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 Trasformar los siguientes coordenadas rectangulares a coordenadas polares. 
A) (2,8) primero lo hacemos por definición 
휃 = tan−1( 
푦 
푥 
) 푅2 = 푋2 + 푌2 
8 
2 
휃 = tan−1( 
) 푅2 = (2)² + (8)² 
휃 = tan−1(4) 푅2 = 4 + 64 
휃 = 75.96 푅 = √68 
푅 = 2√17 
Luego transformando a radiaciones se tiene que 
휋 − 180 
× −75 ∙ 96 
×= 75 ∙ 96휋 
180 
(2√17,75 ∙96휋) 
180 
B) (5, −6) 
휃 = tan−1( 
−6 
−5 
) 
휃 = 50.19° 푅2 = (−5)² + (−6)² 
푅2 = 25 + 36 
푅 = √61 
√61, 50.19 휋 
180 
C) (√2, 
1 
5 
) 푅2 = (√2) ² + (1 
5 
)² 
휃 = tan−1( 
1 
5 
√2 
) 푅2 = 2 + 1 
25 
휃 = tan−1( −1 
5√2 
) 푅2 = 51 
25 
휃 = 8. 04° 푅 = √51 
25 
= 푅 = √51 
5
Tenemos que entonces transformando a radiales 
휋 − 180 
×∙ 8.04 
×= 8.04휋 
180 
(√51 
5 
, 8.04휋) 
180 
 Calcular el área que encierra la curva de ecuación polar r = +푠푒푛휃 
푅 = 1 + 푠푒푛휃 (2, 
휋 
2 
) 
푠푖 푟 = 0 
1 + 푠푒푛휃 = 0 
푠푒푛휃 = −1 (0,0) 
휃 = 푠푒푛−1 (−1) 
휃 = − 휋 
2 
sen 휃 = 푦휋 
2 
= − 휋 
2 
A = 2 
푍 
휋 
2 
0 
Producto notable 
∫ (1 + 푠푒푛휃)2푑표 
휋 
2 
0 
A = ∫ (1 + 2푠푒푛휃휃 + 푠푒푛20)푑표 
Luego por identidad trigonométrica 
1 − 푐표푠2휃 
푠푒푛2휃 = 
2 
A= ∫(1 + 2푠푒푛휃 + 1−푐표푠휃 
2 
) do 
Por 
A = ∫( 
1 + 4푠푒푛휃 + 1 − 푐표푠2휃 
2 
) 푑표 
Se reducen términos semejantes 
A = 
1 
2 
∫ 
(3 + 4푠푒푛휃 − 푐표푠2휃)푑표 
휋 
2 
0 
A = 
1 
2 
[3휃 − 퐴푐표푠휃 − 푠푒푛2휃] ∫ 
휋 
2 
0 
2
퐴 = 
1 
2 
[3 ( 
휋 
2 
) − 4 cos ( 
휋 
2 
) ° − 푠푒푛푡2 ( 
휋 
2 
) ° ] + 4 cos(0) 
2 
퐴 = 
1 
2 
휋3 
2 
[ 
+ 4] 
퐴 = 
1 
2 
3휋 + 8 
[ 
2 
] 
퐴 = 
3휋 + 8 
퐴 
 Transformar los siguientes puntos de coordenadas rectangulares a coordenadas 
polares 
퐴) (2, 
휋 
4 
) 
2 
2 
×= 푟푐표푠휃 ×= 2̕ 
(√ 
푦 = 푟푠푒푛휃 ×= √2 
×= 2푐표푠 휋 
4 
푦 = 2푠푒푛 ( 
휋 
4 
) 푦 = 2√ 
2 
2 
푦 = √2 (√2, √2) 
푏) (−8, 
3휋 
2 
) 
×= −8푐표푠 ( 
3휋 
2 
) 
푦 = −8푠푒푛 ( 
3휋 
2 
) ×= 0 푦 = 8 (0,8) 
퐶) ( 
−1 
2 
, 
5휋 
4 
) 
×= 
−1 
2 
cos ( 
5휋 
4 
) ;×= 
−1 
2 
−√2 
2 
( 
2 
4 
) =×= √ 
푦 = 
−1 
2 
푠푒푛 ( 
5휋 
4 
) ; 푦 = 
−1 
2 
(−√ 
2 
2 
) = 푦 = √ 
2 
4 
2 
4 
(√ 
2 
4 
, √ 
) 
 Calcular el área que encierra la curva de ecuación polar r= 4 cos(200) 
푅 = 4푐표푠(2휃) Cambio de variable 
4푐표푠(2휃) = 0 푢 = 2휃 휃 = 푢 
2 
푐표푠2휃 = 0 
푐표푠푢휃 = 0
푢 = 푐표푠 − ¹(0) 
푚 = 
휋 
2 
휃 = 
5 
4 
휋 휃 = 
7 
4 
휋 휃 = 
휋 
4 
휃 = 
3 
4 
휋 
3휋 
4 
휋 
4 
5휋 
4 
7휋 
4 
퐴 = 0 ( 
1 
2 
휋 
4 
) ∫ [4푐표푠(2휃)] 
−휋 
4 
2 
푑표 
휋 
4 
퐴 = 08 ∫ 푐표푠²2휃 푑표 
−휋 
4 
휋 
4 
퐴 = 2(8) ∫ 푐표푠² 2휃 푑표 
0 
Por identidad trigonométrica 
1 + 퐶푂푆2휃 
퐴 = 16 ∫ [ 
2 
] 
휋 
4 
0 
푑표 
휋 
4 
퐴 = 8 ∫ (1 + 푐표푠 + 2휃) 
0 
푑표 
퐴 = 8 [0+푠푒푛(2휃휃) 
2 
] 
휋 
4 
0 
퐴 = 8 [ 
휋 
4 
+ 푠푒푛 [2 ( 
휋 
4 
)] − (0)] 
퐴 = 2휋푢² 퐴 = 8휋푢² 
Transformar a coordenadas rectangulares 
푅 = 2푐표푠 = (3휃) 푅 = 2푐표푠(2휃 + 휃) 푅 = 2[푐표푠2휃푐표푠휃 − 푠푒푛2휃푠푒푛표]
Por identidad trigonométrica 
푟 = −2[(2푐표푠²휃 − 1)]푐표푠휃 − 2푐표푠0푠푒푛²휃 
푟 = [(2푐표푠²휃 − 1)푐표푠휃 − 2푐표푠0푠푒푛²휃] 
푟 = 2[(2푐표푠²휃 − 1)푐표푠휃 − 2푐표푠휃(1 − 푐표푠²휃)] 
Propiedad distributiva 
푟 = 2[2푐표푠³휃 − 푐표푠휃 − 2푐표푠휃 + 2푐표푠³휃] 
푟 = 2[4푐표푠³휃 − 3푐표푠휃] 
푅2 = 8푐표푠3휃 − 6푐표푠휃 
푅² = 2푐표푠휃[4푐표푠²휃 − 3] 
푅 = 2 ( 
× 
푦 
4 × ² − 3 
) [ 
푟2 
] 
푅4 = 2 × [4 × ² − 3푟²] 
(× ² + 4²)² = 2 × [4 × ² − 3(× ² + 4²)] 
(× ² + 푦²)² = 2 × [퐴 × ² − 3 ×2− 3푦²] 
(× ² + 푦²)² = 2 × [× ² − 3푦²] 
 Transformar las siguientes ecuación de variable 
× ² − 2푦² = 4(× +푦)² 
×= 푐표푠휃 , 푦 = 푟푠푒푛휃 
(푟푐표푠휃)² − 2(푟푠푒푛휃)² = 4[푟푐표푠휃 + 푟푠푒푛휃]² 
푅2푐표푠휃 − 2푟2푠푒푛2 휃 = 4 [푟 ( 
× 
푦 
) + 푟 ( 
푦 
× 
2 
)] 
푅²[푐표푠²휃 − 2푠푒푛²휃] = 4(× +푦)² 
푅² [(× 
푟 
) ² − 2 (푦 
× 
) ²] = 4(× +푦)² 
푅² [ 
× ² − 2푦² 
푟2 
] = 4(× +푦)² × ² − 2푦² = 4(× +푦)² 
×2 − 2푦2 = 4(×2+ 2 × 푦 + 푦2) ×2− 2푦2 = 4 ×2+ 8 × 푦 + 4푦2 
3 × ² + 8 × 푦 + 64² = 0
Isabel

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Isabel

  • 1.  Trasformar los siguientes coordenadas rectangulares a coordenadas polares. A) (2,8) primero lo hacemos por definición 휃 = tan−1( 푦 푥 ) 푅2 = 푋2 + 푌2 8 2 휃 = tan−1( ) 푅2 = (2)² + (8)² 휃 = tan−1(4) 푅2 = 4 + 64 휃 = 75.96 푅 = √68 푅 = 2√17 Luego transformando a radiaciones se tiene que 휋 − 180 × −75 ∙ 96 ×= 75 ∙ 96휋 180 (2√17,75 ∙96휋) 180 B) (5, −6) 휃 = tan−1( −6 −5 ) 휃 = 50.19° 푅2 = (−5)² + (−6)² 푅2 = 25 + 36 푅 = √61 √61, 50.19 휋 180 C) (√2, 1 5 ) 푅2 = (√2) ² + (1 5 )² 휃 = tan−1( 1 5 √2 ) 푅2 = 2 + 1 25 휃 = tan−1( −1 5√2 ) 푅2 = 51 25 휃 = 8. 04° 푅 = √51 25 = 푅 = √51 5
  • 2. Tenemos que entonces transformando a radiales 휋 − 180 ×∙ 8.04 ×= 8.04휋 180 (√51 5 , 8.04휋) 180  Calcular el área que encierra la curva de ecuación polar r = +푠푒푛휃 푅 = 1 + 푠푒푛휃 (2, 휋 2 ) 푠푖 푟 = 0 1 + 푠푒푛휃 = 0 푠푒푛휃 = −1 (0,0) 휃 = 푠푒푛−1 (−1) 휃 = − 휋 2 sen 휃 = 푦휋 2 = − 휋 2 A = 2 푍 휋 2 0 Producto notable ∫ (1 + 푠푒푛휃)2푑표 휋 2 0 A = ∫ (1 + 2푠푒푛휃휃 + 푠푒푛20)푑표 Luego por identidad trigonométrica 1 − 푐표푠2휃 푠푒푛2휃 = 2 A= ∫(1 + 2푠푒푛휃 + 1−푐표푠휃 2 ) do Por A = ∫( 1 + 4푠푒푛휃 + 1 − 푐표푠2휃 2 ) 푑표 Se reducen términos semejantes A = 1 2 ∫ (3 + 4푠푒푛휃 − 푐표푠2휃)푑표 휋 2 0 A = 1 2 [3휃 − 퐴푐표푠휃 − 푠푒푛2휃] ∫ 휋 2 0 2
  • 3. 퐴 = 1 2 [3 ( 휋 2 ) − 4 cos ( 휋 2 ) ° − 푠푒푛푡2 ( 휋 2 ) ° ] + 4 cos(0) 2 퐴 = 1 2 휋3 2 [ + 4] 퐴 = 1 2 3휋 + 8 [ 2 ] 퐴 = 3휋 + 8 퐴  Transformar los siguientes puntos de coordenadas rectangulares a coordenadas polares 퐴) (2, 휋 4 ) 2 2 ×= 푟푐표푠휃 ×= 2̕ (√ 푦 = 푟푠푒푛휃 ×= √2 ×= 2푐표푠 휋 4 푦 = 2푠푒푛 ( 휋 4 ) 푦 = 2√ 2 2 푦 = √2 (√2, √2) 푏) (−8, 3휋 2 ) ×= −8푐표푠 ( 3휋 2 ) 푦 = −8푠푒푛 ( 3휋 2 ) ×= 0 푦 = 8 (0,8) 퐶) ( −1 2 , 5휋 4 ) ×= −1 2 cos ( 5휋 4 ) ;×= −1 2 −√2 2 ( 2 4 ) =×= √ 푦 = −1 2 푠푒푛 ( 5휋 4 ) ; 푦 = −1 2 (−√ 2 2 ) = 푦 = √ 2 4 2 4 (√ 2 4 , √ )  Calcular el área que encierra la curva de ecuación polar r= 4 cos(200) 푅 = 4푐표푠(2휃) Cambio de variable 4푐표푠(2휃) = 0 푢 = 2휃 휃 = 푢 2 푐표푠2휃 = 0 푐표푠푢휃 = 0
  • 4. 푢 = 푐표푠 − ¹(0) 푚 = 휋 2 휃 = 5 4 휋 휃 = 7 4 휋 휃 = 휋 4 휃 = 3 4 휋 3휋 4 휋 4 5휋 4 7휋 4 퐴 = 0 ( 1 2 휋 4 ) ∫ [4푐표푠(2휃)] −휋 4 2 푑표 휋 4 퐴 = 08 ∫ 푐표푠²2휃 푑표 −휋 4 휋 4 퐴 = 2(8) ∫ 푐표푠² 2휃 푑표 0 Por identidad trigonométrica 1 + 퐶푂푆2휃 퐴 = 16 ∫ [ 2 ] 휋 4 0 푑표 휋 4 퐴 = 8 ∫ (1 + 푐표푠 + 2휃) 0 푑표 퐴 = 8 [0+푠푒푛(2휃휃) 2 ] 휋 4 0 퐴 = 8 [ 휋 4 + 푠푒푛 [2 ( 휋 4 )] − (0)] 퐴 = 2휋푢² 퐴 = 8휋푢² Transformar a coordenadas rectangulares 푅 = 2푐표푠 = (3휃) 푅 = 2푐표푠(2휃 + 휃) 푅 = 2[푐표푠2휃푐표푠휃 − 푠푒푛2휃푠푒푛표]
  • 5. Por identidad trigonométrica 푟 = −2[(2푐표푠²휃 − 1)]푐표푠휃 − 2푐표푠0푠푒푛²휃 푟 = [(2푐표푠²휃 − 1)푐표푠휃 − 2푐표푠0푠푒푛²휃] 푟 = 2[(2푐표푠²휃 − 1)푐표푠휃 − 2푐표푠휃(1 − 푐표푠²휃)] Propiedad distributiva 푟 = 2[2푐표푠³휃 − 푐표푠휃 − 2푐표푠휃 + 2푐표푠³휃] 푟 = 2[4푐표푠³휃 − 3푐표푠휃] 푅2 = 8푐표푠3휃 − 6푐표푠휃 푅² = 2푐표푠휃[4푐표푠²휃 − 3] 푅 = 2 ( × 푦 4 × ² − 3 ) [ 푟2 ] 푅4 = 2 × [4 × ² − 3푟²] (× ² + 4²)² = 2 × [4 × ² − 3(× ² + 4²)] (× ² + 푦²)² = 2 × [퐴 × ² − 3 ×2− 3푦²] (× ² + 푦²)² = 2 × [× ² − 3푦²]  Transformar las siguientes ecuación de variable × ² − 2푦² = 4(× +푦)² ×= 푐표푠휃 , 푦 = 푟푠푒푛휃 (푟푐표푠휃)² − 2(푟푠푒푛휃)² = 4[푟푐표푠휃 + 푟푠푒푛휃]² 푅2푐표푠휃 − 2푟2푠푒푛2 휃 = 4 [푟 ( × 푦 ) + 푟 ( 푦 × 2 )] 푅²[푐표푠²휃 − 2푠푒푛²휃] = 4(× +푦)² 푅² [(× 푟 ) ² − 2 (푦 × ) ²] = 4(× +푦)² 푅² [ × ² − 2푦² 푟2 ] = 4(× +푦)² × ² − 2푦² = 4(× +푦)² ×2 − 2푦2 = 4(×2+ 2 × 푦 + 푦2) ×2− 2푦2 = 4 ×2+ 8 × 푦 + 4푦2 3 × ² + 8 × 푦 + 64² = 0