣ d t = ∫ 0 2 π 1 d t = 2 π \int_{x^2+y^2=1} 1 \d \gamma = \int^{2\pi}_{0} 1 \cdot ||g'(t)|| \d t =
\int^{2\pi}_{0} 1 \d t = 2\pi ∫ x 2 + y 2 = 1 1 d γ = ∫ 0 2 π 1 ⋅ ∣∣ g ′ ( t ) ∣∣ d t = ∫ 0 2 π 1 d t = 2 π Considere o segmento que une ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( 0 , 1 , 0 ) (0,1,0) ( 0 , 1 , 0 ) Sabe-se que a densidade de massa deste segmento é dada por z 2 z^2 z 2
Qual a massa do segmento?
Comecemos por parametrizar o segmento, que podemos fazer
através de da expressão vetorial de uma reta.
g ( t ) = ( 1 , 0 , 1 ) + t ( ( 0 , 1 , 0 ) − ( 1 , 0 , 1 ) ⏞ v ) = ( 1 , 0 , 1 ) + t ( − 1 , 1 , − 1 ) = ( 1 − t , t , 1 − t ) t ∈ [ 0 , 1 ] g ′ ( t ) = ( − 1 , 1 , − 1 ) ∣ ∣ g ′ ( t ) ∣ ∣ = 3 \begin{array}{ll}
\begin{aligned}
g(t) &= (1,0,1) + t(\overbrace{(0,1,0) - (1,0,1)}^v)\\
&=(1,0,1) + t(-1,1,-1)\\
&= (1-t, t, 1-t)
\end{aligned} & t \in [0,1]\\
\\
g'(t) = (-1,1,-1) & ||g'(t)|| = \sqrt 3
\end{array} g ( t ) = ( 1 , 0 , 1 ) + t ( ( 0 , 1 , 0 ) − ( 1 , 0 , 1 ) v ) = ( 1 , 0 , 1 ) + t ( − 1 , 1 , − 1 ) = ( 1 − t , t , 1 − t ) g ′ ( t ) = ( − 1 , 1 , − 1 ) t ∈ [ 0 , 1 ] ∣∣ g ′ ( t ) ∣∣ = 3 Massa do segmento = ∫ segmento densidade de massa \text{Massa do segmento} = \int_{\text{segmento}} \text{densidade de massa} Massa do segmento = ∫ segmento densidade de massa Massa = ∫ segmento z 2 d γ = ∫ 0 1 ( 1 − t ) 2 ⋅ 3 d t = 3 [ − ( 1 − t ) 3 3 ] 0 1 = 3 ( 0 − ( − 1 3 ) ) = 3 3 \begin{aligned}
\text{Massa} &= \int_{\text{segmento}} z^2 \d \gamma\\
&= \int^1_0 (1-t)^2 \cdot \sqrt{3} \d t\\
&= \sqrt{3} \left[- \frac{(1-t)^3}{3} \right]^1_0\\
&= \sqrt{3} \left(0 - \left(-\frac{1}{3} \right) \right)\\
&= \frac{\sqrt{3}}{3}
\end{aligned} Massa = ∫ segmento z 2 d γ = ∫ 0 1 ( 1 − t ) 2 ⋅ 3 d t = 3 [ − 3 ( 1 − t ) 3 ] 0 1 = 3 ( 0 − ( − 3 1 ) ) = 3 3 DEFINIÇÃO
Considerando dim M ≥ 1 \dim M \geq 1 dim M ≥ 1 M M M R n \R^n R n
∫ M f d γ = ∫ a b f ( g ( t ) ) ⋅ ∣ d e t D g T D g ∣ d t \int_M f \d \gamma = \int^b_a f(g(t)) \cdot \sqrt{|det Dg^T Dg|} \d t ∫ M f d γ = ∫ a b f ( g ( t )) ⋅ ∣ d e t D g T D g ∣ d t D g T Dg^T D g T D g Dg D g
Devido às propriedades da matriz transposta, sabemos sempre que D g T D g Dg^T Dg D g T D g
