UNIVERSIDAD DE EL SALVADOR
FACULTAD DE INGENIERÍA Y ARQUITECTURA
ESCUELA DE INGENIERÍA INDUSTRIAL
MATEMATICA III
CICLO I 2020
EVALUACIÓN: Tarea de resolución de ejercicios 2.
PROFESOR: Ing. Ricardo Alfaro.
ALUMNO: Jonathan José Reyes Vásquez.
GRUPO: 1.
CARNET: RV19022
Ciudad Universitaria, 18 de mayo de 2020
Tarea de Resolución de ejercicios 2
1) 𝐸𝑙𝑎𝑏𝑜𝑟𝑒 𝑙𝑎 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑑𝑒𝑙 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑝𝑙𝑎𝑛𝑜: 𝑥 − 2𝑦 − 4𝑧 = 8
𝑇𝑟𝑎𝑧𝑎𝑠:
𝑇𝑟𝑎𝑧𝑎 𝑝𝑙𝑎𝑛𝑜 𝑥𝑦 (𝑧 = 0)
𝑥 − 2𝑦 = 8
𝑆𝑖𝑒𝑛𝑑𝑜 𝑥 = 0
0 − 2𝑦 = 8 → 𝑦 = −4
𝑆𝑖𝑒𝑛𝑑𝑜 𝑦 = 0
𝑥 − 2(0) = 8 → 𝑥 = 8
𝐿𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑠𝑜𝑛 (0, −4, 0) 𝑦 (8, 0, 0).
𝑇𝑟𝑎𝑧𝑎 𝑝𝑙𝑎𝑛𝑜 𝑥𝑧 (𝑦 = 0)
𝑥 − 4𝑧 = 8
𝑆𝑖𝑒𝑛𝑑𝑜 𝑥 = 0
0 − 4𝑧 = 8 → 𝑧 = −2
𝑆𝑖𝑒𝑛𝑑𝑜 𝑧 = 0
𝑥 − 4(0) = 8 → 𝑥 = 8
𝐿𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑠𝑜𝑛 (0,0, −2) 𝑦 (8, 0, 0).
𝑇𝑟𝑎𝑧𝑎 𝑝𝑙𝑎𝑛𝑜 𝑦𝑧 (𝑦 = 0)
−2𝑦 − 4𝑧 = 8
𝑆𝑖𝑒𝑛𝑑𝑜 𝑦 = 0
−2(0) − 4𝑧 = 8 → 𝑧 = −2
𝑆𝑖𝑒𝑛𝑑𝑜 𝑧 = 0
− 2𝑦 − 4(0) = 8 → 𝑦 = −4
𝐿𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑠𝑜𝑛 (0,0, −2) 𝑦 (0, 0, −8).
𝑈𝑛𝑎 𝑣𝑒𝑧 𝑡𝑒𝑛𝑖𝑑𝑜 𝑙𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑒𝑛 𝑐𝑜𝑚ú𝑛, 𝑠𝑒 𝑝𝑟𝑜𝑐𝑒𝑑𝑒 𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎𝑟:
𝐴(0, −4,0)
𝐵(8,0,0)
𝐶(0,0, −2)
2) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 𝑙𝑎 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑑𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑃(1,1,1) 𝑎𝑙 𝑝𝑙𝑎𝑛𝑜 𝑑𝑒𝑙 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟
𝐷=
𝐷=
|𝑎𝑥𝑜 + 𝑏𝑦𝑜 + 𝑐𝑧𝑜 + 𝑑|
√𝑎2 + 𝑏 2 + 𝑐 2
→ < 𝑎, 𝑏, 𝑐 > = < 1, −2, −4 > ; 𝑑 = −8 ; 𝑃(𝑥𝑜 , 𝑦𝑜 , 𝑧𝑜 )
|(1)(1) + (−2)(1) + (−4)(1) + (−8)|
√12 + (−2)2 + (−4)2
𝐷=
|−13|
√21
=
|1 − 2 − 4 − 8|
√1 + 4 + 16
= 2.8368 ….
