![Tenemos que entonces transformando a radiales
180
8.04
= 8.04
180
(51
5
, 8.04)
180
Calcular el 叩rea que encierra la curva de ecuaci坦n polar r = +
= 1 + (2,
2
)
= 0
1 + = 0
= 1 (0,0)
= 1 (1)
=
2
sen =
2
=
2
A = 2
2
0
Producto notable
(1 + )2
2
0
A = (1 + 2 + 20)
Luego por identidad trigonom辿trica
1 2
2 =
2
A= (1 + 2 + 1
2
) do
Por
A = (
1 + 4 + 1 2
2
)
Se reducen t辿rminos semejantes
A =
1
2
(3 + 4 2)
2
0
A =
1
2
[3 危 2]
2
0
2](https://image.slidesharecdn.com/isabel-140902102943-phpapp01/85/Isabel-2-320.jpg)
![=
1
2
[3 (
2
) 4 cos (
2
) 属 2 (
2
) 属 ] + 4 cos(0)
2
=
1
2
3
2
[
+ 4]
=
1
2
3 + 8
[
2
]
=
3 + 8
Transformar los siguientes puntos de coordenadas rectangulares a coordenadas
polares
) (2,
4
)
2
2
= = 2
(
= = 2
= 2
4
= 2 (
4
) = 2
2
2
= 2 (2, 2)
) (8,
3
2
)
= 8 (
3
2
)
= 8 (
3
2
) = 0 = 8 (0,8)
) (
1
2
,
5
4
)
=
1
2
cos (
5
4
) ;=
1
2
2
2
(
2
4
) ==
=
1
2
(
5
4
) ; =
1
2
(
2
2
) = =
2
4
2
4
(
2
4
,
)
Calcular el 叩rea que encierra la curva de ecuaci坦n polar r= 4 cos(200)
= 4(2) Cambio de variable
4(2) = 0 = 2 =
2
2 = 0
= 0](https://image.slidesharecdn.com/isabel-140902102943-phpapp01/85/Isabel-3-320.jpg)
![= 孫(0)
=
2
=
5
4
=
7
4
=
4
=
3
4
3
4
4
5
4
7
4
= 0 (
1
2
4
) [4(2)]
4
2
4
= 08 族2
4
4
= 2(8) 族 2
0
Por identidad trigonom辿trica
1 + 躯2
= 16 [
2
]
4
0
4
= 8 (1 + + 2)
0
= 8 [0+(2)
2
]
4
0
= 8 [
4
+ [2 (
4
)] (0)]
= 2族 = 8族
Transformar a coordenadas rectangulares
= 2 = (3) = 2(2 + ) = 2[2 2]](https://image.slidesharecdn.com/isabel-140902102943-phpapp01/85/Isabel-4-320.jpg)
![Por identidad trigonom辿trica
= 2[(2族 1)] 20族
= [(2族 1) 20族]
= 2[(2族 1) 2(1 族)]
Propiedad distributiva
= 2[2続 2 + 2続]
= 2[4続 3]
2 = 83 6
族 = 2[4族 3]
= 2 (
4 族 3
) [
2
]
4 = 2 [4 族 3族]
( 族 + 4族)族 = 2 [4 族 3( 族 + 4族)]
( 族 + β)族 = 2 [ 族 3 2 3β]
( 族 + β)族 = 2 [ 族 3β]
Transformar las siguientes ecuaci坦n de variable
族 2β = 4( +)族
= , =
()族 2()族 = 4[ + ]族
2 222 = 4 [ (
) + (
2
)]
族[族 2族] = 4( +)族
族 [(
) 族 2 (
) 族] = 4( +)族
族 [
族 2β
2
] = 4( +)族 族 2β = 4( +)族
2 22 = 4(2+ 2 + 2) 2 22 = 4 2+ 8 + 42
3 族 + 8 + 64族 = 0](https://image.slidesharecdn.com/isabel-140902102943-phpapp01/85/Isabel-5-320.jpg)
Isabel
- 1. Trasformar los siguientes coordenadas rectangulares a coordenadas polares. A) (2,8) primero lo hacemos por definici坦n = tan1( ) 2 = 2 + 2 8 2 = tan1( ) 2 = (2)族 + (8)族 = tan1(4) 2 = 4 + 64 = 75.96 = 68 = 217 Luego transformando a radiaciones se tiene que 180 75 96 = 75 96 180 (217,75 96) 180 B) (5, 6) = tan1( 6 5 ) = 50.19属 2 = (5)族 + (6)族 2 = 25 + 36 = 61 61, 50.19 180 C) (2, 1 5 ) 2 = (2) 族 + (1 5 )族 = tan1( 1 5 2 ) 2 = 2 + 1 25 = tan1( 1 52 ) 2 = 51 25 = 8. 04属 = 51 25 = = 51 5
- 2. Tenemos que entonces transformando a radiales 180 8.04 = 8.04 180 (51 5 , 8.04) 180 Calcular el 叩rea que encierra la curva de ecuaci坦n polar r = + = 1 + (2, 2 ) = 0 1 + = 0 = 1 (0,0) = 1 (1) = 2 sen = 2 = 2 A = 2 2 0 Producto notable (1 + )2 2 0 A = (1 + 2 + 20) Luego por identidad trigonom辿trica 1 2 2 = 2 A= (1 + 2 + 1 2 ) do Por A = ( 1 + 4 + 1 2 2 ) Se reducen t辿rminos semejantes A = 1 2 (3 + 4 2) 2 0 A = 1 2 [3 危 2] 2 0 2
- 3. = 1 2 [3 ( 2 ) 4 cos ( 2 ) 属 2 ( 2 ) 属 ] + 4 cos(0) 2 = 1 2 3 2 [ + 4] = 1 2 3 + 8 [ 2 ] = 3 + 8 Transformar los siguientes puntos de coordenadas rectangulares a coordenadas polares ) (2, 4 ) 2 2 = = 2 ( = = 2 = 2 4 = 2 ( 4 ) = 2 2 2 = 2 (2, 2) ) (8, 3 2 ) = 8 ( 3 2 ) = 8 ( 3 2 ) = 0 = 8 (0,8) ) ( 1 2 , 5 4 ) = 1 2 cos ( 5 4 ) ;= 1 2 2 2 ( 2 4 ) == = 1 2 ( 5 4 ) ; = 1 2 ( 2 2 ) = = 2 4 2 4 ( 2 4 , ) Calcular el 叩rea que encierra la curva de ecuaci坦n polar r= 4 cos(200) = 4(2) Cambio de variable 4(2) = 0 = 2 = 2 2 = 0 = 0
- 4. = 孫(0) = 2 = 5 4 = 7 4 = 4 = 3 4 3 4 4 5 4 7 4 = 0 ( 1 2 4 ) [4(2)] 4 2 4 = 08 族2 4 4 = 2(8) 族 2 0 Por identidad trigonom辿trica 1 + 躯2 = 16 [ 2 ] 4 0 4 = 8 (1 + + 2) 0 = 8 [0+(2) 2 ] 4 0 = 8 [ 4 + [2 ( 4 )] (0)] = 2族 = 8族 Transformar a coordenadas rectangulares = 2 = (3) = 2(2 + ) = 2[2 2]
- 5. Por identidad trigonom辿trica = 2[(2族 1)] 20族 = [(2族 1) 20族] = 2[(2族 1) 2(1 族)] Propiedad distributiva = 2[2続 2 + 2続] = 2[4続 3] 2 = 83 6 族 = 2[4族 3] = 2 ( 4 族 3 ) [ 2 ] 4 = 2 [4 族 3族] ( 族 + 4族)族 = 2 [4 族 3( 族 + 4族)] ( 族 + β)族 = 2 [ 族 3 2 3β] ( 族 + β)族 = 2 [ 族 3β] Transformar las siguientes ecuaci坦n de variable 族 2β = 4( +)族 = , = ()族 2()族 = 4[ + ]族 2 222 = 4 [ ( ) + ( 2 )] 族[族 2族] = 4( +)族 族 [( ) 族 2 ( ) 族] = 4( +)族 族 [ 族 2β 2 ] = 4( +)族 族 2β = 4( +)族 2 22 = 4(2+ 2 + 2) 2 22 = 4 2+ 8 + 42 3 族 + 8 + 64族 = 0