Podemos observar que se m = 1 m=1 m = 1
D g ( t ) = g ′ ( t ) D g T ⋅ D g = g ′ ( t ) ⋅ g ′ ( t ) = ∣ ∣ g ′ ( t ) ∣ ∣ 2 \begin{array}{ll}
Dg(t) = g'(t) & Dg^T \cdot Dg = g'(t) \cdot g'(t) = ||g'(t)||^2
\end{array} D g ( t ) = g ′ ( t ) D g T ⋅ D g = g ′ ( t ) ⋅ g ′ ( t ) = ∣∣ g ′ ( t ) ∣ ∣ 2 Exemplo Considerando a variedade P P P
P = { ( x , y , z ) ∈ R 3 : x 2 + y 2 = z , z < 1 } P = \{(x,y,z) \in \R^3: x^2+y^2=z, z < 1\} P = {( x , y , z ) ∈ R 3 : x 2 + y 2 = z , z < 1 } Qual a área de P P P
Podemos utilizar coordenadas cilíndricas:
{ x = r cos θ y = r sin θ z = z = r 2 \begin{cases}
x=r \cos \theta\\
y=r \sin \theta\\
z=z=r^2
\end{cases} ⎩ ⎨ ⎧ x = r cos θ y = r sin θ z = z = r 2 E assim parametrizar:
g ( r , θ ) = ( r cos θ , r sin θ , r 2 ) θ ∈ ] 0 , 2 π [ , r ∈ ] 0 , 1 [ \begin{array}{ll}
g(r, \theta) = (r \cos \theta, r \sin \theta, r^2) & \theta \in ]0, 2 \pi[, r \in ]0, 1[
\end{array} g ( r , θ ) = ( r cos θ , r sin θ , r 2 ) θ ∈ ] 0 , 2 π [ , r ∈ ] 0 , 1 [ ∫ P 1 d S = ∫ 0 2 π ∫ 0 1 1 ⋅ ∣ det ( D g t D g ) ∣ d r d θ \int_P 1 \d S = \int^{2 \pi}_0 \int^1_0 1 \cdot \sqrt{|\det (Dg^t Dg ) | } \d r \d \theta ∫ P 1 d S = ∫ 0 2 π ∫ 0 1 1 ⋅ ∣ det ( D g t D g ) ∣ d r d θ D g ( r , θ ) = [ cos θ − r sin θ sin θ r cos θ 2 r 0 ] Dg(r, \theta) = \begin{bmatrix}
\cos \theta & - r \sin \theta\\
\sin \theta & r \cos \theta\\
2 r & 0
\end{bmatrix} D g ( r , θ ) = cos θ sin θ 2 r − r sin θ r cos θ 0 D g t ⋅ D g = [ cos θ sin θ 2 r − r sin θ r cos θ 0 ] [ cos θ − r sin θ sin θ r cos θ 2 r 0 ] = [ 1 + 4 r 2 0 0 r 2 ] Dg^t \cdot Dg = \begin{bmatrix}
\cos \theta & \sin \theta & 2r\\
-r \sin \theta & r \cos \theta & 0
\end{bmatrix}
\begin{bmatrix}
\cos \theta & - r \sin \theta\\
\sin \theta & r \cos \theta\\
2 r & 0
\end{bmatrix}
=
\begin{bmatrix}
1+4r^2 & 0\\
0 & r^2
\end{bmatrix} D g t ⋅ D g = [ cos θ − r sin θ sin θ r cos θ 2 r 0 ] cos θ sin θ 2 r − r sin θ r cos θ 0 = [ 1 + 4 r 2 0 0 r 2 ] det = ( 4 r 2 + 1 ) r 2 det = r 4 r 2 + 1 \begin{array}{ll}
\det = (4r^2 + 1 ) r^2 & \sqrt{\det} = r \sqrt{4r^2 +1}
