C´ALCULOvolume 1
A. Carlos & J. Adonai
UFAL-2007
C
Conte´udo
ii Conte´udo
1 Vetores e Func¸˜oes Vetoriais . . . . . . . . . . . . . . . . . . . . 1
1.1 O Espac¸o Euclidiano Rn
. . . ....
Conte´udo iii
2.3 Geometria das Curvas Parametrizadas . . . . . . . . . . . . . 68
2.3.8 Curvatura e Torc¸˜ao . . . . . . ...
iv Conte´udo
4.5.3 Interpretac¸˜ao Geom´etrica . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
4 Derivadas Parci...
Conte´udo v
6.4 O Teorema da Func¸˜ao Impl´ıcita . . . . . . . . . . . . . . . . . . . 236
6.4.1 O Caso f : D ⊂ R2
−→ R . ...
1
Vetores
e
Func¸˜oes Vetoriais
rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
rrrrrrrr...
1.1
O Espac¸o Euclidiano Rn
Nesta se¸c˜ao, introduziremos a no¸c˜ao de espa¸co euclidiano Rn
, estabelecendo suas propri-
...
Vetores e Func¸˜oes Vetoriais 3
1.1.8
Exemplo Em R7
, considere X = (0, 1, 2, −1, 0,
√
2,
√
3) e Y = (2,
√
3, π, 1, 0, 2, ...
4 O Espac¸o Euclidiano Rn
onde usamos a propriedade distributiva de R junto com a defini¸c˜ao 1.1.5, e obtemos (i). Para (i...
Vetores e Func¸˜oes Vetoriais 5
Figura 1: Os N´umeros Reais R
r
0
r
1
r√
3
r
π +∞−∞
r
−
√
2
Para visualizar o R2
, tomamos...
6 O Espac¸o Euclidiano Rn
Pronto, agora ´e s´o subir (se x3  0), ou descer (se x3  0), paralelamente ao eixo-z at´e atingi...
Vetores e Func¸˜oes Vetoriais 7
Agora podemos descrever geometricamente a adi¸c˜ao de n-uplas e a multiplica¸c˜ao de uma n...
8 Produto Interno e Norma
1.2.2
Proposic¸˜ao Se X, Y, Z ∈ Rn
e a ∈ R s˜ao arbitr´arios, ent˜ao valem as seguintes propried...
Vetores e Func¸˜oes Vetoriais 9
v) X − Y 2
= X 2
− 2X · Y + Y 2
.
Demonstrac¸˜ao: Temos que
aX = (aX) · (aX) = a2(X · X) =...
10 Produto Interno e Norma
Tomemos agora X, Y dois vetores em R2
(ou R3
) que fazem entre si um ˆangulo θ, como
mostra a fi...
Vetores e Func¸˜oes Vetoriais 11
´E conveniente notar aqui que a express˜ao que define PXY pode muito bem ser usada para
o ...
12 Produto Interno e Norma
1.2.16
Proposic¸˜ao Sejam X, Y ∈ Rn
com X = O. Ent˜ao, Y − PXY ´e perpendicular a X.
Portanto, ...
Vetores e Func¸˜oes Vetoriais 13
(iii) [Desigualdade triangular] X + Y ≤ X + Y ;
(iv) [Desigualdade triangular] | X − Y | ...
14 Produto Interno e Norma
1.2.21
Exemplo Sejam X = (1, 2, 1, 0) e Y = (1, 1, 3, 1) dois elementos do R4
. Ent˜ao, X =
√
6...
Vetores e Func¸˜oes Vetoriais 15
(iii) d(X, Z) ≤ d(X, Y ) + d(Y, Z).
Demonstrac¸˜ao: Para (ii) basta observar que Y − X = ...
16 Retas e Planos
Logo, Q ∈ l, o que implica que l ´e a reta que passa por P e Q. Deixando t percorrer o intervalo
fechado...
Vetores e Func¸˜oes Vetoriais 17
1.3.6
Exemplo Sejam l1 = P + [V ] e l2 = Q + [W], onde P = (1, 0, 0), Q = (0, 1, 0), V = ...
18 Retas e Planos
cont´em os pontos Q e R. Para ver isto, ponha s = 1 e t = 0, para obter Q, e s = 0 e t = 1, para
encontr...
Vetores e Func¸˜oes Vetoriais 19
Logo,
z − p3 = (v3 w3)
v1 w1
v2 w2
−1
x − p1
y − p2
=
1
v1w2 − v2w1
(v3 w3)
w2 −w1
−v2 v1...
20 Retas e Planos
1.3.13
Definic¸˜ao Sejam X = (x1, x2, x3) e Y = (y1, y2, y3) duas triplas em R3
. O produto vetorial
de ...
Vetores e Func¸˜oes Vetoriais 21
1.3.16
Corol´ario Se X, Y ∈ R3
, ent˜ao
X × Y = X Y sen ∠(X, Y ).
(Geometricamente, isto ...
22 Retas e Planos
1.3.19
Corol´ario Se X, Y ∈ R3
s˜ao ortonormais (unit´arios e ortogonais), ent˜ao {X, Y, X × Y } ´e
uma ...
Vetores e Func¸˜oes Vetoriais 23
PXY . Da proposi¸c˜ao 1.2.16 vem que PXY = λX, onde λ = (X · Y )/ X 2
, e que Y − PXY ´e
...
24 Retas e Planos
1.3.23
Corol´ario Dados vetores X, Y e Z em R3
, ent˜ao X × (Y × Z) = (X × Y ) × Z se, e
somente se, (X ...
Vetores e Func¸˜oes Vetoriais 25
1.3.25
Proposic¸˜ao Seja l = P +[V ] a reta do Rn
que passa por P e ´e paralela a V . Dad...
26 Retas e Planos
onde N = (a, b) = (−v2, v1) ´e normal a l e c = ax1 + bx2. (Neste caso, a equa¸c˜ao cartesiana de
l ´e: ...
Vetores e Func¸˜oes Vetoriais 27
1.3.30
Proposic¸˜ao Seja H o hiperplano do Rn
que ´e perpendicular a N e passa por P. Dad...
28 Func¸˜oes Vetoriais
1.4
Func¸˜oes Vetoriais
Nesta se¸c˜ao, estudaremos as no¸c˜oes b´asicas relacionadas com aplica¸c˜o...
Vetores e Func¸˜oes Vetoriais 29
1.4.4
Exemplo Seja f(t) = (x0 + a cos t, y0 + b sen t), t ∈ [0, 2π], onde a  0, b  0, x0 ...
30 Func¸˜oes Vetoriais
1.4.5
Exemplo A fun¸c˜ao vetorial
g : R2
−−−−−→ R3
(u, v) −−−−−→ g(u, v) = (1 + u, 2 + v, u + v)
te...
Vetores e Func¸˜oes Vetoriais 31
1.4.9 [Parabol´oide de Revoluc¸˜ao]
Exemplo Seja f : R2
−→ R, f(x, y) = x2
+ y2
, a qual
...
32 Func¸˜oes Vetoriais
Procedimento an´alogo, agora usando planos a ≤ 0, d´a a figura 21-(b). Os cortes de G(f) por
planos ...
Vetores e Func¸˜oes Vetoriais 33
Observac¸˜ao ´E muito comum na pr´atica nos referirmos a uma equa¸c˜ao Y = f(x1, x2, . . ...
34 Func¸˜oes Vetoriais
(iii) f−1
(a) = S1
(
√
a) = {(x, y); x2
+ y2
= a}, se a  0.
Na figura 25, temos o gr´afico de f e alg...
Vetores e Func¸˜oes Vetoriais 35
que, claro, n˜ao tem solu¸c˜oes, se a  0. Se a ≥ 0, f−1
(a, b) ´e a interse¸c˜ao de S2
(
...
36 Func¸˜oes Vetoriais Especiais
´e a geratriz de f−1
(−1), que, por isso, tem duas folhas, como mostramos na figura 28-(c)...
Vetores e Func¸˜oes Vetoriais 37
Agora, se a ∈ R, vem que
T(aX) = T(ax1, ax2)
= (ax1 + ax2, ax2 − ax1)
= a(x1 + x2, x2 − x...
38 Func¸˜oes Vetoriais Especiais
Como Im(T) ⊂ Rm
, vem que
T(e1) =






a11
a21
...
am1






, T(e2) =


...
Vetores e Func¸˜oes Vetoriais 39
1.5.4
Definic¸˜ao A matriz M(T) = (aij) ´e conhecida como a matriz de T com rela¸c˜ao `as...
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  1. 1. C´ALCULOvolume 1 A. Carlos & J. Adonai UFAL-2007
  2. 2. C Conte´udo
  3. 3. ii Conte´udo 1 Vetores e Func¸˜oes Vetoriais . . . . . . . . . . . . . . . . . . . . 1 1.1 O Espac¸o Euclidiano Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.4 Operac¸˜oes com n-uplas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.13 Interpretac¸˜oes Geom´etricas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Produto Interno e Norma . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2.7 Interpretac¸˜oes Geom´etricas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2.11 A Desigualdade de Cauchy-Schwarz . . . . . . . . . . . . . . . . . . . 11 1.3 Retas e Planos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.3.12 Produto Vetorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.3.24 Distˆancia de um Ponto a uma Reta . . . . . . . . . . . . . . . . . . . . 24 1.3.29 Distˆancia de um Ponto a um Hiperplano . . . . . . . . . . . . . . 26 1.4 Func¸˜oes Vetoriais . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.4.6 Conjuntos Associados a Func¸˜oes Vetoriais . . . . . . . . . . . 30 1.5 Func¸˜oes Vetoriais Especiais . . . . . . . . . . . . . . . . . . . . . . . . 36 1.5.22 Superf´ıcies de Revoluc¸˜ao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 1 Vetores e Func¸˜oes Vetoriais – Exerc´ıcios . . . . . 50 2 C´alculo das Curvas Parametrizadas . . . . . . . . . . . 57 2.1 Limite e Continuidade . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 2.2 Derivadas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2.2.3 Interpretac¸˜ao Geom´etrica . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 2.2.14 Derivadas de Ordem Superior . . . . . . . . . . . . . . . . . . . . . . . . . 65 2.2.19 Interpretac¸˜ao F´ısica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
  4. 4. Conte´udo iii 2.3 Geometria das Curvas Parametrizadas . . . . . . . . . . . . . 68 2.3.8 Curvatura e Torc¸˜ao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 2.3.27 Curvas Planas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 2.3.31 C´ırculos no R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 2.3.36 Comprimento de Arco . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 2 Curvas Parametrizadas – Exerc´ıcios . . . . . . . . . . . 84 3 Func¸˜oes Cont´ınuas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 3.1 Limite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.1.15 Propriedades dos Limites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 3.2 Continuidade . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 3 Func¸˜oes Cont´ınuas – Exerc´ıcios . . . . . . . . . . . . . . 111 4 Derivadas Parciais . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 4.1 Derivadas Parciais em R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 4.1.17 Interpretac¸˜ao Geom´etrica . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.2 Derivadas Parciais de Ordem Superior . . . . . . . . . . . . 123 4.2.3 O Teorema de Schwarz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 4.3 Derivadas Parciais em Rn . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.4 Derivadas Parciais Vetoriais . . . . . . . . . . . . . . . . . . . . . . 135 4.4.10 Interpretac¸˜ao Geom´etrica . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 4.5 Derivadas Direcionais . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
  5. 5. iv Conte´udo 4.5.3 Interpretac¸˜ao Geom´etrica . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 4 Derivadas Parciais – Exerc´ıcios . . . . . . . . . . . . . . . 147 5 Aplicac¸˜oes Diferenci´aveis . . . . . . . . . . . . . . . . . . . . . 152 5.1 A Derivada . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 5.1.21 Aplicac¸˜oes de Classe C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 5.1.30 Aproximac¸˜ao Afim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 5.2 Operac¸˜oes com Aplicac¸˜oes Diferenci´aveis . . . . . . . . 173 5.2.6 A Regra da Cadeia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 5.3 O Teorema do Valor M´edio . . . . . . . . . . . . . . . . . . . . . . . 187 5.4 Algumas Aplicac¸˜oes do Gradiente . . . . . . . . . . . . . . . . 194 5.4.6 Superf´ıcies Definidas Implicitamente . . . . . . . . . . . . . . . . . 196 5 Aplicac¸˜oes Diferenci´aveis – Exerc´ıcios . . . . . . . 199 6 Func¸˜oes Inversa e Impl´ıcita . . . . . . . . . . . . . . . . . . . 206 6.1 Preliminares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.1.1 Seq¨uˆencias em Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.1.24 Func¸˜oes Cont´ınuas em Conjuntos Compactos . . . . . . . . 214 6.1.39 Norma de Uma Aplicac¸˜ao Linear . . . . . . . . . . . . . . . . . . . . . 217 6.2 Contrac¸˜oes, Pontos Fixos e Perturbac¸˜oes . . . . . . . . 222 6.3 O Teorema da Func¸˜ao Inversa . . . . . . . . . . . . . . . . . . . . . 229
  6. 6. Conte´udo v 6.4 O Teorema da Func¸˜ao Impl´ıcita . . . . . . . . . . . . . . . . . . . 236 6.4.1 O Caso f : D ⊂ R2 −→ R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 6.4.9 O Caso f : D ⊂ Rn+m −→ Rm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 6.5 Superf´ıcies Regulares em R3 . . . . . . . . . . . . . . . . . . . . . . . 249 6 Func¸˜oes Inversa e Impl´ıcita – Exerc´ıcios . . . . . 254 S Sugest˜oes e Respostas . . . . . . . . . . . . . . . . . . . . . . . . 264 I ´Indice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 R Referˆencias Bibliogr´aficas . . . . . . . . . . . . . . . . . . 290