𝐷 = 2.84 𝑢𝑙
3) 𝐸𝑙𝑎𝑏𝑜𝑟𝑒 𝑙𝑎 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑠𝑢𝑝𝑒𝑟𝑓𝑖𝑐𝑖𝑒 𝑐𝑖𝑙í𝑛𝑑𝑟𝑖𝑐𝑎 𝑧 = 𝐿𝑛 (𝑦)
Curva generatriz en el plano yz y las rectas directrices se encuentran en el eje x.
4) 𝐺𝑟𝑎𝑓𝑖𝑞𝑢𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑐𝑢á𝑑𝑟𝑖𝑐𝑎: 𝑧 =
𝑇𝑟𝑎𝑧𝑎 𝑐𝑜𝑛 𝑒𝑙 𝑝𝑙𝑎𝑛𝑜 𝑥𝑧: 𝑧 =
𝑥2 𝑦2
−
9
4
𝑥2
→ 𝐶𝑢𝑎𝑛𝑑𝑜 𝑧 = 1
9
1(9) = 𝑥 2 → 𝑥 = 3 𝑦 𝑥 = −3
𝑇𝑟𝑎𝑧𝑎 𝑐𝑜𝑛 𝑒𝑙 𝑝𝑙𝑎𝑛𝑜 𝑦𝑧: 𝑧 = −
𝑦2
→ 𝐶𝑢𝑎𝑛𝑑𝑜 𝑧 = −1
4
(−1)(−4) = 𝑦 2 → 𝑦 = 2 𝑦 𝑦 = −2
𝐶𝑜𝑟𝑡𝑒 𝑒𝑛 𝑙𝑜𝑠 𝑒𝑥𝑡𝑟𝑒𝑚𝑜𝑠 𝑑𝑒 𝑙𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎:
𝐶𝑢𝑎𝑛𝑑𝑜 𝑥 = ±3
𝑧=
(±3)2 𝑦 2
𝑦2
−
→𝑧 =1−
9
4
4
𝑧−1=−
𝑦2
→ 𝑉𝑒𝑟𝑡𝑖𝑐𝑒 𝑡𝑟𝑎𝑠𝑙𝑎𝑑𝑎𝑑𝑜 (±3, 0, 1)
4
𝑃𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑒𝑛 𝑙𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 𝑑𝑒 𝑙𝑎 𝑖𝑧𝑞𝑢𝑖𝑒𝑟𝑑𝑎
(3, (√2)(2), −1) 𝑦 (3, (−√2)(2), −1) → (3,2√2, −1) 𝑦 (3, −2√2, −1)
𝑃𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑒𝑛 𝑙𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 𝑑𝑒 𝑙𝑎 𝑑𝑒𝑟𝑒𝑐ℎ𝑎
(−3, (√2)(2), −1) 𝑦 (−3, (−√2)(2), −1) → (−3,2√2, −1) 𝑦 (−3, −2√2, −1)
Grafica:
5) 𝐺𝑟𝑎𝑓𝑖𝑞𝑢𝑒 𝑒𝑙 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑠ó𝑙𝑖𝑑𝑜: 𝑥 = 𝑦 2 + 𝑧² 𝑦 𝑒𝑙 𝑝𝑟𝑖𝑚𝑒𝑟 𝑜𝑐𝑡𝑎𝑛𝑡𝑒.
𝐿𝑎 𝑓𝑢𝑛𝑐𝑖𝑜𝑛 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎 𝑢𝑛 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑜𝑖𝑑𝑒 𝑒𝑙𝑖𝑝𝑡𝑖𝑐𝑜 𝑒𝑛 𝑒𝑙 𝑒𝑗𝑒 𝑥, 𝑝𝑎𝑟𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎𝑟𝑙𝑜 𝑠𝑒 𝑐𝑜𝑙𝑜𝑐𝑎𝑟𝑎
𝑒𝑙 𝑒𝑗𝑒 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑒𝑛 𝑥 = 9 (𝑜 𝑐𝑢𝑎𝑛𝑑𝑜 𝑦 = 𝑧 = 3 ).