\end{array} det = ( 4 r 2 + 1 ) r 2 det = r 4 r 2 + 1 ∫ P 1 d S = ∫ 0 2 π ∫ 0 1 1 ⋅ r 4 r 2 + 1 d r d θ = ∫ 0 2 π ∫ 0 1 1 8 ⋅ 8 r 4 r 2 + 1 d r d θ = 1 8 ∫ 0 2 π [ ( 4 r 2 + 1 ) 3 2 3 2 ] 0 1 d θ = 1 8 ∫ 0 2 π 2 3 ( 5 3 2 − 1 ) d θ = 1 8 ⋅ 2 3 ⋅ 2 π ⋅ ( 5 3 2 − 1 ) = π 6 ( 5 3 2 − 1 ) \begin{aligned}
\int_P 1 \d S &= \int^{2 \pi}_0 \int^1_0 1 \cdot r \sqrt{4r^2 + 1} \d r \d \theta\\
&= \int^{2 \pi}_0 \int^1_0 \frac{1}{8} \cdot 8 r \sqrt{4r^2 + 1} \d r \d \theta\\
&= \frac{1}{8} \int^{2\pi}_0 \left[\frac{(4r^2 + 1)^{\frac{3}{2}}}{\frac{3}{2}} \right]^1_0 \d \theta\\
&= \frac{1}{8} \int^{2\pi}_0 \frac{2}{3} (5^{\frac{3}{2}} - 1) \d \theta\\
&= \frac{1}{8} \cdot \frac{2}{3} \cdot 2\pi \cdot (5^{\frac{3}{2}} - 1)\\
&= \frac{\pi}{6} (5^{\frac{3}{2}} - 1)
\end{aligned} ∫ P 1 d S = ∫ 0 2 π ∫ 0 1 1 ⋅ r 4 r 2 + 1 d r d θ = ∫ 0 2 π ∫ 0 1 8 1 ⋅ 8 r 4 r 2 + 1 d r d θ = 8 1 ∫ 0 2 π [ 2 3 ( 4 r 2 + 1 ) 2 3 ] 0 1 d θ = 8 1 ∫ 0 2 π 3 2 ( 5 2 3 − 1 ) d θ = 8 1 ⋅ 3 2 ⋅ 2 π ⋅ ( 5 2 3 − 1 ) = 6 π ( 5 2 3 − 1 ) Podemos definir o produto externo de u , v ∈ R 3 u, v \in \R^3 u , v ∈ R 3 u = ( u 1 , u 2 , u 3 ) u = (u_1, u_2, u_3) u = ( u 1 , u 2 , u 3 ) v = ( v 1 , v 2 , v 3 ) v = (v_1, v_2, v_3) v = ( v 1 , v 2 , v 3 ) e 1 , e 2 , e 3 e_1, e_2, e_3 e 1 , e 2 , e 3
u × v = det [ e 1 e 2 e 3 u 1 u 2 u 3 v 1 v 2 v 3 ] u \times v = \det \begin{bmatrix}
e_1 & e_2 & e_3\\
u_1 & u_2 & u_3\\
v_1 & v_2 & v_3
\end{bmatrix} u × v = det e 1 u 1 v 1 e 2 u 2 v 2 e 3 u 3 v 3 Efetuando a regra de Laplace na primeira linha:
= ( u 2 v 3 − v 2 u 3 ) e 1 − ( u 1 v 3 − v 1 u 3 ) e 2 + ( u 1 v 2 − v 1 u 2 ) e 3 = ( u 2 v 3 − v 2 u 3 , − u 1 v 3 + v 1 u 3 , u 1 v 2 − v 1 u 2 ) \begin{array}{l}
= (u_2 v_3 - v_2 u_3) e_1 - (u_1v_3 - v_1u_3) e_2 + (u_1 v_2 - v_1u_2) e_3\\
= (u_2 v_3 - v_2 u_3, -u_1 v_3 + v_1u_3 , u_1v_2-v_1u_2)
\end{array} = ( u 2 v 3 − v 2 u 3 ) e 1 − ( u 1 v 3 − v 1 u 3 ) e 2 + ( u 1 v 2 − v 1 u 2 ) e 3 = ( u 2 v 3 − v 2 u 3 , − u 1 v 3 + v 1 u 3 , u 1 v 2 − v 1 u 2 ) Propriedades do produto externo :
u × v u \times v u × v u u u v v v u ⋅ ( v × w ) = v ⋅ ( w × u ) u \cdot (v \times w) = v\cdot (w \times u) u ⋅ ( v × w ) = v ⋅ ( w × u ) u × ( v × w ) = ( u ⋅ w ) v − ( u ⋅ v ) w u \times (v \times w ) = (u \cdot w) v - (u \cdot v) w u × ( v × w ) = ( u ⋅ w ) v − ( u ⋅ v ) w
