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VECTORES UNITARIOS UNI FIM
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20/03/2015
1
VECTORES UNITARIOS
PARA UNA CURVA REGULAR
CAPÍTULO ICÁLCULO VECTORIAL
SESIÓN 6 Vectores Unitarios
ROSA NIQUE ALVAREZ 2
X
Z
Y
i
j
k b = (2, 7, 4)
kjib 472 ++=
[ ] 3,: ℜ→ba:C r
ROSA NIQUE ALVAREZ 3
CURVA REGULAR
con rapidez arbitraria, denotada por )(tr'
VECTORES UNITARIOS
Sea una curva
VECTOR TANGENTE
)()()(
ttt
r'r'T = )(tr'
0P
)()()( ttt Tr'v =
ROSA NIQUE ALVAREZ 4
ROSA NIQUE ALVAREZ 5
VECTORES TANGENTES A UNA CURVA REGULAR
VECTOR NORMAL PRINCIPAL
0)(;)()()( ≠′
′′
= tttt T
TTN
0P
)()()( ttt NTT ′=′
ROSA NIQUE ALVAREZ 6
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VECTOR BINORMAL
0P
)()()( ttt NTB ×=
ROSA NIQUE ALVAREZ 7
EJEMPLO 1
Considere la curva regular C definida por:
Determine el vector T(t) en el punto:
ROSA NIQUE ALVAREZ 8
kjir )()cos()()cos()( tsenttsentt ++=
( )1,2,221 −−=P
Solución
Cuando t =7π/4 se obtiene el punto
ROSA NIQUE ALVAREZ 9
( )1,2,221 −−=P
( )
=
−+
==
0,22,
22
47
2cos,cos,2cos1
1)()()(
2
πT
r'r'T tttsen
tttt
kjir )()cos()()cos()( tsenttsentt ++=
ROSA NIQUE ALVAREZ 10
-1-0.5
00.5
11.5
-1
-0.5
0
0.5
1-0.5
0
0.5
X
Curva Cerrada
Y
Z
curvaCERRADA
kjir )()cos()()cos()( tsenttsentt:C ++=
TRIEDRO MOVIL
ROSA NIQUE ALVAREZ 11
0P
El triedro móvil esuna estructura quejuega un importantepapel en la rama delas matemáticasconocida comogeometría diferencialy en susaplicaciones almovimiento de navesespaciales.
ROSA NIQUE ALVAREZ 12
TRIEDRO MOVIL
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ROSA NIQUE ALVAREZ 13
VIDEO: Triedro Espiral
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
10
5
10
15
20
X
Hélice Circular
Y
Z( ) [ ]π∈= 6,0;,,cos)( ttsentttr
ROSA NIQUE ALVAREZ 14
VECTORES : VELOCIDAD Y ACELERACIÓN
)()(naceleració
)()(velocidad
tt
tt
'r'a
r'v
=
=
ROSA NIQUE ALVAREZ 15
VECTOR VELOCIDAD
)()()(ttt
r'r'T = )()()( ttt Tr'r' =
)()(')(velocidad ttt Trv =
ROSA NIQUE ALVAREZ 16
VECTOR ACELERACION
[ ])()()()( ttdtdt
dtdt Tr'r'a ==
)()()(ttt
r'r'T = )()()( ttt Tr'r' =
ROSA NIQUE ALVAREZ 17
VECTOR ACELERACION
)()()()(
)( tttdt
tdt Tr'T
r'a ′+=
[ ])()()()( ttdtdt
dtdt Tr'r'a ==
ROSA NIQUE ALVAREZ 18
VECTOR ACELERACION
→′′
=)()()(
ttt
TTN )()()( ttt NTT ′=′
)()()()(
)( tttdt
tdt Tr'T
r'a ′+=
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ROSA NIQUE ALVAREZ 19
VECTORA ACELERACION
)()()()()()( tttsttst NTTa ′′+′′=
)()()()(
)( tttdt
tdt Tr'T
r'a ′+=
,)((t)pero: ts r'=′ )()()( ttt NTT ′=′
ROSA NIQUE ALVAREZ 20
COMPONENTES DE LA ACELERACION
{ )()()()()()( tttsttstNT aa
NTTa 43421′′+′′=
)()()()()()( tttsttst NTTa ′′+′′=
COMPONENTES DE LA ACELERACION
ROSA NIQUE ALVAREZ 21
)(tsaT ′′=
)()( ttsaN T ′′=
Componente tangencial de la aceleración
Componente normal de la aceleración
COMPONENTES DE LA ACELERACION
)()()( tatat NT NTa +=
ROSA NIQUE ALVAREZ 22
)()()()()()( tttsttst NTTa ′′+′′=
COMPONENTES DE LA ACELERACION
Ta
T
Na N
ROSA NIQUE ALVAREZ 23
)()()( tatat NT NTa +=
)(tsaT ′′=
)()( ttsaN T ′′=
22
222
222
)(
)(
)()()(
TN
NT
NT
ata
aat
tatat
−=
+=
+=
a
a
NTa
COMPONENTES DE LA ACELERACION
ROSA NIQUE ALVAREZ 24
)()()( tatat NT NTa +=
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'r'r ×'
r'
'r'
ROSA NIQUE ALVAREZ 25
Vector Binormal
)()()()()(
ttttt
'r'r''r'r'B
××
=
ROSA NIQUE ALVAREZ 26
[ ]batttt ,);()()( ∈∀= TBN x
VECTORES UNITARIOS
ROSA NIQUE ALVAREZ 27
)()()(
ttt
r'r'T =
)()()()()( tt
ttt TB
TTN x=
′′
=
)()()()()()()(
ttttttt
'r'r''r'r'NTB
××
== x
PARA UNA CURVA REGULAR Y DE RAPIDEZ ARBITRARIAEJEMPLO 2
Considerando la siguiente función vectorial
Determine los vectores unitarios T(t), B(t), N(t) para t = 0, 1, 2.