𝑄𝑢𝑒𝑑𝑎𝑛𝑑𝑜: 9 = 𝑦 2 + 𝑧 2 𝑞𝑢𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎 𝑢𝑛 𝑐𝑖𝑟𝑐𝑢𝑙𝑜 𝑑𝑒 𝑟 = 3
𝐿𝑢𝑒𝑔𝑜 𝑠𝑒 𝑡𝑟𝑎𝑧𝑎𝑛 𝑙𝑎𝑠 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎𝑠 𝑒𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑠 𝑥𝑦 𝑦 𝑥𝑧 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑎𝑚𝑒𝑛𝑡𝑒:
𝐶𝑜𝑚𝑜 𝑒𝑠 𝑒𝑛 𝑒𝑙 𝑝𝑟𝑖𝑚𝑒𝑟 𝑜𝑐𝑡𝑎𝑛𝑡𝑒 𝑑𝑜𝑛𝑑𝑒 𝑙𝑜𝑠 3 𝑒𝑗𝑒𝑠 𝑠𝑜𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑜𝑠 𝑞𝑢𝑒𝑑𝑎𝑟í𝑎:
1
+𝑦
𝑫𝒐𝒎𝒊𝒏𝒊𝒐: 𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑛𝑜 𝑞𝑢𝑒𝑑𝑒 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑒𝑠 𝑥 2 + 𝑦 ≠ 0 ó 𝑥 2 ≠ −𝑦
𝑃𝑜𝑟 𝑙𝑜 𝑡𝑎𝑛𝑡𝑜 𝑒𝑙 𝑑𝑜𝑚𝑖𝑛𝑖𝑜 𝑞𝑢𝑒𝑑𝑎𝑟í𝑎:
6) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑒𝑙 𝑑𝑜𝑚𝑖𝑛𝑖𝑜, 𝑟𝑎𝑛𝑔𝑜 𝑦 𝑔𝑟𝑎𝑓𝑖𝑞𝑢𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑓𝑢𝑛𝑐𝑖ó𝑛: 𝑓(𝑥, 𝑦) =
𝑥2
𝐷𝑓 = {(𝑥, 𝑦) ∈ 𝑹𝟐 /𝑥 2 + 𝑦 2 ≠ 0}
𝑹𝒂𝒏𝒈𝒐: 𝑦𝑎 𝑞𝑢𝑒 𝑐𝑜𝑛𝑜𝑐𝑒𝑚𝑜𝑠 𝑒𝑙 𝑑𝑜𝑚𝑖𝑛𝑖𝑜 𝑠𝑒 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟𝑎𝑛 𝑣𝑎𝑟𝑎𝑖𝑎𝑏𝑙𝑒𝑠 𝑐𝑒𝑟𝑐𝑎𝑛𝑎𝑠 𝑎 𝑐𝑒𝑟𝑜 𝑐𝑜𝑛
𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑙𝑒𝑠 𝑝𝑎𝑟𝑎 𝑖𝑛𝑑𝑖𝑐𝑎𝑟 𝑙𝑎 𝑡𝑒𝑛𝑑𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖𝑜𝑛:
1
= −∞
(𝑥,𝑦)→(0,0)− 𝑥 2 + 𝑦
lim
1
= +∞
(𝑥,𝑦)→(0,0)+ 𝑥 2 + 𝑦
lim
𝑆𝑖𝑛 𝑒𝑚𝑏𝑎𝑟𝑔𝑜 𝑎𝑙 𝑒𝑣𝑎𝑙𝑢𝑎𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑒𝑙 𝑙𝑖𝑚𝑖𝑡𝑒 𝑒𝑛 𝑒𝑙 𝑑𝑜𝑚𝑖𝑛𝑖𝑜:
lim
(𝑥,𝑦)→(0,0) 𝑥 2
1
=∄
+𝑦
𝑇𝑎𝑙 𝑞𝑢𝑒 𝑎𝑙 𝑒𝑣𝑎𝑙𝑢𝑎𝑟𝑙𝑎 𝑒𝑛 0 𝑝𝑒𝑟𝑡𝑒𝑛𝑒𝑐𝑒 𝑎𝑙 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜 𝑑𝑒𝑙 𝑣𝑎𝑐í𝑜.