Alternativamente ao fator de escala ∣ det ( D g T D g ) ∣ \sqrt{ | \det (Dg^T Dg) | } ∣ det ( D g T D g ) ∣
∣ det ( D g t D g ) ∣ = ∣ ∣ ∂ g ∂ u × ∂ g ∂ v ∣ ∣ \sqrt{ | \det \left(Dg^t Dg \right) | } = \left|\left| \frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v} \right|\right| ∣ det ( D g t D g ) ∣ = ∂ u ∂ g × ∂ v ∂ g Exemplo Considerando a variedade
S = { x 2 + y 2 = z 2 + 1 , 0 < z < 1 } S = \{x^2 + y^2 = z^2 + 1, 0 < z < 1\} S = { x 2 + y 2 = z 2 + 1 , 0 < z < 1 }
Sabe-se que a densidade de massa é ϕ ( x , y , z ) = 1 2 z 2 + 1 \phi (x,y,z) = \frac{1}{\sqrt{2z^2+1}} ϕ ( x , y , z ) = 2 z 2 + 1 1
Qual a massa de S S S
A massa é dada por ∫ S ϕ \int_S \phi ∫ S ϕ
Começamos por parametrizar S S S
{ x = r cos θ y = r sin θ z = z \begin{cases}
x=r \cos \theta\\
y=r \sin \theta\\
z=z
\end{cases} ⎩ ⎨ ⎧ x = r cos θ y = r sin θ z = z r 2 = z 2 + 1 r = z 2 + 1 0 < z < 1 \begin{array}{lll}
r^2 = z^2 + 1 & r = \sqrt{z^2 + 1} & 0 < z < 1
\end{array} r 2 = z 2 + 1 r = z 2 + 1 0 < z < 1 g ( θ , z ) = ( z 2 + 1 cos θ , z 2 + 1 sin θ , z ) g(\theta, z) = (\sqrt{z^2 + 1} \cos \theta, \sqrt{z^2+1} \sin \theta, z) g ( θ , z ) = ( z 2 + 1 cos θ , z 2 + 1 sin θ , z ) D g ( θ , z ) = [ − z 2 + 1 sin θ z z 2 + 1 cos θ z 2 + 1 cos θ z z 2 + 1 sin θ 0 1 ] Dg (\theta, z) = \begin{bmatrix}
-\sqrt{z^2+1} \sin \theta & \frac{z}{\sqrt{z^2 + 1}}\cos \theta\\
\sqrt{z^2+1} \cos \theta & \frac{z}{\sqrt{z^2 + 1}}\sin \theta\\
0 & 1
\end{bmatrix} D g ( θ , z ) = − z 2 + 1 sin θ z 2 + 1 cos θ 0 z 2 + 1 z cos θ z 2 + 1 z sin θ 1 ∂ g ∂ θ × ∂ g ∂ z = ∣ e 1 e 2 e 3 − z 2 + 1 sin θ z 2 + 1 cos θ 0 z z 2 + 1 cos θ z z 2 + 1 sin θ 1 ∣ = ( z 2 + 1 cos θ , z 2 + 1 sin θ , − z sin 2 θ − z cos 2 θ ) = ( z 2 + 1 cos θ , z 2 + 1 sin θ , − z ) \begin{aligned}
\frac{\partial g}{\partial \theta} \times \frac{\partial g}{\partial z} &= \begin{vmatrix}
e_1 & e_2 & e_3\\
-\sqrt{z^2 + 1} \sin \theta & \sqrt{z^2 + 1}\cos \theta & 0\\
\frac{z}{\sqrt{z^2 + 1}}\cos \theta & \frac{z}{\sqrt{z^2 + 1}}\sin \theta & 1
\end{vmatrix}\\
&=(\sqrt{z^2+1} \cos \theta, \sqrt{z^2 + 1} \sin \theta, -z \sin ^2 \theta - z \cos^2 \theta)\\
&=(\sqrt{z^2+1} \cos \theta, \sqrt{z^2 + 1} \sin \theta, -z)