ROSA NIQUE ALVAREZ 28
=
3,,2)(
32 ttttr
Solución
Vector Tangente
ROSA NIQUE ALVAREZ 29
( )( )2
,2,2)()()( 2
2
+==
′′
ttt
ttt
rrT
=
3,,2)(
32 ttttr Solución: vector Binormal
( )( )2
2,2,)()()()()( 2
2
+−
=××
=t
ttttttt
'r'r''r'r'B
ROSA NIQUE ALVAREZ 30
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Solución
Vector Normal Principal
ROSA NIQUE ALVAREZ 31
( )( )
( )( )
( )( )22
343
2
2
2
2
242,4,42)(
2,2,2
22,2,)()()(
+
+−−−=
++−==
ttttttt
ttt
tttttt
N
TBN xx
ROSA NIQUE ALVAREZ 32
( )( )2
,2,2)( 2
2
+=
ttttT ( )
( )22,2,)( 2
2
+−
=t
tttB
( )( )22
343
242,4,42)(
+
+−−−=
tttttttN
Solución: VECTORES UNITARIOS
Solución
ROSA NIQUE ALVAREZ 33
t r(t) T(t) N(t) B(t)
0 P0(0, 0, 0) (1,0,0) (0,1,0) (0,0,1)
1 P1(2, 1, 1/3)
2 P2(4, 4, 8/3)
)1,2,2(31 )2,1,2(
31
− )2,2,1(31
−
)2,1,2(31
−− )1,2,2(31
−)2,2,1(31
=
3,,2)(
32 ttttC r:
ROSA NIQUE ALVAREZ 34curvaregular6
-2 0 2 4 6 8
-5
0
5
10-2
0
2
4
6
8
10
X
Y
Z
Curvavector tangentevector normalbinormal
=
3,,2)(
32 ttttC r:
EJEMPLO 3
Determine el vector Binormal de la curva plana
ROSA NIQUE ALVAREZ 35
[ ]π4,0;3)cos()(cos)( ∈+−++= ttttsentsenttt kjir
Solución
ROSA NIQUE ALVAREZ 36
[ ]π4,0;3)cos()(cos)( ∈+−++= ttttsentsenttt kjir
[ ]
=∈−=
+=
34,0,cos
cos:
ztttsenty
tsenttxC π
La curva C esta contenida en el plano z = 3
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Solución
ROSA NIQUE ALVAREZ 37
( ) 0)1,0,0(,0,0)(
)()()()()(
2
2
≠===
××
=
tt
tt
ttttt
k;B
'r'r''r'r'B
[ ]π4,0;3)cos()(cos)( ∈+−++= ttttsentsenttt kjir
Solución
ROSA NIQUE ALVAREZ 38
kB =)(tVector Binormal constante
[ ]π4,0;3)cos()(cos)( ∈+−++= ttttsentsenttt kjir
0=′= )(ttd
d BBNOTA:
-15 -10-5
0 5 10
-15-10
-50
510
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
XY
Z
CURVA PLANA
ROSA NIQUE ALVAREZ 39
CurvaPlana
[ ]π4,0;3)cos()(cos)( ∈+−++= ttttsentsenttt kjir
VECTORES UNITARIOS
CURVA REGULAR DE RAPIDEZ UNITARIA
1)( =su'
ROSA NIQUE ALVAREZ 40
CURVA REGULAR DE RAPIDEZ UNITARIA
ROSA NIQUE ALVAREZ 41
[ ] 3,:: ℜ→dcC u
1)( =′ surapidez unitaria
s : longitud de arco
Sea una curva
CURVA REGULAR DE RAPIDEZ UNITARIA
ROSA NIQUE ALVAREZ 42
)()( ssT u'=
VECTOR TANGENTE
1)( =su'
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CURVA REGULAR DE RAPIDEZ UNITARIA
ROSA NIQUE ALVAREZ 43
)()(
)()()(
ss
sss
'u''u'
TTN =
′′
=
VECTOR NORMAL PRINCIPAL
1)( =su'
CURVA REGULAR DE RAPIDEZ UNITARIA
ROSA NIQUE ALVAREZ 44
)()()( sss NTB ×=
VECTOR BINORMAL
1)( =su'