𝐶𝑜𝑛 𝑒𝑠𝑡𝑜 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑞𝑢𝑒 𝑝𝑢𝑒𝑑𝑒𝑛 𝑠𝑒𝑟 𝑡𝑜𝑑𝑜𝑠 𝑙𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠 𝑠𝑖𝑒𝑚𝑝𝑟𝑒 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜
𝑙𝑎 𝑓𝑢𝑛𝑐𝑖𝑜𝑛 𝑠𝑒𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒 𝑑𝑒 𝑐𝑒𝑟𝑜.
𝑅𝑎𝑛𝑔𝑜: ] − ∞, 0[ 𝑈 ] 0, +∞[
ó
{𝑅𝑓(𝑥, 𝑦) ∈ 𝑹 / 𝑓(𝑥, 𝑦) ≠ 0}
𝑮𝒓𝒂𝒇𝒊𝒄𝒂: 𝑃𝑎𝑟𝑎 𝑒𝑙𝑙𝑜 𝑒𝑠 𝑛𝑒𝑐𝑒𝑠𝑎𝑟𝑖𝑜 𝑣𝑒𝑟 𝑙𝑎𝑠 𝑡𝑟𝑎𝑧𝑎𝑠 𝑑𝑒 𝑙𝑜𝑠 𝑒𝑗𝑒𝑠.
𝑇𝑟𝑎𝑧𝑎 𝑥𝑧 (𝑦 = 0)
𝑧=
1
𝑥2
𝑇𝑟𝑎𝑧𝑎 𝑦𝑧 (𝑥 = 0)
𝑇𝑟𝑎𝑧𝑎 𝑥𝑦 (𝑧 ≠ 0)
𝐶𝑜𝑚𝑜 𝑦𝑎 𝑠𝑒 𝑠𝑎𝑏𝑒 𝑞𝑢𝑒 𝑞𝑢𝑒𝑑𝑎 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎 𝑎 𝑐𝑒𝑟𝑜, 𝑠𝑒 𝑡𝑜𝑚𝑎𝑟𝑎𝑛 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 − 4 𝑦 4.
±4 =
1
𝑥2 + 𝑦
𝑆𝑖𝑒𝑛𝑑𝑜 𝑧 = 4
4(𝑥 2 + 𝑦) = 1 → 4𝑥 2 + 4𝑦 = 1
1 − 4𝑥 2
2
4𝑦 = 1 − 4𝑥 → 𝑦 =
4
𝑆𝑖𝑒𝑛𝑑𝑜 𝑧 = −4
−4(𝑥 2 + 𝑦) = 1 → −4𝑥 2 − 4𝑦 = 1
1 + 4𝑥 2
2
−4𝑦 = 1 + 4𝑥 → 𝑦 = − (
)
4
𝑇𝑒𝑛𝑖𝑒𝑛𝑑𝑜 𝑡𝑜𝑑𝑎𝑠 𝑙𝑎𝑠 𝑔𝑟𝑎𝑓𝑖𝑐𝑎𝑠 𝑒𝑛 𝑅 3 :
7) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑒𝑙 𝑑𝑜𝑚𝑖𝑛𝑖𝑜, 𝑟𝑎𝑛𝑔𝑜 𝑦 𝑔𝑟𝑎𝑓𝑖𝑞𝑢𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑓𝑢𝑛𝑐𝑖ó𝑛:
𝑓(𝑥, 𝑦) = 𝑠𝑒𝑛(𝑥 + 𝑦)
𝑫𝒐𝒎𝒊𝒏𝒊𝒐: 𝐶𝑜𝑚𝑜 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒𝑛 𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑎𝑟𝑎 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑐𝑖𝑜𝑛 𝑠𝑜𝑛 𝑡𝑜𝑑𝑜𝑠 𝑙𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠:
𝐷𝑓 = {(𝑥, 𝑦) ∈ 𝑹𝟐 }
𝑹𝒂𝒏𝒈𝒐: 𝐷𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒 𝑓(𝑥, 𝑦) 𝑒𝑠 𝑢𝑛𝑎 𝑓𝑢𝑛𝑐𝑖𝑜𝑛 𝑠𝑒𝑛𝑜, 𝑑𝑒𝑝𝑒𝑛𝑑𝑒 𝑑𝑒 𝑠𝑢 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑 𝑞𝑢𝑒 𝑒𝑠
𝑑𝑒 − 1 𝑎 1.
𝑅𝑎𝑛𝑔𝑜: [−1 , 1 ]
ó
{𝑅𝑓(𝑥, 𝑦) ∈ 𝑹/−1 ≤ 𝑓(𝑥, 𝑦) ≤ +1}
𝑮𝒓𝒂𝒇𝒊𝒄𝒂: 𝐸𝑠 𝑛𝑒𝑐𝑒𝑠𝑎𝑟𝑖𝑜 𝑐𝑜𝑛𝑜𝑐𝑒𝑟 𝑠𝑢𝑠 𝑡𝑟𝑎𝑧𝑎𝑠
𝑧 = 𝑠𝑒𝑛(𝑥 + 𝑦)
𝑇𝑟𝑎𝑧𝑎 𝑥𝑧 (𝑦 = 0)
𝑧 = 𝑠𝑒𝑛(𝑥)
𝑇𝑟𝑎𝑧𝑎 𝑦𝑧 (𝑥 = 0)
𝑧 = 𝑠𝑒𝑛(𝑦)
𝑇𝑟𝑎𝑧𝑎 𝑥𝑦 (𝑧 = 0)
0 = 𝑠𝑒𝑛(𝑥 + 𝑦) → 𝑎𝑟𝑐𝑠𝑒𝑛(0) = 𝑥 + 𝑦 → −𝑥 = 𝑦
𝐺𝑟𝑎𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑙𝑎𝑠 𝑡𝑟𝑎𝑧𝑎𝑠 𝑒𝑛 𝑒𝑙 𝑒𝑠𝑝𝑎𝑐𝑖𝑜:
8) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎𝑠 𝑝𝑟𝑖𝑚𝑒𝑟𝑎𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑝𝑎𝑟𝑐𝑖𝑎𝑙𝑒𝑠 𝑝𝑎𝑟𝑎
𝑓(𝑥, 𝑦) = 𝑒 2𝑥−𝑦 − 𝑙𝑛(𝑥 + 𝑦)
1
𝑥+𝑦
1
𝑓′𝑦 = −𝑒 2𝑥−𝑦 −
𝑥+𝑦
𝑓′𝑥 = 2𝑒 2𝑥−𝑦 −
9) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎𝑠 𝑠𝑒𝑔𝑢𝑛𝑑𝑎𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑝𝑎𝑟𝑐𝑖𝑎𝑙𝑒𝑠 𝑝𝑎𝑟𝑎
𝑦
𝑓(𝑥, 𝑦) = 𝑥²𝑦² − 𝐿𝑛 [𝑆𝑒𝑛 ( )]
𝑥
𝑓′𝑥 = 2𝑥𝑦 2 −