\end{aligned} ∂ θ ∂ g × ∂ z ∂ g = e 1 − z 2 + 1 sin θ z 2 + 1 z cos θ e 2 z 2 + 1 cos θ z 2 + 1 z sin θ e 3 0 1 = ( z 2 + 1 cos θ , z 2 + 1 sin θ , − z sin 2 θ − z cos 2 θ ) = ( z 2 + 1 cos θ , z 2 + 1 sin θ , − z ) ∣ ∣ ∂ g ∂ θ × ∂ g ∂ z ∣ ∣ 2 = ( z 2 + 1 ) cos 2 θ + ( z 2 + 1 ) sin 2 θ + z 2 = 2 z 2 + 1 \left|\left|\frac{\partial g}{\partial \theta} \times \frac{\partial g}{\partial z}\right|\right|^2 =
(z^2 + 1) \cos^2 \theta + (z^2 + 1) \sin^2 \theta + z^2 = 2z^2 + 1 ∂ θ ∂ g × ∂ z ∂ g 2 = ( z 2 + 1 ) cos 2 θ + ( z 2 + 1 ) sin 2 θ + z 2 = 2 z 2 + 1 ϕ ( x , y , z ) = 1 2 z 2 + 1 ϕ ( g ( θ , z ) ) = 1 2 z 2 + 1 \begin{array}{ll}
\phi(x,y,z) = \frac{1}{\sqrt{2z^2 + 1}} & \phi (g(\theta,z )) = \frac{1}{\sqrt{2z^2 + 1}}
\end{array} ϕ ( x , y , z ) = 2 z 2 + 1 1 ϕ ( g ( θ , z )) = 2 z 2 + 1 1 ∫ S ϕ = ∫ 0 1 ∫ 0 2 π 1 2 z 2 + 1 ⋅ 2 z 2 + 1 d θ d z = ∫ 0 1 ∫ 0 2 π 1 d θ d z = 2 π \begin{aligned}
\int_S \phi &= \int^1_0 \int^{2\pi}_0 \frac{1}{\sqrt{2z^2 + 1}} \cdot \sqrt{2z^2 + 1} \d \theta \d z\\
&= \int^1_0 \int^{2\pi}_0 1 \d \theta \d z\\
&= 2\pi
\end{aligned} ∫ S ϕ = ∫ 0 1 ∫ 0 2 π 2 z 2 + 1 1 ⋅ 2 z 2 + 1 d θ d z = ∫ 0 1 ∫ 0 2 π 1 d θ d z = 2 π Tal como referido em Aplicações do Integral ,
é possível calcular o centroide e momento de inércia de um sólido através do
cálculo de um integral.
Esses cálculos são também aplicáveis em campos escalares, pelo que voltamos a ter a mesma definição:
x ‾ = ∫ M x ∫ M 1 y ‾ = ∫ M y ∫ M 1 \begin{darray}{ll}
\overline x = \frac{\int_M x}{\int_M 1} &
\overline y = \frac{\int_M y}{\int_M 1}
\end{darray} x = ∫ M 1 ∫ M x y = ∫ M 1 ∫ M y As definições de quantidades físicas em regiões de R 2 \R^2 R 2 R 3 \R^3 R 3
Exemplos Considerando a variedade definida por
M = { y = x 2 , x ∈ ] − 1 , 1 [ } M = \{ y = x^2, x \in ]-1, 1[\} M = { y = x 2 , x ∈ ] − 1 , 1 [ } Qual o centroide de M M M
Começamos por fazer a sua parametrização:
g ( t ) = ( t , t 2 ) t ∈ ] − 1 , 1 [ g ′ ( t ) = ( 1 , 2 t ) ∣ ∣ g ′ ( t ) ∣ ∣ = 1 + 4 t 2 \begin{array}{cc}
g(t) = (t, t^2) & t \in ]-1,1[\\
g'(t) = (1,2t) & ||g'(t)|| = \sqrt{1 + 4t^2}