  7. 7. 1 Vetores e Func¸˜oes Vetoriais rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr π = P + [{V, W}] T z E y         ©x ¨¨¨¨ ¨¨¨ ¨ d d d d d d d d ¨¨¨¨B d d d ¨ ¨ ¨ d d d d‚ ¨¨¨B ˆˆˆˆˆˆˆˆz d d d‚ V sV W tW sX X − PssP O s Verso Preliminar por A. Carlos & J. Adonai
  8. 8. 1.1 O Espac¸o Euclidiano Rn Nesta se¸c˜ao, introduziremos a no¸c˜ao de espa¸co euclidiano Rn , estabelecendo suas propri- edades alg´ebricas e geom´etricas b´asicas. Come¸camos com sua defini¸c˜ao. 1.1.1 Definic¸˜ao Dado n ∈ N, o espa¸co euclidiano Rn ´e definido como sendo o conjunto de todas as n-uplas de n´umeros reais X = (x1, x2, . . . , xn), isto ´e, Rn = {X = (x1, x2, . . . , xn); xi ∈ R, i = 1, 2, . . . n}. 1.1.2 Definic¸˜ao Dada uma n-upla X = (x1, x2, . . . , xn), os n´umeros reais x1, x2, . . . , xn s˜ao chamados coordenadas de X. 1.1.3 Definic¸˜ao Dadas n-uplas X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn), diremos que X = Y se x1 = y1, x2 = y2, . . . , xn = yn. 1.1.4 Operac¸˜oes com n-uplas As estruturas aditiva e multiplicativa do corpo R induzem, naturalmente, uma estrutura de espa¸co vetorial sobre Rn . A adi¸c˜ao de n-uplas e a multiplica¸c˜ao de uma n-upla por um n´umero real s˜ao definidas a seguir. 1.1.5 Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). A soma de X com Y , indicada por X + Y , ´e a n-upla X + Y = (x1 + y1, x2 + y2, . . . , xn + yn). 1.1.6 Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e a ∈ R. O produto de X pelo n´umero real a, indicada por aX, ´e a n-upla aX = (ax1, ax2, . . . , axn). 1.1.7 Exemplo No espa¸co R4 , considere X = (1, 2, −π, √ 2), Y = (2, √ 3, π, 1) e a = √ 2. Ent˜ao, X + Y = (3, 2 + √ 3, 0, 1 + √ 2) e aX = ( √ 2, 2 √ 2, −π √ 2, 2). 2
  9. 9. Vetores e Func¸˜oes Vetoriais 3 1.1.8 Exemplo Em R7 , considere X = (0, 1, 2, −1, 0, √ 2, √ 3) e Y = (2, √ 3, π, 1, 0, 2, − √ 3). Ent˜ao, X + Y = (2, 1 + √ 3, 2 + π, 0, 0, 2 + √ 2, 0). As proposi¸c˜oes que seguem mostram que as opera¸c˜oes com n-uplas rec´em-definidas satis- fazem os axiomas de espa¸co vetorial. Tal fato justifica a terminologia que consiste em chamar uma n-upla, de vetor no Rn . 1.1.9 Proposic¸˜ao Se X, Y, Z ∈ Rn , ent˜ao valem as seguintes propriedades: (i) [Comutatividade] X + Y = Y + X; (ii) [Associatividade] (X + Y ) + Z = X + (Y + Z); (iii) [Elemento Neutro] a n-upla O = (0, 0, . . . , 0), chamada n-upla nula (ou zero), ´e a ´unica n-upla tal que X + O = X; (iv) [Sim´etrico] a n-upla −X = (−x1, −x2, . . . , −xn), chamada sim´etrico da n-upla X, ´e a ´unica n-upla tal que X + (−X) = O. Demonstrac¸˜ao: Vejamos a demonstra¸c˜ao de (ii). As demais s˜ao igualmente simples, e ser˜ao deixadas como exerc´ıcio para o leitor. Temos que (X + Y ) + Z = (x1 + y1, x2 + y2, . . . , xn + yn) + (z1, z2, . . . , zn) = ((x1 + y1) + z1, (x2 + y2) + z2, . . . , (xn + yn) + zn)) = (x1 + (y1 + z1), x2 + (y2 + z2), . . . , xn + (yn + zn)) = X + (Y + Z), onde, na passagem da segunda para a terceira equa¸c˜ao, foi usada a propriedade associativa dos n´umeros reais. Os demais pontos envolvem apenas a defini¸c˜ao 1.1.5. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.1.10 Proposic¸˜ao Se X, Y ∈ Rn e a, b ∈ R, ent˜ao valem as seguintes propriedades: (i) [Distributividade] a(X + Y ) = aX + aY ; (ii) [Distributividade] (a + b)X = aX + bX; (iii) [Associatividade] (ab)X = a(bX); (iv) 1 X = X. Demonstrac¸˜ao: Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). Temos que a(X + Y ) = a(x1 + y1, x2 + y2, . . . , xn + yn) = (a(x1 + y1), a(x2 + y2), . . . , a(xn + yn)) = (ax1, ax2, . . . , axn) + (ay1, ay2, . . . , ayn) = aX + aY,
  10. 10. 4 O Espac¸o Euclidiano Rn onde usamos a propriedade distributiva de R junto com a defini¸c˜ao 1.1.5, e obtemos (i). Para (ii), a propriedade distributiva de R e a defini¸c˜ao 1.1.6 s˜ao usadas: (a + b)X = ((a + b)x1, (a + b)x2, . . . , (a + b)xn) = (ax1 + bx1, ax2 + bx2, . . . , axn + bxn) = aX + bX, como quer´ıamos. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.1.11 Corol´ario Rn ´e um espa¸co vetorial de dimens˜ao n. Demonstrac¸˜ao: As proposi¸c˜oes 1.1.9 e 1.1.10 mostram que Rn ´e uma espa¸co vetorial. Falta mostrar que dim Rn = n. Para isto, sejam e1, e2, . . . , en, definidos por e1 = (1, 0, 0, . . . , 0) e2 = (0, 1, 0, . . . , 0) ... ... ... en = (0, 0, . . . , 0, 1). Note que se X = (x1, x2, . . . , xn), ent˜ao X = x1e1 + x2e2 + · · · + xnen. Logo, {e1, e2, . . . , en} gera Rn . Agora se c1, c2, . . ., cn s˜ao n´umeros reais tais que c1e1 + c2e2 + · · · + cnen = (0, 0, . . . , 0), vem que c1 = c2 = · · · = cn = 0, o que completa a prova. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.1.12 Definic¸˜ao A base {e1, e2, . . . , en} constru´ıda acima ´e chamada base canˆonica do espa¸co Rn . 1.1.13 Interpretac¸˜oes Geom´etricas A interpreta¸c˜ao geom´etrica que descreve R como uma reta orientada, sobre a qual se esco- lhe um ponto O, o qual corresponde ao n´umero zero, uma unidade de medida, que corresponde ao n´umero 1, pode ser estendida a uma interpreta¸c˜ao geom´etrica dos espa¸cos euclidianos R2 e R3 . Para Rn , n ≥ 4, fica por conta da imagina¸c˜ao de cada um.
  11. 11. Vetores e Func¸˜oes Vetoriais 5 Figura 1: Os N´umeros Reais R r 0 r 1 r√ 3 r π +∞−∞ r − √ 2 Para visualizar o R2 , tomamos duas c´opias de R, as quais chamamos de eixos coordenados. Estes eixos s˜ao denotados por eixo-x e eixo-y. O passo seguinte consiste em dispor os eixos coordenados em um plano euclidiano de modo que eles se interceptem ortogonalmente ao longo de suas origens, o que produz o ponto O, que ser´a associado `a dupla (2-upla) nula (0, 0), conforme figura 2. Feito isso, uma dupla de R2 , digamos X = (x1, x2), ´e olhada como aquele ponto do plano, tamb´em indicado por X, que se projeta ortogonalmente sobre eixo-x e eixo-y naqueles pontos que correspondem a x1 e x2, respectivamente. Isto ´e feito tra¸cando-se por x1, uma reta perpendicular ao eixo-x, e por x2, uma reta perpendicular ao eixo-y. A interse¸c˜ao destas perpendiculares ´e exatamente o ponto do plano que representar´a X. Deste modo, fica estabelecida uma bije¸c˜ao entre o plano euclidiano que fixamos e o espa¸co R2 . H´a situa¸c˜oes em que ´e conveniente representar uma dupla X como um segmento orientado localizado em O. A figura 2 exibe duas duplas X = (x1, x2) e Y = (y1, y2), de modo que cada uma delas aparece ora como ponto, ora como segmento orientado. Neste ponto, observamos que a no¸c˜ao geom´etrica de ˆangulo entre X e Y ´e mais adequada `a figura 2-(c), onde ambas s˜ao olhadas como segmentos orientados localizados em O. J´a a figura 2-(b) ´e perfeita para motivar a no¸c˜ao de reta que passa por Y e ´e paralela a X, conforme defini¸c˜ao 1.3.1. O s E T x y Figura 2-(a) s Y y2 r y1 r sXx2 r x1 r O s E T x y Figura 2-(b) s Y y2 r y1 r¨ ¨¨¨¨¨¨¨B Xx2 r x1 r O s E T x y Figura 2-(c) ¢ ¢ ¢ ¢ ¢ ¢¢ Y y2 r y1 r¨ ¨¨¨¨¨¨¨B Xx2 r x1 r Analogamente, para fazermos geometria em R3 , recorremos a trˆes retas, que formar˜ao os eixos coordenados, indicados, respectivamente, por eixo-x, eixo-y e eixo-z, e as colocamos no espa¸co euclidiano tridimensional de modo que elas se interceptem ortogonalmente em suas ori- gens, produzindo o ponto O, que corresponder´a `a tripla (3-upla) nula (0, 0, 0). Feito isso, temos em m˜aos trˆes planos especiais, chamados coordenados, e denotados por plano-xy, plano-xz e plano-yz. Agora a uma tripla X = (x1, x2, x3) fazemos corresponder o ponto do espa¸co cujas proje¸c˜oes ortogonais sobre os eixos coordenados eixo-x, eixo-y e eixo-z coincidem, respectiva- mente, com os pontos destes eixos que est˜ao associados aos n´umeros reais x1, x2, x3. A figura 3 mostra como isso ´e feito: inicialmente, marcamos x1 no eixo-x, x2 no eixo-y e x3 no eixo-z. Depois, a partir de x1, caminhamos paralelamente ao eixo-y at´e atingir a medida x2, onde en- contramos o ponto que representa a proje¸c˜ao de X no plano-xy, que corresponde a (x1, x2, 0).
  12. 12. 6 O Espac¸o Euclidiano Rn Pronto, agora ´e s´o subir (se x3 0), ou descer (se x3 0), paralelamente ao eixo-z at´e atingir uma altura x3, e encontramos o ponto do espa¸co que representar´a X. Note que na figura 3-(b), a tripla X ´e mostrada como um segmento orientado localizado na origem. O T z E y             ©x X (x1, x2, 0) x2 rx3 r (x1, 0, x3) x1 s (0, x2, x3) s E E E E E E E                        T T T T T d d d d d d d d Figura 3-(a) s r s s O T z E y             ©x ¨¨¨¨¨B X (x1, x2, 0) x2 rx3 r (x1, 0, x3) x1 s (0, x2, x3) s E E E E E E E                        T T T T T d d d d d d d d Figura 3-(b) s r s O s E T x y Figura 4-(a) s Y y2 r y1 r sXx2 r x1 r s X + Y x2 + y2 r x1 + y1 r O s E T x y Figura 4-(b) ¢ ¢ ¢ ¢ ¢ ¢¢ Y y2 r y1 r¨¨¨¨ ¨¨¨ ¨B Xx2 r x1 r                    X + Y x2 + y2 r x1 + y1 r ¨ ¨ ¨ ¨ ¨ ¨B ¢¢ ¢¢ ¢¢ ¢¢ O s E T x y Figura 4-(c) sYy2 r y1 r sXx2 r ¨¨ ¨¨¨¨¨¨B x1 r s X + Y x2 + y2 r x1 + y1 r ¨ ¨ ¨ ¨ ¨ ¨B
  13. 13. Vetores e Func¸˜oes Vetoriais 7 Agora podemos descrever geometricamente a adi¸c˜ao de n-uplas e a multiplica¸c˜ao de uma n-upla por um n´umero real. Para isso, usamos a figura 4, onde marcamos as duplas X = (x1, x2) e Y = (y1, y2) juntamente com sua soma. Olhando atentamente o conjunto de figuras 4, observamos que a (b), que mostra X e Y como segmentos orientados localizados em O, nos d´a uma regra geom´etrica evidente: o segmento orientado X + Y ´e a diagonal do paralelogramo com arestas X e Y . Portanto, a dupla X + Y ´e o ponto final deste segmento. A regra geom´etrica contida na figura 4-(c) ´e a mais simples: localizamos em Y , o segmento orientado X. O ponto final obtido ´e a dupla X + Y . F´acil, n˜ao? Para finalizar, consideremos a figura abaixo que ilustra geometricamente como funciona a multiplica¸c˜ao de uma n-upla por um escalar: a n-upla aX, a ∈ R, tem comprimento igual ao comprimento de X multiplicado pelo valor absoluto de a. Seu sentido ´e o mesmo de X, quando a 0, e lhe ´e contr´ario, quando a 0. O s E T x y Figura 5 ¨¨¨ ¨¨¨ ¨¨B ¨¨¨ ¨¨¨ ¨¨% X −X x2 r x1 r¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨B ¨¨¨ ¨B r aX (a 1) aX (0 a 1) ax1 r ax2 1.2 Produto Interno e Norma Vimos, recentemente, que o espa¸co Rn possui uma estrutura de espa¸co vetorial. Em v´arias situa¸c˜oes, precisamos das no¸c˜oes de comprimento, ˆangulo, ortogonalidade, n˜ao presentes nas opera¸c˜oes de espa¸co vetorial. Estas no¸c˜oes s˜ao obtidas a partir de um produto escalar (ou interno), que introduziremos agora. 1.2.1 Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). O produto escalar (ou in- terno) de X por Y , indicado por X · Y , ´e o n´umero real dado por X · Y = x1y1 + x2y2 + · · · + xnyn. A seguinte proposi¸c˜ao descreve as propriedades deste produto.
  14. 14. 8 Produto Interno e Norma 1.2.2 Proposic¸˜ao Se X, Y, Z ∈ Rn e a ∈ R s˜ao arbitr´arios, ent˜ao valem as seguintes proprieda- des: (i) [Positividade] X · X ≥ 0, e X · X = 0 se, e somente se, X = O = (0, 0, . . . , 0); (ii) [Comutatividade] X · Y = Y · X; (iii) [Distributividade] X · (Y + Z) = X · Y + X · Z; (iv) [Homogeneidade] (aX) · Y = X · (aY ) = a(X · Y ). Demonstrac¸˜ao: Com X = (x1, x2, . . . , xn), Y = (y1, y2, . . . , yn) e Z = (z1, z2, . . . , zn), temos que X · (Y + Z) = x1(y1 + z1) + x2(y2 + z2) + · · · + xn(yn + zn) = x1y1 + x1z1 + x2y2 + x2z2 + · · · + xnyn + xnzn = (x1y1 + x2y2 + · · · + xnyn) + (x1z1 + x2z2 + · · · + xnzn) = X · Y + X · Z. Assim, fica provado (iii). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp A proposi¸c˜ao 1.2.2, em seu item (i), permite-nos dar a seguinte defini¸c˜ao. 1.2.3 Definic¸˜ao Seja X = (x1, x2, . . . , xn). A norma (ou comprimento) de X ´e dada por X = √ X · X = x2 1 + x2 2 + · · · + x2 n. 1.2.4 Definic¸˜ao X ∈ Rn ´e dito unit´ario se X = 1. 1.2.5 Exemplo Se X = (1, 2, −1, 0) e Y = (3, 1, 0, − √ 2), ent˜ao X · Y = 5, X = √ 6 e Y = 2 √ 3. 1.2.6 Proposic¸˜ao Dados X, Y ∈ Rn e a ∈ R, temos que (i) X ≥ 0, e X = 0 se, e somente se, X = O; (ii) aX = |a| X , onde |a| ´e o valor absoluto de a; (iii) se X = O, o vetor uX = X/ X ´e unit´ario (uX ´e conhecido como vetor unit´ario na dire¸c˜ao de X); (iv) X + Y 2 = X 2 + 2X · Y + Y 2 ;
  15. 15. Vetores e Func¸˜oes Vetoriais 9 v) X − Y 2 = X 2 − 2X · Y + Y 2 . Demonstrac¸˜ao: Temos que aX = (aX) · (aX) = a2(X · X) = |a| X , o que prova (ii). Agora, usando (ii), vem que uX = X X = 1 X X = 1, e segue-se (iii). Para (iv), simplesmente expandimos X + Y 2 , usando a distributividade e a comutatividade do produto escalar. X + Y 2 = (X + Y ) · (X + Y ) = X · X + X · Y + Y · X + Y · Y = X 2 + 2X · Y + Y 2 . Os demais itens s˜ao, tamb´em, de prova simples. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.2.7 Interpretac¸˜oes Geom´etricas Estudaremos agora os aspectos geom´etricos envolvidos pelo produto escalar e pela norma. Consideremos a figura 6 que segue. Note que o triˆangulo de v´ertices O, X e (x1, x2, 0) ´e retˆangulo no v´ertice (x1, x2, 0) e seus catetos medem x2 1 + x2 2 e x3. Logo, sua hipotenusa mede x2 1 + x2 2 + x2 3, o que coincide com X . rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrr rrrrrrrrr rrrrrrrrr rrrrrrrrrr rrrrrrrrrrr rrrrrrrrrrrr rrrrrrrrrrrr rrrrrrrrrrrrr rrrrrrrrrrrrrr rrrrrrrrrrrrrrr rrrrrrrrrrrrrrr rrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrr qqqqqqqqqqqqqq qqqqqqq qqqqqqqqq qqqqqqqqqq qqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqqqq qqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq T z E y             ©x ¨¨¨¨¨B X s(x1, x2, 0) x2 rx3 r r x1         d d d d d d d d d Figura 6 x2 1 + x2 2 x3 d   O s Deste modo, podemos visualizar X como o comprimento (euclidiano) do segmento orientado X. Isto tamb´em acontece no R2 , como o leitor pode verificar facilmente.
  16. 16. 10 Produto Interno e Norma Tomemos agora X, Y dois vetores em R2 (ou R3 ) que fazem entre si um ˆangulo θ, como mostra a figura 7. (Note a interpreta¸c˜ao geom´etri- ca para a diferen¸ca Y − X: o segmento orientado, localizado em X, que representa Y − X termina em Y .) Aplicando a lei dos cossenos ao triˆangulo OXY , obtemos que Y − X 2 = X 2 + Y 2 − 2 X Y cos θ, o que comparado com (v) da proposi¸c˜ao 1.2.6 d´a que X · Y = X Y cos θ. O s E T x y Figura 7 ¢ ¢ ¢ ¢ ¢ ¢¢ Yy2 r y1 r¨¨ ¨¨ ¨¨ ¨¨B Xx2 r x1 r€€€€€€i Y − X y2 − x2 r y1 − x1 r €€ €€€i θppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Assim vemos que a no¸c˜ao de produto interno est´a bem ligada `a no¸c˜ao de ˆangulo entre vetores, e colhemos a seguinte proposi¸c˜ao, onde ∠(X, Y ) indica o ˆangulo, no intervalo [0, π], entre os vetores X e Y . 1.2.8 Proposic¸˜ao Se X, Y ∈ Rn (n = 2, 3) e θ = ∠(X, Y ), ent˜ao X · Y = X Y cos θ. Em particular, vale a desigualdade de Cauchy-Schwarz: |X · Y | ≤ X Y , a igualdade ocorrendo apenas quando X e Y s˜ao linearmente dependentes. 1.2.9 Exemplo A proposi¸c˜ao 1.2.8 mostra que dois vetores X e Y em Rn , n = 2, 3, s˜ao perpen- diculares se, e somente se, X · Y = 0. De fato, X e Y s˜ao perpendiculares se, e somente se, ∠(X, Y ) = π/2. Como caso particular disto, note que dado X = (x1, x2), o vetor Y = (−x2, x1) ´e perpendicular a X. O vetor Y ´e obtido de X por uma rota¸c˜ao em torno de O no sentido anti-hor´ario. Como exerc´ıcio, o leitor deve esbo¸car X e Y , para se convencer deste fato. Continuando com a nossa discuss˜ao geom´e- trica construiremos, agora, o que chamamos de proje¸c˜ao ortogonal de um vetor na dire¸c˜ao de ou- tro n˜ao-nulo dado. Sejam, ent˜ao, X = O e Y como na figura 8, onde uX = X/ X ´e o vetor unit´ario na dire¸c˜ao de X, θ = ∠(X, Y ) e PXY ´e o vetor obtido pela proje¸c˜ao ortogonal de Y sobre X. Assim PXY = auX, onde a = Y cos θ = uX · Y = X · Y X . O s e ¨ Figura 8: Proje¸c˜ao de Y sobre X ¨ ¨ e e¨¨ a ¢ ¢ ¢ ¢ ¢ ¢ ¢¢ e e e e eu ¨ ¨¨ ¨¨B Y ¨ ¨¨ ¨¨¨ ¨¨ ¨¨B ¨ ¨¨B uX PXY Y − PXY X θppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Logo, PXY = X · Y X 2 X. Da constru¸c˜ao de PXY decorre facilmente que o vetor Y − PXY ´e ortogonal a X, o que pode ser verificado, tamb´em, analiticamente: X · (Y − PXY ) = X · (Y − X · Y X 2 X) = X · Y − X · Y = 0.
  17. 17. Vetores e Func¸˜oes Vetoriais 11 ´E conveniente notar aqui que a express˜ao que define PXY pode muito bem ser usada para o espa¸co Rn , visto que ela n˜ao cont´em nenhum apelo geom´etrico expl´ıcito. Isto ser´a parte do conte´udo da pr´oxima subse¸c˜ao. 1.2.10 Exemplo Considere, em R2 , o triˆangulo ABC, onde A = (1, 1), B = (3, 2) e C = (0, 4). Os veto- res X = C − B e Y = A − B aparecem na figura 9 localiza- dos em B. Temos que X = (−3, 2) e Y = (−2, −1). Assim, X · Y = 4, X 2 = 13 e PXY = 4 13 X = 4 13 (−3, 2). Agora fica f´acil calcular a altura relativa ao lado BC, hBC, do triˆangulo ABC. De fato, temos hBC = Y − PXY = (− 14 13 , − 21 13 ) = 7 √ 13 13 . O s E T x y Figura 9 sBhBC PXY r 3 r2 s A X Y 1 1 r r ssC4      k  k ¨¨¨¨%f f f f f ff r 1.2.11 A Desigualdade de Cauchy-Schwarz Inicialmente, nos inspiramos nas no¸c˜oes geom´etricas que usamos h´a pouco, para definir ortogonalidade entre n-uplas e construir a proje¸c˜ao ortogonal de uma n-upla sobre outra. 1.2.12 Definic¸˜ao Dados X, Y ∈ Rn , diremos que X ´e ortogonal (perpendicular) a Y se X · Y = 0. 1.2.13 Definic¸˜ao Um subconjunto {v1, v2, . . . , vk} ⊂ Rn ´e dito ortogonal se vi · vj = 0, para 1 ≤ i, j ≤ k, i = j. {v1, v2, . . . , vk} ´e ortonormal se ´e ortogonal e seus elementos s˜ao vetores unit´arios. 1.2.14 Exemplo Seja {e1, e2, . . . , en} a base canˆonica do espa¸co Rn (veja defini¸c˜ao 1.1.12). ´E claro que e1 = e2 = · · · = en = 1. Al´em disto, dados i, j ∈ {1, 2, . . . , n}, i = j, temos que ei · ej = 0. Logo, a base canˆonica ´e uma base ortonormal do espa¸co Rn . 1.2.15 Definic¸˜ao Dados dois vetores X, Y ∈ Rn , X = O, o vetor PXY = X · Y X 2 X ´e chamado proje¸c˜ao de Y sobre X.
  18. 18. 12 Produto Interno e Norma 1.2.16 Proposic¸˜ao Sejam X, Y ∈ Rn com X = O. Ent˜ao, Y − PXY ´e perpendicular a X. Portanto, ´e tamb´em perpendicular a PXY . Demonstrac¸˜ao: X · (Y − PXY ) = X · (Y − X · Y X 2 X) = X · Y − X · Y = 0. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp Agora podemos estender o teorema de Pit´agoras para o Rn . 1.2.17 [Pit´agoras] Proposic¸˜ao Sejam X, Y ∈ Rn com X = O. Ent˜ao, X ´e perpendicular a Y se, e somente se, X + Y 2 = X 2 + Y 2 . Demonstrac¸˜ao: Resulta imediatamente de X + Y 2 = X 2 + Y 2 + 2X · Y , como indica a proposi¸c˜ao 1.2.6, item (iv). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp Enfim a desigualdade de Cauchy-Schwarz, j´a obtida via argumento geom´etricos para os espa¸cos R2 e R3 . 1.2.18 [Cauchy-Schwarz] Teorema Sejam X, Y ∈ Rn . Ent˜ao, |X · Y | ≤ X Y , e a igual- dade ´e atingida se, e somente se, X e Y s˜ao linearmente dependentes. Demonstrac¸˜ao: Inicialmente notamos que se X = O, a desigualdade ´e facilmente verificada. Portanto, podemos supor X = O. Seja PXY a proje¸c˜ao ortogonal de Y sobre X. Segue-se da proposi¸c˜ao 1.2.16 que Y − PXY ´e perpendicular a PXY . Usando a proposi¸c˜ao 1.2.17, obtemos que Y 2 = (Y − PXY ) + PXY 2 = (Y − PXY ) 2 + PXY 2 ≥ PXY 2 , (¶1) e a igualdade ocorre se, e somente se, Y = PXY = X · Y X 2 X. Mas PXY 2 = X · Y X 2 X 2 = (X · Y )2 X 2 , o que combinado com a desigualdade (¶1) d´a (X · Y )2 ≤ X 2 Y 2 , como quer´ıamos. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp Como corol´ario da desigualdade de Cauchy-Schwarz, obtemos as propriedades da norma que faltavam ser apresentadas: as desigualdades triangulares. 1.2.19 Corol´ario Se X, Y ∈ Rn e a ∈ R, ent˜ao (i) X ≥ 0, e X = 0 se, e somente se, X = O; (ii) aX = |a| X ;
  19. 19. Vetores e Func¸˜oes Vetoriais 13 (iii) [Desigualdade triangular] X + Y ≤ X + Y ; (iv) [Desigualdade triangular] | X − Y | ≤ X − Y . Demonstrac¸˜ao: Note que (i) e (ii) aparecem na proposi¸c˜ao 1.2.6. Daremos uma prova para (iii) e (iv). Temos que X + Y 2 = X 2 + 2X · Y + Y 2 ≤ X 2 + 2|X · Y | + Y 2 ≤ X 2 + 2 X Y + Y 2 = ( X + Y )2 , o que implica (iii). A segunda desigualdade triangular resulta da primeira. De fato, X = (X − Y ) + Y ≤ X − Y + Y . Logo, X − Y ≤ X − Y . (¶2) Trocando X por Y , vem que Y − X ≤ Y − X = X − Y . (¶3) Agora, juntando (¶2) e (¶3), segue-se que − X − Y ≤ X − Y ≤ X − Y , o que ´e equivalente a | X − Y | ≤ X − Y , e est´a pronto o corol´ario. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp Neste ponto, tomamos duas n-uplas n˜ao-nulas X e Y . Da desigualdade de Cauchy- Schwarz, obtemos que −1 ≤ X · Y X Y ≤ 1. Logo, existe um ´unico n´umero real θ ∈ [0, π] tal que cos θ = X · Y X Y . Posto isto, temos a seguinte defini¸c˜ao. 1.2.20 Definic¸˜ao Dadas as n-uplas n˜ao-nulas X e Y , o n´umero real ∠(X, Y ) = arccos X · Y X Y . ´e chamado ˆangulo entre X e Y .
  20. 20. 14 Produto Interno e Norma 1.2.21 Exemplo Sejam X = (1, 2, 1, 0) e Y = (1, 1, 3, 1) dois elementos do R4 . Ent˜ao, X = √ 6, Y = 2 √ 3 e X · Y = 6. Logo, ∠(X, Y ) = arccos 6 2 √ 18 = arccos √ 2 2 = π 4 . Observac¸˜ao Seja V um espa¸co vetorial qualquer sobre R, de dimens˜ao finita ou n˜ao. Um produto interno em V ´e definido como sendo uma forma bilinear sim´etrica e positiva definida, que indicamos por , . Isto significa que se X, Y, Z ∈ V e a ∈ R, ent˜ao devem valer: (i) X, X ≥ 0, e X, X = 0 se, e somente se, X ´e o vetor nulo de V; (ii) X, Y = Y, X ; (iii) X, Y + Z = X, Y + X, Z ; (iv) a X, Y = aX, Y = X, aY . Note que o produto escalar · que definimos para o Rn satisfaz estas propriedades, como indica a proposi¸c˜ao 1.2.2. O que queremos chamar a aten¸c˜ao aqui ´e que todo o conte´udo desta subse¸c˜ao poderia ser aplicado para o espa¸co V, com apenas uma mudan¸ca, a saber: a troca do produto escalar · por , . Em particular, ter´ıamos a desigualdade de Cauchy-Schwarz: | X, Y | ≤ X Y , onde, ´e claro, X = X, X . Esta norma tamb´em satisfaz as propriedades do corol´ario 1.2.19. Para finalizar esta subse¸c˜ao, consideraremos em Rn a distˆancia induzida por sua norma. 1.2.22 Definic¸˜ao A distˆancia entre as n-uplas X e Y , indicada por d(X, Y ), ´e o n´umero real d(X, Y ) = Y − X = (y1 − x1)2 + (y2 − x2)2 + · · · + (yn − xn)2. (A figura 7 sugere, tamb´em, esta defini¸c˜ao.) 1.2.23 Exemplo Se X = (1, 2, 3, −1, 2), Y = (1, 1, 2, 0, 1), ent˜ao d(X, Y ) = 2. 1.2.24 Proposic¸˜ao Sejam X, Y, Z ∈ Rn . A distˆancia tem as seguintes propriedades. (i) d(X, Y ) ≥ 0, e d(X, Y ) = 0 se, e somente se, X = Y ; (ii) d(X, Y ) = d(Y, X);
  21. 21. Vetores e Func¸˜oes Vetoriais 15 (iii) d(X, Z) ≤ d(X, Y ) + d(Y, Z). Demonstrac¸˜ao: Para (ii) basta observar que Y − X = X − Y . Vejamos (iii). d(X, Z) = Z − X = (Z − Y ) + (Y − X) ≤ Z − Y + Y − X ≤ d(X, Y ) + d(Y, Z), onde a desigualdade obtida vem do corol´ario 1.2.19, item (iii). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3 Retas e Planos Como vimos fazendo at´e aqui, para definirmos reta e plano no Rn , usaremos alguns argu- mentos geom´etricos no espa¸co euclidiano R2 . A figura 10 ao lado mostra a dupla P, os vetores V = O e N (perpendicular a V ) e a reta l que passa por P e ´e paralela a V . Se X ´e um ponto qualquer de l, ent˜ao o vetor X −P deve ser um m´ultiplo de V , isto ´e, existe t ∈ R tal que X − P = tV, ou X = P + tV, equa¸c˜ao que descreve os pontos de l, e motiva a seguinte defini¸c˜ao. O s E T x y Figura 10: Reta passando por P e paralela a V s¨¨¨ ¨¨¨ ¨¨¨ ¨ ¨¨¨¨¨¨¨¨ ¨¨¨ ¨¨¨ ¨ ¨¨¨B ¨¨¨ ¨¨Bs X ¨¨ ¨¨¨ P X − P ¨¨¨B e e eu V N l = P + [V ] [V ] 1.3.1 Definic¸˜ao Dados P, V ∈ Rn , V = O, o subconjunto l = P + [V ], onde [V ] indica o subespa¸co gerado por V , ´e chamado reta que passa por P e ´e paralela ao vetor V . Assim, l = {X ∈ Rn ; X = P + tV, t ∈ R}. A equa¸c˜ao X = P + tV ´e a equa¸c˜ao param´etrica de l. 1.3.2 Exemplo Dados P, Q ∈ Rn , P = Q, a reta que passa por P (ou Q) e ´e paralela ao vetor Q − P ´e a reta lPQ = P + [Q − P]. Para t = 1, obtemos X = P + t(Q − P) = Q.
  22. 22. 16 Retas e Planos Logo, Q ∈ l, o que implica que l ´e a reta que passa por P e Q. Deixando t percorrer o intervalo fechado [0, 1], obtemos o subconjunto [P, Q] ⊂ lPQ, o qual chamaremos de segmento de reta ligando P a Q. Assim, [P, Q] = {X = P + t(Q − P); 0 ≤ t ≤ 1}. Para t = 1/2, obtemos M = P + 1 2 (Q − P) = P + Q 2 ∈ [P, Q], ¨¨¨ ¨¨ Figura 11: Segmento [P, Q] P QM s s s o ponto m´edio de [P, Q]. Observe que d(M, P) = d(M, Q) = M − P = M − Q = d(P, Q)/2. 1.3.3 Exemplo Tomemos, em R2 , P = (x0, y0) e V = (v1, v2) = (0, 0). Se X = (x, y) ∈ l = P + [V ] = {X = (x, y) = (x0, y0) + t(v1, v2), t ∈ R} ´e um ponto qualquer de l, ent˜ao x = x0 + tv1 e y = y0 + tv2, t ∈ R. Donde v2x = v2x0 + tv2v1 e v1y = v1y0 + tv1v2 e, portanto, ax + by = c, onde a = −v2, b = v1 e c = ax0 + by0. Esta ´e a equa¸c˜ao cartesiana de l, forma usual nos textos elementares de Geometria Anal´ıtica, e que pode ser reescrita como (X − P) · N = 0, onde N = (a, b) = (−v2, v1) ´e perpendicular a V e, portanto, a l (veja o exemplo 1.2.9). Observac¸˜ao Uma reta l = P + [V ] n˜ao determina unicamente P e V . De fato, se Q ∈ l ´e um ponto qualquer de l e W = λV , λ = 0, ent˜ao l = Q + [W]. 1.3.4 Definic¸˜ao Duas retas no Rn , l1 = P + [V ] e l2 = Q + [W], s˜ao ditas paralelas se V e W s˜ao linearmente dependentes. 1.3.5 Exemplo Sejam l1 = P + [V ] e l2 = Q + [W] duas retas no R2 que n˜ao s˜ao paralelas. Logo, como nossa intui¸c˜ao espera, l1 e l2 devem se tocar num ´unico ponto (o que pode n˜ao ocorrer em dimens˜oes maiores que 2, como mostra o exemplo 1.3.6). Com efeito, {V, W} ´e uma base do R2 (por quˆe?) e, portanto, devem existir ´unicos t1, t2 ∈ R tais que Q − P = t1V + t2W. Donde, Q − t2W = P + t1V . Mas P + t1V ∈ l1 e Q − t2W ∈ l2. Logo, l1 e l2 se interceptam em R = Q − t2W = P + t1V .
  23. 23. Vetores e Func¸˜oes Vetoriais 17 1.3.6 Exemplo Sejam l1 = P + [V ] e l2 = Q + [W], onde P = (1, 0, 0), Q = (0, 1, 0), V = (1, 1, 1) e W = (1, 1, 0). Os vetores V e W s˜ao linearmente independentes, o que resulta de uma simples observa¸c˜ao de suas terceiras coordenadas. Assim, l1 e l2 n˜ao s˜ao paralelas. Entretanto, ao contr´ario do que ocorre no plano (exemplo 1.3.5), l1 e l2 n˜ao se interceptam. De fato, se R ´e um ponto onde estas retas se interceptam, ent˜ao R = P + t1V = (1 + t1, t1, t1) e R = Q + t2W = (t2, 1 + t2, 0), para alguns t1, t2 ∈ R. Isto implica que t2 = 1 = −1, um absurdo. Portanto, devemos mesmo ter l1 ∩ l2 = ∅. Nosso objetivo agora ´e construir planos no Rn . Come¸caremos trabalhando em R3 . Sejam V e W dois vetores linearmente independentes em R3 , localizados em P, como mostra a figura 12. qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq T z E y π = P + [{V, W}]          ©x ¨ ¨¨¨ ¨¨¨ ¨ d d d d d d d d ¨ ¨¨¨B d d d ¨ ¨ ¨ d d d d‚ ¨ ¨¨B ˆˆˆˆˆˆˆˆz d d d‚ V sV W tW sX X − PssP O s Figura 12: Plano que passa por P e ´e paralelo a V e W Seja π = P + [{V, W}] o plano que passa por P e ´e paralelo aos vetores V e W. Dado X ∈ π, o vetor X − P est´a no subespa¸co gerado pelos vetores V e W. Logo, existem escalares s e t tais que X − P = sV + tW, donde X = P + sV + tW. 1.3.7 Definic¸˜ao Dados P, V, W ∈ Rn com {V, W} linearmente independente, o subconjunto π = P + [{V, W}], onde [{V, W}] ´e o subespa¸co gerado por {V, W}, ´e chamado plano que passa por P e ´e paralelo aos vetores V e W. Em outras palavras, π = {X ∈ Rn ; X = P + sV + tW, s, t ∈ R}. A equa¸c˜ao X = P + sV + tW ´e a equa¸c˜ao param´etrica de π. 1.3.8 Exemplo Sejam P, Q, R ∈ Rn trˆes pontos tais que o triˆangulo PQR seja n˜ao-degenerado, isto ´e, os vetores V = Q−P e W = R−P s˜ao linearmente independentes. Ent˜ao, o plano π = P + [{V, W}] = {X ∈ Rn ; X = P + s(Q − P) + t(R − P), s, t ∈ R}
  24. 24. 18 Retas e Planos cont´em os pontos Q e R. Para ver isto, ponha s = 1 e t = 0, para obter Q, e s = 0 e t = 1, para encontrar R. Este ´e o plano que passa pelos pontos P, Q, R, que indicaremos por πPQR. Como caso particular, tomemos, em R3 , os pontos P = (0, 0, 2), Q = (4, 1, 0) e R = (1, 1, 1). Ent˜ao, V = (4, 1, −2) e W = (1, 1, −1), e πPQR fica πPQR = {(x, y, z) = (0, 0, 2) + s(4, 1, −2) + t(1, 1, −1), s, t ∈ R} = {(x, y, z) = (4s + t, s + t, 2 − 2s − t), s, t ∈ R}, Eliminando s e t na equa¸c˜ao param´etrica obtida, obtemos que πPQR = {(x, y, z); x + 2y + 3z = 6}, que ´e a forma cartesiana de πPQR. Note que os coeficientes desta ´ultima equa¸c˜ao, a saber, 1, 2 e 3, d˜ao origem ao vetor N = (1, 2, 3) que, como ´e f´acil de ver, ´e perpendicular aos vetores V e W. Portanto, N ´e tamb´em perpendicular a πPQR. A seguinte proposi¸c˜ao generaliza esta situa¸c˜ao. 1.3.9 Proposic¸˜ao Seja π = P + [{V, W}] um plano do R3 . Ent˜ao existe N = (a, b, c), n˜ao-nulo, perpendicular a V e W (e portanto perpendicular a π) tal que π = {X ∈ R3 ; (X − P) · N = 0} = {(x, y, z); ax + by + cz = d}, onde d = N · P. Demonstrac¸˜ao: Sejam P = (p1, p2, p3), V = (v1, v2, v3) e W = (w1, w2, w3). Assim, π = {(x, y, z) = (p1 + sv1 + tw1, p2 + sv2 + tw2, p3 + sv3 + tw3), s, t ∈ R}. (¶4) Como V e W s˜ao linearmente independentes, a matriz   v1 w1 v2 w2 v3 w3   tem posto 2. Resulta da´ı, que pelo menos uma das matrizes v1 w1 v2 w2 , v1 w1 v3 w3 e v2 w2 v3 w3 tem determinante n˜ao-nulo. Logo, podemos supor, sem perda de generalidade, que a primeira destas matrizes tem inversa, a qual ´e dada por v1 w1 v2 w2 −1 = 1 v1w2 − v2w1 w2 −w1 −v2 v1 . Seja X = (x, y, z) ∈ π um ponto qualquer. De (¶4) vem que x − p1 y − p2 = v1 w1 v2 w2 s t z − p3 = (v3 w3) s t .
  25. 25. Vetores e Func¸˜oes Vetoriais 19 Logo, z − p3 = (v3 w3) v1 w1 v2 w2 −1 x − p1 y − p2 = 1 v1w2 − v2w1 (v3 w3) w2 −w1 −v2 v1 x − p1 y − p2 . Donde, (v2w3 − v3w2)(x − p1) + (v3w1 − v1w3)(y − p2) + (v1w2 − v2w1)(z − p3) = 0, ou (X − P) · N = 0, onde N = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1). A equa¸c˜ao (X − P) · N = 0 implica, em particular, que N ´e perpendicular aos vetores V e W. De fato, tomando X = P + V ∈ π, temos que (X − P) · N = V · N = 0. Da mesma forma, vemos que W · N = 0. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp A equa¸c˜ao (X −P)·N = 0, obtida para planos no R3 , serve, como vimos no exemplo 1.3.3, tamb´em para retas em R2 . Isto sugere a seguinte defini¸c˜ao. 1.3.10 Definic¸˜ao Sejam P, N ∈ Rn , onde N ´e um vetor n˜ao-nulo. O subconjunto H = {X ∈ Rn ; (X − P) · N = 0} ´e chamado hiperplano que passa por P e ´e perpendicular a N. Observac¸˜ao Os hiperplanos de R2 s˜ao retas; os do R3 s˜ao os planos. Por analogia a estes casos ´e de se esperar que a “dimens˜ao” destes objetos dependa do ambiente no qual eles habitam. Um hiperplano H ⊂ Rn deve ter dimens˜ao n − 1. 1.3.11 Exemplo Seja H = {X = (x1, x2, x3, x4) ∈ R4 ; x1 + x2 − 2x3 − x4 = 1}. Temos que H ´e o hiperplano do R4 que ´e perpendicular a N = (1, 1, −2, −1) e passa, por exemplo, por P = (0, 0, 0, −1). Agora observe que X ∈ H se, e somente se, X = (x1, x2, x3, x1 + x2 − 2x3 − 1) = P + x1(1, 0, 0, 1) + x2(0, 1, 0, 1) + x3(0, 0, 1, −2), o que mostra que os pontos de H s˜ao descritos por uma equa¸c˜ao param´etrica a trˆes parˆametros. Isto basta para sentir que a dimens˜ao de H ´e 3. 1.3.12 Produto Vetorial Devido ao seu valor geom´etrico, o vetor N constru´ıdo na proposi¸c˜ao 1.3.9 merece destaque especial. Nesta subse¸c˜ao colocaremos as propriedades b´asicas deste vetor.
  26. 26. 20 Retas e Planos 1.3.13 Definic¸˜ao Sejam X = (x1, x2, x3) e Y = (y1, y2, y3) duas triplas em R3 . O produto vetorial de X por Y , denotado por X × Y (ou X ∧ Y ), ´e definido por X × Y = (x2y3 − x3y2, x3y1 − x1y3, x1y2 − x2y1), que pode ser facilmente lembrado expandindo o determinante abaixo ao longo da primeira linha: X × Y = e1 e2 e3 x1 x2 x3 y1 y2 y3 = (x2y3 − x3y2)e1 + (x3y1 − x1y3)e2 + (x1y2 − x2y1)e3, onde {e1, e2, e3} ´e a base canˆonica do R3 . 1.3.14 Exemplo Sejam X = (1, 1, 2) e Y = (3, −1, 1). O produto vetorial de X por Y ´e o vetor X × Y = e1 e2 e3 1 1 2 3 −1 1 = 3e1 + 5e2 − 4e3 = (3, 5, −4). Note que (X × Y ) · X = (3, 5, −4) · (1, 1, 2) = 0. Tamb´em (X × Y ) · Y = 0, o que diz que X × Y ´e perpendicular a X e a Y . Esta propriedade ´e verdadeira em geral, como veremos a seguir. 1.3.15 Proposic¸˜ao Sejam X, Y, Z ∈ R3 e a ∈ R. As seguintes propriedades s˜ao verificadas. (i) X × Y = −Y × X; (ii) a(X × Y ) = (aX) × Y = X × (aY ); (iii) X × (Y + Z) = X × Y + X × Z; (iv) (X × Y ) · Z = det (X, Y, Z); (v) X × Y 2 = X 2 Y 2 − (X · Y )2 , onde (X, Y, Z) indica a matriz cujas colunas (ou linhas) s˜ao as triplas X, Y e Z, olhadas como matrizes 3 × 1. Assim, det (X, Y, Z) = x1 y1 z1 x2 y2 z2 x3 y3 z3 = x1 x2 x3 y1 y2 y3 z1 z2 z3 . Demonstrac¸˜ao: A demonstra¸c˜ao destas propriedades ´e feita via computa¸c˜ao direta, usando a defini¸c˜ao 1.3.13. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
  27. 27. Vetores e Func¸˜oes Vetoriais 21 1.3.16 Corol´ario Se X, Y ∈ R3 , ent˜ao X × Y = X Y sen ∠(X, Y ). (Geometricamente, isto significa que a ´area do paralelogramo gerado por X e Y ´e X × Y .) Demonstrac¸˜ao: O item (v) da proposi¸c˜ao 1.3.15, junto com a equa¸c˜ao da proposi¸c˜ao 1.2.8, implica que X × Y = X 2 Y 2 − (X · Y )2 = X 2 Y 2 (1 − cos2 ∠(X, Y )) = X Y sen ∠(X, Y ) , o que quer´ıamos. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrr O s e ¨ Figura 13: Paralelogramo gerado por X e Y ¢ ¢ ¢ ¢ ¢ ¢ ¢¢ ¨¨¨¨¨¨¨¨¨¨B e e e e e Y ¨¨¨¨¨¨¨¨¨¨B¢ ¢ ¢ ¢ ¢ ¢ ¢¢ h = Y sen θ X θppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp 1.3.17 Corol´ario Os vetores X, Y ∈ R3 s˜ao linearmente dependentes se, e somente se, X×Y = O. Demonstrac¸˜ao: X e Y s˜ao linearmente dependentes se, e somente se, sen ∠(X, Y ) = 0. Agora ´e s´o aplicar o corol´ario 1.3.17. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.18 Corol´ario Se X, Y ∈ R3 , ent˜ao det (X, Y, X × Y ) = X × Y 2 . Em particular, se X e Y s˜ao linearmente independentes, ent˜ao {X, Y, X ×Y } ´e uma base com a mesma orienta¸c˜ao da base canˆonica. Demonstrac¸˜ao: Usando o item (iv) da proposi¸c˜ao 1.3.15, vem que det (X, Y, X × Y ) = (X × Y ) · (X × Y ) = X × Y 2 . Agora, como X e Y s˜ao linearmente inde- pendentes, temos que X × Y = O, o que vem do corol´ario 1.3.17. Portanto, det (X, Y, X × Y ) = X × Y 2 0, e obtemos que (X, Y, X × Y ), que ´e a ma- triz de passagem da base {X, Y, X × Y } para a base canˆonica, tem determinante positivo, isto ´e, tem a mesma orienta¸c˜ao que a base canˆonica. Geometricamente, isto significa que {X, Y, X × Y } est´a posi- cionada no espa¸co de modo an´alogo `a base {e1, e2, e3}, como mostra a figura 14. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp g g g g d T e3 E e2     © e1 sg g g g g g g gyX × Y I Y d d d d d‚ X s Figura 14: Produto Vetorial
  28. 28. 22 Retas e Planos 1.3.19 Corol´ario Se X, Y ∈ R3 s˜ao ortonormais (unit´arios e ortogonais), ent˜ao {X, Y, X × Y } ´e uma base ortonormal do R3 . Demonstrac¸˜ao: Falta verificar que X × Y ´e tamb´em unit´ario. Isto segue-se facilmente de (v) da proposi¸c˜ao 1.3.15. Com efeito, X × Y 2 = X 2 Y 2 − (X · Y )2 = 1 − 0 = 1. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.20 [Duplo Produto Vetorial] Corol´ario Se X, Y, Z ∈ R3 , ent˜ao vale a f´ormula do duplo produto vetorial: (X × Y ) × Z = (X · Z) Y − (Y · Z) X. Demonstrac¸˜ao: Suponhamos, inicialmente, que X e Y sejam ortonormais. Logo, {X, Y, X × Y } ´e uma base ortonormal do R3 , e isto implica que existem (e s˜ao ´unicos) n´umeros reais c1, c2 e c3 tais que Z = c1X + c2Y + c3X × Y . Na realidade, c1 = X · Z, c2 = Y · Z e c3 = (X ×Y )·Z. Como (X ×Y )×Z ´e perpendicular a X ×Y , vem que ele deve ser combina¸c˜ao linear de X e Y . Portanto, (X × Y ) × Z = aX + bY, onde a = ((X × Y ) × Z) · X e b = ((X × Y ) × Z) · Y . Mas ((X × Y ) × Z) · X = det (X × Y, Z, X) = det (X × Y, c1X + c2Y + c3X × Y, X) = c2 det (X × Y, Y, X) = −c2 det (X, Y, X × Y ) = −c2 X × Y 2 = −c2 = −Y · Z. Analogamente, vemos que ((X × Y ) × Z) · Y = c1 = X · Z. Logo, a = −Y · Z e b = X · Z, o que prova a f´ormula do duplo produto vetorial para o caso onde X e Y s˜ao ortonormais. Para estendˆe-la para o caso onde X e Y s˜ao apenas ortogonais, tomamos os unit´arios uX e uY . Logo (uX × uY ) × Z = (uX · Z) uY − (uY · Z) uX, ou, equivalentemente, 1 X Y (X × Y ) × Z = 1 X Y (X · Z) Y − 1 X Y (Y · Z) X, que simplificada d´a (X × Y ) × Z = (X · Z) Y − (Y · Z) Y, e a f´ormula do duplo produto vetorial funciona tamb´em quando X e Y s˜ao ortogonais. Para o caso geral, onde X e Y s˜ao linearmente independentes, recorremos `a proje¸c˜ao de Y sobre X,
  29. 29. Vetores e Func¸˜oes Vetoriais 23 PXY . Da proposi¸c˜ao 1.2.16 vem que PXY = λX, onde λ = (X · Y )/ X 2 , e que Y − PXY ´e perpendicular a X. Logo, (X × (Y − PXY )) × Z = (X · Z) (Y − PXY ) − ((Y − PXY ) · Z) X. (¶5) Mas X × (Y − PXY ) = X × Y − X × PXY = X × Y, visto que PXY ´e paralelo a X, e (X · Z)(Y − PXY ) − ((Y − PXY ) · Z)X = (X · Z)Y − (X · Z)PXY − (Y · Z)X + (PXY · Z)X = (X · Z)Y − (Y · Z)X − λ((X · Z)X − (X · Z)X) = (X · Z)Y − (Y · Z)X. Assim, (¶5) fica: (X × Y ) × Z = (X · Z) Y − (Y · Z) X. A f´ormula do duplo produto vetorial agora vale sempre que X e Y s˜ao linearmente independentes. O caso onde X e Y s˜ao linearmente dependente (Y = kX) ´e trivial: a f´ormula tem ambos os membros nulos, como pode ser facilmente verificado pelo leitor. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.21 Corol´ario Sejam T e N dois vetores ortonormais do R3 . Se B = T ×N, ent˜ao B × T = N e N × B = T. Demonstrac¸˜ao: A f´ormula do duplo produto vetorial d´a que B × T = (T × N) × T = (T · T)N − (N · T)T = N, visto que T = 1 e T · N = 0. Agora convidamos o leitor a provar que N × B = T. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.22 Exemplo Sejam e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Um c´alculo direto mostra que e1 × e2 = e3. Logo, e2 × e3 = e1 e e3 × e1 = e2. Em particular, (e1 × e2) × e2 = e3 × e2 = −e1 e1 × (e2 × e2) = e1 × O = O, o que implica que (e1 × e2) × e2 = e1 × (e2 × e2) e mostra que o produto vetorial, em geral, n˜ao ´e associativo. O pr´oximo corol´ario mostra quando o produto vetorial ´e associativo.
  30. 30. 24 Retas e Planos 1.3.23 Corol´ario Dados vetores X, Y e Z em R3 , ent˜ao X × (Y × Z) = (X × Y ) × Z se, e somente se, (X · Y ) Z = (Y · Z) X. Demonstrac¸˜ao: Usando o corol´ario 1.3.20 temos que X × (Y × Z) = −(Y × Z) × X = −((X · Y ) Z − (X · Z) Y ) = (X · Z) Y − (X · Y ) Z, que comparado com (X × Y ) × Z = (X · Z) Y − (Y · Z) X, mostra que X × (Y × Z) = (X × Y ) × Z se, e somente se, (X · Y ) Z = (Y · Z) X. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.24 Distˆancia de um Ponto a uma Reta Seja l = P + [V ] uma reta no Rn . Dado Q ∈ Rn um ponto qualquer, seja Y = Q − P, como mostra a figura 15. A proje¸c˜ao de Y sobre V ´e dada por PV Y = λV, onde λ = Y · V V 2 . Como a figura 15 mostra ´e bastante razo´avel se esperar que a distˆancia de Q a l, que indicaremos por d(Q, l), definida como sendo o m´ınimo das distˆancias de Q a pontos de l, isto ´e, d(Q, l) = min{d(Q, X), X ∈ l}, ¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨ ¨ ¨¨¨ Figura 15 ¢ ¢ ¢ ¢ ¢ ¢ ¢¢ e e e e eu ¨¨ ¨¨¨B Y = Q − P Qs ¨¨ ¨¨¨ ¨¨¨ ¨¨B PV Y Y − PV Y V l = P + [V ] sQ P s seja atingida no ponto Q ∈ l, a proje¸c˜ao ortogonal de Q sobre l, dado por Q = P + PV Y = P + λV = P + (Q − P) · V V 2 V. Portanto, (d(Q, l))2 = Y − PV Y 2 = (Q − P) − (Q − P) · V V 2 V 2 = Q − P 2 − 2(Q − P) · ( (Q − P) · V V 2 V ) + ((Q − P) · V )2 V 4 V 2 = Q − P 2 V 2 − ((Q − P) · V )2 V 2 , o que produz o seguinte resultado.
  31. 31. Vetores e Func¸˜oes Vetoriais 25 1.3.25 Proposic¸˜ao Seja l = P +[V ] a reta do Rn que passa por P e ´e paralela a V . Dado Q ∈ Rn a distˆancia de Q a l ´e dada por d(Q, l) = Q − P 2 V 2 − ((Q − P) · V )2 V . Al´em disto, o ponto Q ∈ l onde esta distˆancia ´e atingida ´e dado por Q = P + (Q − P) · V V 2 V. A partir desta proposi¸c˜ao obtemos as f´ormulas usuais da distˆancia de um ponto a uma reta, em R3 e R2 , como mostram os corol´arios 1.3.27 e 1.3.28. 1.3.26 Exemplo Sejam P = (1, 0, −2, 3), Q = (1, 1, , 0, 2) e V = (1, 1, −1, 1), e consideremos a reta l = P + [V ]. Temos que Q − P 2 = 6, (Q − P) · V = −2 e V = 2. Logo, usando a proposi¸c˜ao 1.3.25, a distˆancia de Q a l ´e d(Q, l) = Q − P 2 V 2 − ((Q − P) · V )2 V = √ 5. O ponto Q ∈ l onde d(Q, l) ´e atingida ´e dado por Q = P + (Q − P) · V V 2 V = (1, 0, −2, 3) − 1 2 (1, 1, −1, 1) = 1 2 (1, −1, −3, 5). Sugerimos ao leitor o c´alculo de d(Q, Q ) que, claro, deve produzir √ 5. 1.3.27 Corol´ario Seja l = P + [V ] a reta do R3 que passa por P e ´e paralela a V . Dado Q ∈ R3 a distˆancia de Q a l ´e dada por d(Q, l) = (Q − P) × V V . Demonstrac¸˜ao: Resulta de (v), proposi¸c˜ao 1.3.15, junto com a proposi¸c˜ao 1.3.25. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.28 Corol´ario Seja l = P + [V ] a reta do R2 que passa por P = (x1, x2) e ´e paralela a V = (v1, v2). Dado Q = (x0, y0) a distˆancia de Q a l ´e dada por d(Q, l) = |ax0 + by0 − c| √ a2 + b2 ,
  32. 32. 26 Retas e Planos onde N = (a, b) = (−v2, v1) ´e normal a l e c = ax1 + bx2. (Neste caso, a equa¸c˜ao cartesiana de l ´e: ax + by = c.) Demonstrac¸˜ao: Visando utilizar o corol´ario 1.3.27, mergulharemos R2 em R3 , isto ´e, olharemos uma dupla X = (x1, x2), como sendo a tripla X = (x1, x2, 0). Assim, d(Q, l) = (x0 − x1, y0 − y1, 0) × (v1, v2, 0) (v1, v2, 0) , que expressa em termos de a = −v2, b = v1 e c = ax1 + bx2 fica: d(Q, l) = (0, 0, −ax0 − by0 + c) (b, −a, 0) = |ax0 + by0 − c| √ a2 + b2 . pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.3.29 Distˆancia de um Ponto a um Hiperplano Seja π o plano do R3 que ´e perpendicular a N e passa por P, conforme mostra a figura 16. Dado Q ∈ R3 , a distˆancia de Q a π, d(Q, π), ´e definida como sendo o m´ınimo das distˆancias de Q a pontos de π, isto ´e, d(Q, π) = min{d(Q, X); X ∈ π}. Seja l a reta que passa por Q e ´e paralela a N. Temos que l intercepta (ortogonalmente) π no ponto Q = Q − PN (Q − P) = Q − (Q − P) · N N 2 N, onde PN (Q − P) ´e a proje¸c˜ao de Q − P sobre N. Fixemos X ∈ π um ponto arbitr´ario. Como Q − Q ´e perpendicular a π, ele ´e perpendicular a X − Q . Usando o teorema de Pit´agoras (veja proposi¸c˜ao 1.2.17), vem que X − Q 2 = X − Q − Q + Q 2 = X − Q 2 + Q − Q 2 ≥ Q − Q 2 e, portanto, obtemos Q − Q ≤ X − Q , ∀X ∈ π. Segue-se, ent˜ao, que d(Q, π) ´e atingida em Q e d(Q, π) = Q − Q = (Q − P) · N N 2 N = |(Q − P) · N| N . Isto prova a seguinte proposi¸c˜ao. rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr 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rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr ˆ ¨¨¨¨ ¨¨¨¨¨¨¨ ˆˆˆˆˆˆˆˆ d d d d d d d d g g g g g g g g Q d(Q, π) d(Q, X) Q π Q − PT                     N P s sXs Figura 16
  33. 33. Vetores e Func¸˜oes Vetoriais 27 1.3.30 Proposic¸˜ao Seja H o hiperplano do Rn que ´e perpendicular a N e passa por P. Dado Q ∈ Rn , a distˆancia de Q a H ´e dada por d(Q, H) = |(Q − P) · N| N . Mais ainda, d(Q, H) ´e atingida no ponto de H Q = Q − (Q − P) · N N 2 N. Em particular, se n = 3, Q = (x0, y0, z0), N = (a, b, c) e d = N · P, a distˆancia de Q ao plano H fica: d(Q, H) = |ax0 + by0 + cz0 − d| √ a2 + b2 + c2 . 1.3.31 Exemplo A distˆancia de Q = (3, −2, 1) ao plano π de equa¸c˜ao cartesiana 2x − 2y − z = −9 vale d(Q, π) = |(3, −2, 1) · (2, −2, −1) + 9| 3 = 6. Agora se X = (x, y, z) ´e um ponto qualquer de π, ent˜ao X = (x, y, 2x − 2y + 9) = (0, 0, 9) + x(1, 0, 2) + y(0, 1, −2) = P + xV + yW, onde P = (0, 0, 9), V = (1, 0, 2) e W = (0, 1, −2). Logo, π = P + [{V, W}], e obtemos uma representa¸c˜ao param´etrica para π. 1.3.32 Exemplo Sejam l1 = P + [V ] e l2 = Q + [W] duas retas em Rn . Se V e W s˜ao linearmente independentes, h´a duas possibilidades para a interse¸c˜ao l1 ∩ l2, a saber: (i) l1 ∩ l2 ´e um ponto, digamos l1 ∩ l2 = {R}; (ii) l1 ∩ l2 ´e vazio. No primeiro caso, o plano π = R + [{V, W}] cont´em l1 e l2, e dizemos que l1 e l2 s˜ao retas coplanares. J´a em (ii), n˜ao existe um plano que contenha ambas as retas, e diremos que l1 e l2 s˜ao retas reversas. O leitor deve observar que as retas do exemplo 1.3.6 s˜ao retas reversas em R3 . Suponhamos, agora, que os vetores V e W sejam linearmente dependentes. Obtemos, outra vez, duas alternativas: (iii) l1 e l2 s˜ao coincidentes, isto ´e, l1 = l2; (iv) l1 e l2 s˜ao paralelas e l1 ∩ l2 = ∅. Em (iv), l1 e l2 s˜ao coplanares, visto que o plano π = P + [{V, W}] = Q + [{V, W}], onde W = Q − P, as cont´em.
  34. 34. 28 Func¸˜oes Vetoriais 1.4 Func¸˜oes Vetoriais Nesta se¸c˜ao, estudaremos as no¸c˜oes b´asicas relacionadas com aplica¸c˜oes entre espa¸cos euclidianos de dimens˜oes quaisquer. 1.4.1 Definic¸˜ao Uma fun¸c˜ao vetorial ´e uma fun¸c˜ao com dom´ınio D ⊂ Rn e contradom´ınio Rm , isto ´e, uma fun¸c˜ao do tipo f : D ⊂ Rn −−−−−→ Rm X −−−−−→ f(X) = (f1(X), f2(X), . . . , fm(X)), onde X = (x1, x2, . . . , xn) ∈ D. Quando m = 1, diremos que f ´e uma fun¸c˜ao real. J´a quando n = 1, f ´e dita uma fun¸c˜ao vetorial de uma vari´avel real. A imagem de f, denotada por Im(f), ou por f(D), ´e o conjunto Im(f) = f(D) = {Y ∈ Rm ; Y = f(X), X ∈ D}. Dizemos, tamb´em, que f parametriza o conjunto Im(f), ou que Im(f) ´e o conjunto parametri- zado por f. 1.4.2 Definic¸˜ao Dada uma fun¸c˜ao vetorial f : D ⊂ Rn −−−−−→ Rm X −−−−−→ f(X) = (f1(X), f2(X), . . . , fm(X)), as m fun¸c˜oes reais f1 : D ⊂ Rn −−−−−→ R X −−−−−→f1(X) f2 : D ⊂ Rn −−−−−→ R X −−−−−→f2(X) ... fm : D ⊂ Rn −−−−−→ R X −−−−−→fm(X) s˜ao as fun¸c˜oes coordenadas de f. 1.4.3 Exemplo Seja f : R2 −→ R definida por f(x, y) = x2 + y2 . Temos que f ´e uma fun¸c˜ao real (de duas vari´aveis) cuja imagem coincide com o intervalo [0, ∞).
  35. 35. Vetores e Func¸˜oes Vetoriais 29 1.4.4 Exemplo Seja f(t) = (x0 + a cos t, y0 + b sen t), t ∈ [0, 2π], onde a 0, b 0, x0 e y0 s˜ao n´umeros reais fixados. A imagem de f, Im(f) = {(x, y) ∈ R2 ; x = x0 + a cos t, y = y0 + b sen t, t ∈ R}, (¶6) coincide com a elipse de semi-eixos a e b, centrada em C = (x0, y0), que denotaremos por E(C, a, b). De fato, se x e y s˜ao como em (¶6), ent˜ao (x − x0)2 a2 + (y − y0)2 b2 = cos2 t + sen2 t = 1. E f O C s E T x E(C, a, b) y r q qy0 x0 a b 0 2πt r Figura 17: Elipse (x − x0)2 a2 + (y − y0)2 b2 = 1 ¨¨¨¨B r f(t) r f1(t) rf2(t)           t r P2 rP1 Assim f parametriza E(C, a, b). A figura 17 mostra, em particular, a constru¸c˜ao geom´etrica da fun¸c˜ao f: pelo ponto C = (x0, y0) tra¸camos a semi-reta que faz o ˆangulo t com o eixo-x. Esta semi-reta intercepta os c´ırculos, centrados em C e de raios a e b, nos pontos P1 e P2, respectivamente. Agora, por P1 tra¸camos uma reta paralela ao eixo-y, e por P2 tra¸camos uma reta paralela ao eixo-x. A interse¸c˜ao destas retas ´e exatamente o ponto da elipse (x − x0)2 a2 + (y − y0)2 b2 = 1 que indicamos por f(t). Em particular, se b = a, obtemos que f parametriza o c´ırculo de centro C = (x0, y0) e raio a, denotado por S1 (C, a). Ob- serve que, neste caso, as fun¸c˜oes coordenadas de f ficam assim: f1(t) = x(t) = x0 + a cos t f2(t) = y(t) = y0 + a sen t, onde t ∈ [0, 2π]. O C s E T x S1 (C, a) y r q qy0 x0 a Figura 18: (x − x0)2 + (y − y0)2 = a2 ppppppppppppppppppppppppppppppppp ppppppppppp pppppppppp pppppppppp ppppppppp ppppppppp pppppppp pppppp pppppp ppppp ppppp pppp pppp pppppp ppppp ppppp pppp pppp ppppp pppp ppppp ppppp pppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp ppp pp ppp ppp pp ppp ppp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp ppp ppp pp ppp ppp ppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp pppp ppppp ppppp pppp ppppp pppp pppp ppppp ppppp pppppp pppp pppp ppppp ppppp pppppp pppppp pppppppp ppppppppp ppppppppp ppppppppp ppppppppp pppppppppppp pppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppp ppppppppppp pppppppppp pppppppppp ppppppppp ppppppppp pppppppp pppppp pppppp ppppp ppppp pppp pppp pppppp ppppp ppppp pppp pppp ppppp pppp ppppp ppppp pppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp ppp pp ppp ppp pp ppp ppp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp ppp ppp pp ppp ppp ppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp pppp ppppp ppppp pppp ppppp pppp pppp ppppp ppppp pppppp pppp pppp ppppp ppppp pppppp pppppp pppppppp ppppppppp ppppppppp ppppppppp ppppppppp pppppppppppp pppppppppppppppppppppppppppppppppp         r f1(t) rf2(t) t rf(t)
  36. 36. 30 Func¸˜oes Vetoriais 1.4.5 Exemplo A fun¸c˜ao vetorial g : R2 −−−−−→ R3 (u, v) −−−−−→ g(u, v) = (1 + u, 2 + v, u + v) tem fun¸c˜oes coordenadas g1(u, v) = x(u, v) = 1 + u g2(u, v) = y(u, v) = 2 + v g3(u, v) = z(u, v) = u + v, (u, v) ∈ R2 , e sua imagem coincide com o plano que passa por (1, 2, 0) e ´e paralelo aos vetores (1, 0, 1) e (0, 1, 1). 1.4.6 Conjuntos Associados a Func¸˜oes Vetoriais Estudaremos, agora, alguns subconjuntos do espa¸co euclidiano que desempenham papel de grande relevˆancia na descri¸c˜ao das fun¸c˜oes vetoriais. 1.4.7 [Gr´aficos] Definic¸˜ao Seja f : D ⊂ Rn −→ Rm , f(X) = (f1(X), f2(X), . . . , fm(X)), uma fun- ¸c˜ao vetorial. O gr´afico de f, indicado por G(f), ´e definido por G(f) = {(X, Y ) ∈ Rn+m ; Y = f(X), X ∈ D} = {(x1, x2, . . . , xn, f1(X), f2(X), . . . , fm(X)); X = (x1, x2, . . . , xn) ∈ D}. Diremos, tamb´em, que G(f) ´e o conjunto definido explicitamente por f. Observac¸˜ao Na defini¸c˜ao acima, introduzimos uma nova nota¸c˜ao, que ser´a ´util em outras situa¸c˜oes: dados X = (x1, x2, . . . , xn) ∈ Rn e Y = (y1, y2, . . . , yn) ∈ Rm , escre- vemos (X, Y ) para representar a (n + m)-upla (x1, x2, . . . , xn, y1, y2, . . . , ym). 1.4.8 Exemplo Seja f : [−1, 1] −→ R definida por f(x) = x2 . Temos que f ´e uma fun- ¸c˜ao real de uma vari´avel real cuja imagem coincide com o intervalo [0, 1], e cujo gr´afico ´e o subcon- junto do R2 dado por G(f) = {(x, y); y = x2 , x ∈ [−1, 1]}, que coincide com o arco da par´abola y = x2 que se projeta sobre o intervalo [−1, 1]. −1 1O E T x y Figura 19: Par´abola y = x2 s
  37. 37. Vetores e Func¸˜oes Vetoriais 31 1.4.9 [Parabol´oide de Revoluc¸˜ao] Exemplo Seja f : R2 −→ R, f(x, y) = x2 + y2 , a qual j´a usamos no exemplo 1.4.3. O seu gr´afico, que chamamos parabol´oide de revolu¸c˜ao (ou rota¸c˜ao), ´e dado por G(f) = {(x, y, z); z = x2 + y2 }. Para fazer um esbo¸co deste conjunto usamos o m´etodo dos cortes por planos da forma z = c, isto ´e, planos paralelos ao plano-xy. Neste caso, quando cortamos G(f) pelo plano z = 4, obtemos, neste plano, o c´ırculo de raio 2 e centro (0, 0, 4): {(x, y, z); x2 + y2 = 4, z = 4}. Mais geralmente, se cortamos G(f) com planos z = a 0, obtemos a´ı o c´ırculo de raio √ a e centro (0, 0, a). Note que, quando a 0, a interse¸c˜ao ´e vazia, e coincide com a origem, quando a = 0. Isto sugere que G(f) ´e um subconjunto de R3 obtido pela rota¸c˜ao em torno do eixo-z de alguma curva plana, por exemplo, do plano-xz. Esta curva ´e facilmente obtida, interceptando-se G(f) com o plano y = 0, isto ´e, com o plano-xz, o que produz a par´abola z = x2 . Como resultado, obtemos a figura 20. T z E y       ©x Figura 20: Parabol´oide z = x2 + y2 1.4.10 [Sela] Exemplo A sela ou parabol´oide hiperb´olico ´e o gr´afico da fun¸c˜ao f(x, y) = y2 − x2 , (x, y) ∈ R2 . Por n˜ao ser um subconjunto obtido por rota¸c˜oes, o seu esbo¸co ´e um pouco mais trabalhoso. Come¸cando com cortes por planos z = a ≥ 0, obtemos as hip´erboles {(x, y, z); y2 − x2 = a, z = a}, que se degeneram no par de retas y = ±x, quando a = 0, como mostra a figura 21-(a). T z E y       ©x Figura 21-(a) T z E y       ©x Figura 21-(b) T z E y       ©x Figura 21-(c): A sela z = y2 − x2
  38. 38. 32 Func¸˜oes Vetoriais Procedimento an´alogo, agora usando planos a ≤ 0, d´a a figura 21-(b). Os cortes de G(f) por planos y = c produz par´abolas z = c2 − y2 , que o leitor dever´a esbo¸car. Finalmente, obtemos a sela, como na figura 21-(c). 1.4.11 [H´elice Circular] Exemplo Consideremos, ago- ra, a fun¸c˜ao de uma vari´avel f : R −−−−−→ R2 t −−−−−→ f(t) = (a cos t, a sen t), onde a 0 ´e uma constante. O seu gr´afico, G(f) = {(t, f(t)) = (t, a cos t, a sen t); t ∈ R}, ´e a h´elice circular de raio a, cujo eixo coincide com eixo-x, como mostra a figura 22. Observe que a proje¸c˜ao de G(f) no plano-yz ´e o c´ırculo de raio a. Suas proje¸c˜oes no plano-xy e plano-xz s˜ao os gr´aficos das fun¸c˜oes reais de uma vari´avel real y = a cos x e z = a sen x, respecti- vamente. T z E y       ©         x Figura 22: H´elice Circular Observac¸˜ao O fato de coincidir com um gr´afico imp˜oe restri¸c˜oes `a forma de um subconjunto. Pensando com uma fun¸c˜ao f : D ⊂ R2 −→ R, vem que G(f) = {(x, y, z); z = f(x, y), (x, y) ∈ D}. Isto significa que para cada (x, y) ∈ D, existe um ´unico ponto em G(f), a saber (x, y, f(x, y)). Geometricamente, isto diz que a reta que passa por (x, y, 0) e ´e perpendicular ao plano-xy intercepta G(f) em um ´unico ponto. Por exemplo, a esfera S2 (a) = {(x, y, z); x2 + y2 + z2 = a2 } ⊂ R3 n˜ao pode ser gr´afico de nenhuma fun¸c˜ao do tipo que es- tamos considerando. De fato, h´a retas perpendiculares ao plano-xy que interceptam esta esfera em dois pontos. Entretanto, parte dela, digamos seu hemisf´erio superior, ´e o conjunto definido explicitamente por f : D[a] ⊂ R2 −−−−−→ R (x, y) −−−−−→ f(x, y) = a2 − x2 − y2, onde D[a] = {(x, y); x2 + y2 ≤ a2 } ´e o disco fechado de raio a. J´a o hemisf´erio inferior ´e o gr´afico de g = −f, isto ´e, g(x, y) = −f(x, y), (x, y) ∈ D[a]. Figura 23      © s x Es y T s z r r
  39. 39. Vetores e Func¸˜oes Vetoriais 33 Observac¸˜ao ´E muito comum na pr´atica nos referirmos a uma equa¸c˜ao Y = f(x1, x2, . . . , xn), onde o lado direito indica uma m-upla envolvendo os s´ımbolos x1, x2, . . ., xn, como uma fun¸c˜ao. O que estamos pensando, na realidade, ´e na fun¸c˜ao f : D −→ Rm , onde D ´e o maior subconjunto de Rn , onde tal express˜ao faz sentido. 1.4.12 Exemplo A express˜ao z = 4 − x2 − y2 log (x2 + y2 − 1) define uma fun¸c˜ao f no anel de R2 dado por D = {(x, y); 1 x2 + y2 ≤ 4}. De fato, se x2 +y2 ≤ 1, ent˜ao n˜ao existe log (x2 + y2 − 1), e se x2 + y2 4 n˜ao tem sentido 4 − x2 − y2. pppppppppppppppppppppppppppppppp ppppppppppppp ppppppppppp pppppppppp ppppppppp ppppppppp pppppppp ppppp ppppp pppppp pppppp ppppp ppppp pppppp ppppp ppppp ppppp pppp pppp pppp pppp pppp pppp pppp ppp ppp ppp ppp pppp ppp ppp pppp ppp ppp ppp pp ppp ppp pp pp ppp ppp pp ppp ppp pp pp pp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pppppppppp pp p pppppp pp pppp pppp pp ppp pppp pp pppp pppp pp ppp ppp pp p pp pppp pp p pp ppp pp p pp ppp pp p p pp p pp pp pp ppp pp p pp p pp pp pp p pp p pp p pp pp pp p pp p p pp p p pp pp pp pp pp p pp pp p pp p p pp pp p pp pp p pp pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp pp pp pp p p pp p p pp p p pp p pp p pp p pp p pp p pp ppp pp p pp pp pp p p pp ppp pp pp pp ppp pp p pp ppp pp p pp ppp ppp pp ppppp ppppp pp pppp pppp pp pppp pppp pp pp pppppp pp pppppppppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp pp pp pp ppp ppp pp ppp ppp pp pp ppp ppp pp ppp ppp ppp pppp ppp ppp ppp pp ppp ppp ppp pppp pppp pppp pppp pppp pppp pppp ppppp ppppp ppppp pppppp ppppp ppppp ppppppp ppppppp ppppp ppppp pppppppp ppppppppp ppppppppp pppppppppp ppppppppppp ppppppppppppp pppppppppppppppppppppppppppppppp r 1 r 2 E x T y Figura 24: Anel 1.4.13 [Conjuntos de N´ıvel] Definic¸˜ao Sejam f : D ⊂ Rn −→ Rm uma fun¸c˜ao vetorial e K um vetor em Rm fixado. O conjunto de n´ıvel K de f ´e o subconjunto de D definido por f−1 (K) = {X ∈ D; f(X) = K}. Isto ´e, se f1, f2, . . ., fm s˜ao as fun¸c˜oes coordenadas de f, e K = (k1, k2, . . . , km), ent˜ao f−1 (K) ´e o conjuntos das solu¸c˜oes em D do sistema de m-equa¸c˜oes e n-inc´ognitas:    f1(x1, x2, . . . , xn) = k1 f2(x1, x2, . . . , xn) = k2 ... ... ... fm(x1, x2, . . . , xn) = km. Os conjuntos de n´ıvel f−1 (K) s˜ao, tamb´em, denominados conjuntos definidos implicitamente por f. Observac¸˜ao f−1 (K) = ∅ se, e somente se, K ∈ Im(f). 1.4.14 Exemplo Os conjuntos de n´ıvel f−1 (a), a ∈ R, de z = f(x, y) = x2 + y2 , (x, y) ∈ R2 , s˜ao os seguintes: (i) f−1 (a) = ∅, se a 0, pois Im(f) = [0, +∞); (ii) f−1 (0) = {(0, 0)};
  40. 40. 34 Func¸˜oes Vetoriais (iii) f−1 (a) = S1 ( √ a) = {(x, y); x2 + y2 = a}, se a 0. Na figura 25, temos o gr´afico de f e alguns conjuntos de n´ıvel, que podem ser obtidos via proje¸c˜ao no plano-xy das interse¸c˜oes de G(f) com planos z = a ≥ 0. z E y     ©x Figura 25: Parabol´oide de Revolu¸c˜ao T 1.4.15 Exemplo Estudemos agora os conjuntos definidos implicitamente por z = f(x, y) = y2 − x2 , (x, y) ∈ R2 . Os conjuntos de n´ıvel f−1 (k), k 0, s˜ao as hip´erboles equil´ateras de semi-eixos √ k: y2 − x2 = k. Quando k 0, temos as hip´erboles, tamb´em equil´ateras e com semi-eixos √ −k, x2 − y2 = −k. J´a f−1 (0) coincide com o par de retas y = ±x. Estes conjuntos definidos impli- citamente aparecem quando cortamos a sela por planos paralelos ao plano-xy, como mostra a figura 21. E x T y Figura 26 k 0k 0 r √ k r √ k k 0 k 0 k = 0 1.4.16 Exemplo Seja f(x, y, z) = x2 + y2 + z2 , (x, y, z) ∈ R3 . Os conjuntos de n´ıvel desta fun¸c˜ao s˜ao ou o conjunto vazio, ou {(0, 0, 0)}, ou as esferas do R3 centradas na origem e de raio √ a: S2 ( √ a) = {(x, y, z); x2 + y2 + z2 = a} = f−1 (a), a 0. (¶7) 1.4.17 Exemplo Seja f : R3 −−−−−→ R2 (x, y, z) −−−−−→ f(x, y, z) = (x2 + y2 + z2 , x + y + z). Dada K = (a, b) o conjunto de n´ıvel f−1 (K) ´e o conjunto de solu¸c˜oes do sistema x2 + y2 + z2 = a x + y + z = b.
  41. 41. Vetores e Func¸˜oes Vetoriais 35 que, claro, n˜ao tem solu¸c˜oes, se a 0. Se a ≥ 0, f−1 (a, b) ´e a interse¸c˜ao de S2 ( √ a) (veja (¶7)) com o plano πb de equa¸c˜ao x+y+z = b. Logo, f−1 (a, b) pode ser vazio, se πb est´a longe de S2 ( √ a); coincidir com um ponto, se πb tangencia S2 (a); ou, fi- nalmente, ser um c´ırculo contido em πb. Por exemplo, os pontos P1 = (1, 0, 0) e P2 = (0, 1, 0) pertencem a f−1 (1, 1). Logo, este conjunto de n´ıvel deve coincidir com um c´ırculo no plano x + y + z = 1. Observando que P3 = (0, 0, 1) tamb´em pertence ao c´ırculo f−1 (1, 1), vemos que o seu centro deve coincidir com o baricentro do triˆangulo (equil´atero) de v´ertices P1, P2 e P3, que ´e o ponto C = P1 + P2 + P3 3 = (1/3, 1/3, 1/3). Para obter o raio, ´e s´o calcular a distˆancia de C a P1, que ´e r = √ 6/3.      © s P1 x Es P2 y T sP3 z Figura 27 Vejamos mais trˆes belos conjuntos definidos implicitamente. 1.4.18 Exemplo Seja f : R3 −→ R definida por f(x, y, z) = x2 + y2 − z2 . Observe que a imagem de f coincide com todo R, isto ´e, f ´e sobrejetiva. De fato, f(0, 0, z) = −z2 , o que mostra que f transforma o eixo-z em (−∞, 0]. Agora ´e s´o calcular, por exemplo, f(x, 0, 0) = x2 , x ∈ R, para cobrir [0, +∞). Neste exemplo, esbo- ¸caremos trˆes conjuntos de n´ıvel de f, a saber: f−1 (0), f−1 (1) e f−1 (−1). Os demais possuem a mesma forma que f−1 (1) ou f−1 (−1), como o leitor pode facilmente verificar. Temos que f−1 (0) = {(x, y, z); z2 = x2 + y2 }, que produz o cone de duas folhas, como mostra a figura 28-(a) ao lado. A t´ecnica para obten- ¸c˜ao desta figura ´e aquela que temos usado: corta- mos o conjunto com planos z = a, o que produz, neste plano, o c´ırculo de raio |a| centrado no ponto (0, 0, a). Quandoa = 0,obtemosapenasumponto, z = x z = −x Tz E y       ©         x Figura 28-(a): Cone de Duas Folhas x2 + y2 − z2 = 0 ponto, a origem. Isto mostra que f−1 (0) ´e de revolu¸c˜ao. A curva perfil, a geratriz do conjunto, ´e obtida fazendo, por exemplo, a interse¸c˜ao com o plano y = 0, o que d´a origem ao par de retas z = ±x. Donde podemos concluir que, de fato, f−1 (0) ´e o cone de duas folhas. Para o esbo¸co dos outros dois n´ıveis, a mesma t´ecnica mostra que eles tamb´em s˜ao de revolu¸c˜ao: o conjunto f−1 (1), mostrado na figura 28-(b), tem como geratriz a hip´erbole H1 = {(x, 0, z); x2 − z2 = 1}. A hip´erbole H2 = {(x, 0, z); z2 − x2 = 1}
  42. 42. 36 Func¸˜oes Vetoriais Especiais ´e a geratriz de f−1 (−1), que, por isso, tem duas folhas, como mostramos na figura 28-(c). T    © E z y x z2 − x2 = 1 Figura 28-(b): Hiperbol´oide de Uma Folha x2 + y2 − z2 = 1 x2 − z2 = 1 T z E y       ©x Figura 28-(c): Hiperbol´oide de Duas Folhas x2 + y2 − z2 = −1 1.5 Func¸˜oes Vetoriais Especiais Reservamos esta se¸c˜ao para destacar algumas fun¸c˜oes vetoriais que s˜ao de grande inte- resse pr´atico, para o C´alculo e para a Geometria. Inicialmente, faremos uma breve exposi¸c˜ao das fun¸c˜oes lineares, que, certamente, constituem a pedra fundamental das fun¸c˜oes do C´alculo Diferencial. 1.5.1 [Func¸˜oes Lineares] Definic¸˜ao Uma fun¸c˜ao (ou aplica¸c˜ao, ou transforma¸c˜ao) linear ´e uma fun¸c˜ao vetorial do tipo T : Rn −−−−−→ Rm X −−−−−→ T(X) = (T1(X), T2(X), . . . , Tn(X)), satisfazendo as seguintes propriedades: (i) T(X + Y ) = T(X) + T(Y ), ∀X, Y ∈ Rn ; (ii) T(aX) = aT(X), ∀a ∈ R e ∀X ∈ Rn . 1.5.2 Exemplo Consideremos a seguinte fun¸c˜ao T : R2 −→ R2 dada por T(x, y) = (x + y, y − x). Sejam X = (x1, x2) e Y = (y1, y2). Temos que T(X + Y ) = T(x1 + y1, x2 + y2) = ((x1 + y1) + (x2 + y2), (x2 + y2) − (x1 + y1)) = ((x1 + x2) + (y1 + y2), (x2 − x1) + (y2 − y1)) = (x1 + x2, x2 − x1) + (y1 + y2, y2 − y1) = T(X) + T(Y ).
  43. 43. Vetores e Func¸˜oes Vetoriais 37 Agora, se a ∈ R, vem que T(aX) = T(ax1, ax2) = (ax1 + ax2, ax2 − ax1) = a(x1 + x2, x2 − x1) = aT(X). Logo, T ´e linear. Um modo eficiente de ver que T ´e linear, ´e introduzindo a seguinte identifica- ¸c˜ao: uma dupla X = (x1, x2) passar´a a ser olhada como a matriz-coluna X = x1 x2 . Isto posto, vem que T(X) = T(x1, x2) = x1 + x2 x2 − x1 = 1 1 −1 1 x1 x2 , ou T(X) = M(T)X, onde M(T) = 1 1 −1 1 . Agora, usando propriedades da multiplica¸c˜ao de matrizes, segue-se facilmente a linearidade de T. De fato, e T(X + Y ) = M(T)(X + Y ) = M(T)X + M(T)Y = T(X) + T(Y ) T(aX) = M(T)(aX) = aM(T)X = aT(X). A identifica¸c˜ao feita no exemplo anterior pode ser usada com uma k-upla qualquer, o que facilitar´a a compreens˜ao das fun¸c˜oes lineares definidas em Rn . Dado X = (x1, x2, . . . , xk) ∈ Rk , identificaremos, sempre que for preciso, X com a matriz-coluna (vetor-coluna) X =       x1 x2 ... xk       . Assim sendo, sejam T : Rn −→ Rm uma aplica¸c˜ao linear, e X ∈ Rn . Temos que X = (x1, x2, . . . , xn) =       x1 x2 ... xn       = x1       1 0 ... 0       + x2       0 1 ... 0       + · · · + xn       0 0 ... 1       . Logo, T(X) = x1T(e1) + x2T(e2) + · · · + xnT(en) = x1T       1 0 ... 0       + x2T       0 1 ... 0       + · · · + xnT       0 0 ... 1       . (¶8)
  44. 44. 38 Func¸˜oes Vetoriais Especiais Como Im(T) ⊂ Rm , vem que T(e1) =       a11 a21 ... am1       , T(e2) =       a12 a22 ... am2       , . . . , T(en) =       a1n a2n ... amn       , para alguns n´umeros reais aij, 1 ≤ i ≤ m e 1 ≤ j ≤ n, o que posto em (¶8) d´a que T(X) = x1       a11 a21 ... am1       + x2       a12 a22 ... am2       + · · · + xn       a1n a2n ... amn       =       x1a11 x1a21 ... x1am1       +       x2a12 x2a22 ... x2am2       + · · · +       xna1n xna2n ... xnamn       =       a11x1 + a12x2 + · · · + a1nxn a21x1 + a22x2 + · · · + a2nxn ... am1x1 + am2x2 + · · · + amnxn       =       a11 a12 . . . a1n a21 a22 . . . a2n ... ... ... am1 am2 . . . amn             x1 x2 ... xn       . Isto prova o seguinte teorema. 1.5.3 Teorema Sejam T : Rn −→ Rm uma aplica¸c˜ao linear, e M(T) a matriz de ordem m × n cujas colunas s˜ao os vetores T(e1), T(e2), . . ., T(en), nesta ordem. Temos que (i) T(X) = M(T)X; (ii) Im(T) ´e gerado pelas colunas de M(T); (iii) posto T = posto M(T), onde posto T indica a dimens˜ao de Im(T), e posto M(T) indica o posto da matriz M(T), isto ´e, o n´umero m´aximo de colunas linearmente independentes que ela possui.
  45. 45. Vetores e Func¸˜oes Vetoriais 39 1.5.4 Definic¸˜ao A matriz M(T) = (aij) ´e conhecida como a matriz de T com rela¸c˜ao `as bases canˆonicas do Rn e Rm . Por simplicidade, chamaremos M(T) de matriz de T. Um conjunto de n´ıvel especial de uma fun¸c˜ao linear T : Rn −→ Rm ´e o seu n´ucleo, N(T) = T−1 (0, 0, . . . , 0) = {X ∈ Rn ; T(X) = (0, 0, . . . , 0)}. Os outros conjuntos definidos implicitamente por T, quando n˜ao-vazios, s˜ao determinados a partir dele, como mostra seguinte proposi¸c˜ao. 1.5.5 Proposic¸˜ao Seja T : Rn −→ Rm uma aplica¸c˜ao linear. Se K = T(P), P ∈ Rn , ent˜ao T−1 (K) = P + N(T) = {X ∈ Rn ; X = P + V, V ∈ N(T)}. Demonstrac¸˜ao: Seja X ∈ T−1 (K). Logo, T(X) = K = T(P). Como T ´e linear, vem que T(X − P) = O, isto ´e, V = X − P ∈ N(T). Assim, X = P + V , o que prova que T−1 (K) ⊂ P + N(T). Por outro lado, se X = P + V , para algum V ∈ N(T), ent˜ao T(X) = T(P) + T(V ) = T(P) = K. Donde, P + N(T) ⊂ T−1 (K)). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp 1.5.6 Exemplo Seja T : R3 −−−−−→ R3 (x, y, z) −−−−−→ T(x, y, z) = (x + y, x + 2y + z, −x + 3y + 4z). Temos que T(e1) = (1, 1, −1), T(e2) = (1, 2, 3) e T(e3) = (0, 1, 4). Logo, a matriz de T ´e M(T) =    1 1 0 1 2 1 −1 3 4    , e, usando o teorema 1.5.3, obtemos T    x y z    =    1 1 0 1 2 1 −1 3 4       x y z   . Sugerimos ao leitor que verifique diretamente esta identidade. Como det M(T) = 0, segue-se que posto M(T) ≤ 2. Como, por exemplo, as duas primeiras colunas de M(T) s˜ao linearmente independentes, devemos ter posto T = 2. (Conv´em observar, que esta informa¸c˜ao pode ser obtida, tamb´em, usando opera¸c˜oes elementares sobre as linhas (colunas) de M(T), o que ´e mais conveniente para matrizes de ordem alta.) Logo, Im(T) tem dimens˜ao dois e ´e gerado pelos

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