)()()()(
ssss
'u''u'u'B x
=
VECTORES UNITARIOS
ROSA NIQUE ALVAREZ 45
)()( ss u'T =
)()(
)()()(
ss
sss
uu
TTN
′′′′
=′′
=
)()()()()()(
ssssss
'u''u'u'NTB ×
== x
PARA UNA CURVA REGULAR Y DE RAPIDEZ UNITARIAEJEMPLO 4
ROSA NIQUE ALVAREZ 46
Sea C la curva suave de rapidez unitaria, definida por
( )
−=
+=
=
=
sszsysx
sarctan
1ln
arctan
)( 222u
Determine los vectores unitarios Tangente, Normal y Binormal.
Solución: vector tangente
ROSA NIQUE ALVAREZ 47
)()( ss u'T =
( )22 ,2,1
11)( sss
s+
=T
( )
−=
+=
=
=
sszsysx
sarctan
1ln
arctan
)( 222u
Solución: vector normal principal
ROSA NIQUE ALVAREZ 48
)()(
)()()(
ss
sss
'u''u'
TTN =
′′
=
( ) ( )( )ssss
s 2,12,21
1)( 222
−−+
='u'
212)(s
s+
=′′u
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Solución: vector normal principal
ROSA NIQUE ALVAREZ 49
( ) ( )( )ssss
s 2,1,21
1)( 22 −−
+=N
)()(
)()()(
ss
sss
'u''u'
TTN =
′′
=
Solución: vector Binormal
ROSA NIQUE ALVAREZ 50
( )( )1,2,1
1)( 22 ss
ss −
+=B
)()()()(
ssss
'u''u'u'B x
=
Solución: vectores unitarios
ROSA NIQUE ALVAREZ 51
( )( )1,2,1
1)( 22 ss
ss −
+=B
( ) ( )( )ssss
s 2,1,21
1)( 22 −−
+=N
( )22 ,2,1
11)( sss
s+
=T
( )
−=+=
=
=ssz
sysx
sarctan
1ln
arctan
)( 222u s: longitud de arco
Solución: vector Binormal
ROSA NIQUE ALVAREZ 52
( )( )1,2,1
1)( 22 ss
ss −
+=B
212)(s
s+
=B'
( ) ( )( )ssss
s 2,12,21
1)( 222
−−+
=B'
s: longitud de arco
0)(lim =∞→
ss
B'
Solución
ROSA NIQUE ALVAREZ 53
s P B(s) B’(s) ||B’(s)||
0 (0,0,0) (0, 0, 1) (0,-√2,0) 2
1 (0.78,0.49,0.21) ½(1,-√2,1) ¼(2,0,-2) 1
2 (1.11,1.14,0.89) 1/5(4,-2√2,1) 1/25(4,3√2,-4) 2/5=0.4
3 (1.25,1.63,1.75) 1/10(9,-3√2,1) 1/100(6,8√2,-6) 2/10=0.2
( )( )1,2,1
1)( 22 ss
ss −
+=B 21
2)(s
s+
=B' Solución
ROSA NIQUE ALVAREZ 54
-1 -0.5 0 0.5 1 1.5 2 2.5
-0.50
0.51
1.52
2.5-0.5
0
0.5
1
1.5
2
2.5
3
XY
Z
Curvau(0)u(1)u(2)u(3)Vector Binormal
curvaregular5
||B’(0)||=2||B’(1) || =1||B’(2) ||=0,4||B’(3) ||=0.2
212)(s
s+
=B'
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ROSA NIQUE ALVAREZ 55
RECTAS
Recta Tangente Recta Normal
B
TN
Recta Binormal
0P
C
RECTAS
RECTA TANGENTE A C EN P0 (x0, y0, z0) esta dado por
ROSA NIQUE ALVAREZ 56
{ }ℜ∈+= λλ /)(: 00 tPPT Tl
RECTA
RECTA NORMAL A C EN P0 (x0, y0, z0) esta dado por:
ROSA NIQUE ALVAREZ 57
{ }ℜ∈+= λλ /)(: 00 tPPN Nl
PLANOS
B
TN
PlanoOsculador
PlanoNormal
Plano
Rectificante
Recta TangenteRecta Normal
Recta Binormal
0P
C
ROSA NIQUE ALVAREZ 58
PLANOS
• Plano Osculador en el punto P0 (x0, y0, z0)
• Plano Normal Principal en el punto P0 (x0, y0, z0)
• Plano Rectificante en el punto P0 (x0, y0, z0)
ROSA NIQUE ALVAREZ 59
( )[ ] 0)(,,),,(: 0000 =⋅−Ρ tzyxzyx B
( )[ ] 0)(,,),,(: 0000 =⋅−Ρ tzyxzyx T
( )[ ] 0)(,,),,(: 0000 =⋅−Ρ tzyxzyx N
ROSA NIQUE ALVAREZ 60
PLANO OSCULADOR A LA CURVA REGULAR C
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ROSA NIQUE ALVAREZ 61
( )[ ] 0)(,,),,(: 0000 =⋅− tzyxzyxΡ B
PLANO OSCULADOR A LA CURVA REGULAR C EJEMPLO 5
Determine el vector binormal y la ecuación del plano osculador a la curva regular
en el punto
ROSA NIQUE ALVAREZ 62
π20)()cos()()cos()(:
≤≤++=
ttsenttsenttC kjir
( )1,2,221 −−=P
Solución: vector Binormal
ROSA NIQUE ALVAREZ 63
×
×
=
47
47
47
47
47
ππ
πππ
'r'r'
'r'r'B
( )1,2,2,4
721 −−== Pt π
Solución
ROSA NIQUE ALVAREZ 64
( )1,2,25
1)4/7(
BinormalVector
−=πB
( )1,2,2,4
721 −−== Pt π
Solución: Plano Osculador
ROSA NIQUE ALVAREZ 65
( )1,2,2,4
721 −−== Pt π
04
721,
22,
22),,(: =
⋅
−−− πBzyxP
( )1,2,25
1)4/7( −=πB
Solución
ROSA NIQUE ALVAREZ 66
2322
OsculadorPlano
=+− zyx
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Gráfica
ROSA NIQUE ALVAREZ 67
curvaCERRADA1-1
-0.5
0
0.5
1
-1-0.5
0
0.5
1-0.5
0
0.5
X
Curva Cerrada
Y
Z
π20)()cos()()cos()(:
≤≤++=
ttsenttsenttC kjir
EJEMPLO 6
68Rosa Ñique Alvarez
Dada la siguiente curva regular
Determine el vector binormal y el plano osculador para cada punto de C cuando t = 0,1,2.
=
3,,2)(:
32 ttttC r
Solución
69Rosa Ñique Alvarez
=
3,,2)(:
32 ttttC r
( )2
2,2,)( 2
2
+−
=t
tttB
)2,2,1(31)1(,1 −== Bt
)1,2,2(31)2(,2 −== Bt
kB === )1,0,0()0(,0t
)()()()()(
ttttt
'r'r''r'r'B
××
=
Solución Plano osculador
Rosa Ñique Alvarez 70
[ ] 0)0()0(),,(:,0 0 =⋅−= BrzyxPt
[ ] 0)1()1(),,(:,1 1 =⋅−= BrzyxPt
[ ] 0)2()2(),,(:,2 2 =⋅−= BrzyxPt
=
3,,2)(:
32 ttttC r
Solución Plano osculador
Rosa Ñique Alvarez 71
[ ] 0)0()0,0,0(),,(:,0 =⋅−= BzyxPt
[ ] 0)1()3/1,1,2(),,(:,1 =⋅−= BzyxPt
[ ] 0)2()3/8,4,4(),,(:,2 =⋅−= BzyxPt
=
3,,2)(:
32 ttttC r
Solución Plano osculador
Rosa Ñique Alvarez 72
0:,0 0 == zPt
yxzzyxPt +−=→=+−=21
312663:,1 1
yxzzyxPt 22388366:,2 2 +−=→=+−=
=
3,,2)(:
32 ttttC r
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Plano osculador a C para t =0,1,2
Rosa Ñique Alvarez 73
=
3,,2)(:
32 ttttC r
-20
24
68
-20
24
68-2
0
2
4
6
PlanoOsculador2
kB =)0(
)2,2,1(31)1( −=B
)1,2,2(31)2( −=B
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