\end{array} g ( t ) = ( t , t 2 ) g ′ ( t ) = ( 1 , 2 t ) t ∈ ] − 1 , 1 [ ∣∣ g ′ ( t ) ∣∣ = 1 + 4 t 2 ∫ M 1 d γ = ∫ − 1 1 1 + 4 t 2 d t \int_M 1 \d \gamma = \int^1_{-1} \sqrt{1 + 4t^2} \d t ∫ M 1 d γ = ∫ − 1 1 1 + 4 t 2 d t x ‾ = ∫ M x d γ ∫ M 1 d γ = ∫ − 1 1 t 1 + 4 t 2 d t ∫ − 1 1 1 + 4 t 2 d t \overline x = \frac{\int_M x \d \gamma}{\int_M 1 \d \gamma} =
\frac{\int^1_{-1} t \sqrt{1+4t^2} \d t}{\int^1_{-1} \sqrt{1+4t^2} \d t} x = ∫ M 1 d γ ∫ M x d γ = ∫ − 1 1 1 + 4 t 2 d t ∫ − 1 1 t 1 + 4 t 2 d t y ‾ = ∫ M y d γ ∫ M 1 d γ = ∫ − 1 1 t 2 1 + 4 t 2 d t ∫ − 1 1 1 + 4 t 2 d t \overline y = \frac{\int_M y \d \gamma}{\int_M 1 \d \gamma} =
\frac{\int^1_{-1} t^2 \sqrt{1+4t^2} \d t}{\int^1_{-1} \sqrt{1+4t^2} \d t} y = ∫ M 1 d γ ∫ M y d γ = ∫ − 1 1 1 + 4 t 2 d t ∫ − 1 1 t 2 1 + 4 t 2 d t Considerando a variedade definida por
M = { x 2 + y 2 = 1 , y > 0 } M = \{ x^2 +y^2 = 1, y > 0\} M = { x 2 + y 2 = 1 , y > 0 } Determine o seu centroide
Começamos por parametrizar a variedade:
g ( t ) = ( cos t , sin t ) t ∈ ] 0 , π [ g ′ ( t ) = ( − sin t , cos t ) ∣ ∣ g ′ ( t ) ∣ ∣ = 1 \begin{array}{ll}
g(t) = (\cos t, \sin t) & t \in ]0, \pi[\\
g'(t) = (-\sin t, \cos t) & || g'(t) || = 1
\end{array} g ( t ) = ( cos t , sin t ) g ′ ( t ) = ( − sin t , cos t ) t ∈ ] 0 , π [ ∣∣ g ′ ( t ) ∣∣ = 1 Esta variedade representa um semi-círculo de raio 1.∫ M 1 = π \int_M 1 = \pi ∫ M 1 = π
Centroide de M M M ( x ‾ , y ‾ ) (\overline x, \overline y) ( x , y )
x ‾ = ∫ M x ∫ M 1 = 1 π ∫ M x d γ = 1 π ∫ 0 π cos t ⋅ 1 d t = 1 π [ sin t ] 0 π = 0 \overline x = \frac{\int_M x}{\int_M 1} = \frac{1}{\pi} \int_M x \d \gamma = \frac{1}{\pi} \int^{\pi}_0 \cos t \cdot 1 \d t =
\frac{1}{\pi} \left[\sin t\right]^{\pi}_0 = 0 x = ∫ M 1 ∫ M x = π 1 ∫ M x d γ = π 1 ∫ 0 π cos t ⋅ 1 d t = π 1 [ sin t ] 0 π = 0 y ‾ = ∫ M y ∫ M 1 = 1 π ∫ M y d γ = 1 π ∫ 0 π sin t ⋅ 1 d t = 1 π [ − cos t ] 0 π = 2 π \overline y = \frac{\int_M y}{\int_M 1} = \frac{1}{\pi} \int_M y \d \gamma = \frac{1}{\pi} \int^{\pi}_0 \sin t \cdot 1 \d t =
\frac{1}{\pi} \left[-\cos t\right]^{\pi}_0 = \frac{2}{\pi} y = ∫ M 1 ∫ M y = π 1 ∫ M y d γ = π 1 ∫ 0 π sin t ⋅ 1 d t = π 1 [ − cos t ] 0 π = π 2 Centroide: ( 0 , 2 π ) \left(0, \frac{2}{\pi}\right) ( 0 , π 2 )
Slides: