Trabajo 5_Asqui B_8076 - Parvs BJ

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ESCUELA SUPERIOR POLITÉCNICA DE CHIMBORAZO 
 
 
 
 
 
 
 
 
 
 
 
 
PRIMER SEMESTRE 
PARALELO ¨A¨ 
ANÁLISIS MATEMÁTICO I 
 
BORIS JOSUE ASQUI VACA 
2020-2021
 
 
 
 
 
 
 
𝐸𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜𝑠 𝑑𝑒𝑙 1 𝑎𝑙 12, 𝑐𝑎𝑙𝑐𝑢𝑙𝑒 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 
1. 𝑓(𝑥) = (2𝑥 + 1)3 
𝑓′(𝑥) = 2 ∗ 3(2𝑥 + 1)2 
𝑓′(𝑥) = 6(2𝑥 + 1)2 
𝑓′(𝑥) = 24𝑥2 + 24𝑥 + 6 
 
2. 𝑓(𝑥) = (10 − 5𝑥)4 
𝑓′(𝑥) = 4 ∗ (−5)(10 − 5𝑥)3 
𝑓′(𝑥) = −20(10 − 5𝑥)3 
 
3. 𝐹(𝑥) = (𝑥2 + 4𝑥 − 5)4 
𝑓′(𝑥) = 4(2𝑥 + 4)(𝑥2 + 4𝑥 − 5)3 
𝑓′(𝑥) = (8𝑥 + 16)(𝑥2 + 4𝑥 − 5)3 
 
4. 𝑔(𝑟) = (2𝑟4 + 8𝑟2 + 1)5 
𝑔′(𝑟) = 5(8𝑟3 + 16𝑟)(2𝑟4 + 8𝑟2 + 1)4 
𝑔′(𝑟) = (40𝑟3 + 80𝑟)(2𝑟4 + 8𝑟2 + 1)4 
 
5. 𝑓(𝑡) = (2𝑡4 − 7𝑡3 + 2𝑡 − 1)2 
𝑓′(𝑡) = 2(8𝑡3 − 21𝑡2 + 2)(2𝑡4 − 7𝑡3 + 2𝑡 − 1) 
𝑓′(𝑡) = (16𝑡3 − 42𝑡2 + 4)(2𝑡4 − 7𝑡3 + 2𝑡 − 1) 
 
6. 𝐻(𝑧) = (𝑧3 − 3𝑧2 + 1)−3 
𝐻′(𝑧) = −3(3𝑧2 − 6𝑧)(𝑧3 − 3𝑧2 + 1)−4 
𝐻′(𝑧) = −
3(3𝑧2 − 6𝑧)
(𝑧3 − 3𝑧2 + 1)4
 
 
7. 𝑓(𝑥) = (𝑥2 + 4)−2 
𝑓′(𝑥) = −2(2𝑥)(𝑥2 + 4)−3 
 
 
 
 
 
 
𝑓′(𝑥) =
−4𝑥
(𝑥2 + 4)3
 
 
8. 𝑔(𝑥) = sin 𝑥2 
𝑔′(𝑥) = cos 𝑥2 (2𝑥) 
 
9. 𝑓(𝑥) = 4 cos 3𝑥 − 3 sin 4𝑥 
𝑓′(𝑥) = 4 ∗ 3(− sin 3𝑥) − 3 ∗ 4(cos 4𝑥) 
𝑓′(𝑥) = −12 sin 3𝑥 − 12 cos 4𝑥 
 
10. 𝐺(𝑥) = sec2 𝑥 
𝐺′(𝑥) = 2 sec 𝑥 ∗ (sec 𝑥 ∗ tan 𝑥) 
𝐺′(𝑥) = 2 sec2 𝑥 ∗ tan 𝑥 
 
11. ℎ(𝑡) =
1
3
sec3 2𝑡 − sec 2𝑡 
ℎ′(𝑡) =
1
3
(3) sec2(2𝑡) ∗ (sec 2𝑡 tan 2𝑡) ∗ (2) − 2 ∗ sec 2𝑡 tan 2𝑡 
ℎ′(𝑡) = 2 ∗ sec3(2𝑡) ∗ tan 2𝑡 − 2 ∗ sec 2𝑡 tan 2𝑡 
 
12. 𝑓(𝑥) = cos(3𝑥2 + 1) 
𝑓′(𝑥) = − sin(3𝑥2 + 1) (6𝑥) 
𝐸𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜𝑠 13 𝑎 16, 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 
13.
𝑑
𝑑𝑥
(sec2 𝑥 tan2 𝑥). 
𝑑
𝑑𝑥
(sec2 𝑥 tan2 𝑥) = tan2 𝑥
𝑑
𝑑𝑥
sec2 𝑥 + sec2 𝑥
𝑑
𝑑𝑥
tan2 𝑥 
𝑑
𝑑𝑥
(sec2 𝑥 tan2 𝑥) = tan2 𝑥 (2 sec2 𝑥 tan 𝑥) + sec2 𝑥 (2 tan 𝑥 sec2 𝑥) 
𝑑
𝑑𝑥
(sec2 𝑥 tan2 𝑥) = 2 sec2 𝑥 tan3 𝑥 + 2 tan 𝑥 sec4 𝑥 
 
 
 
 
 
14.
𝑑
𝑑𝑡
(2 sin3 𝑡 cos2 𝑡) 
2
𝑑
𝑑𝑡
(sin3 𝑡 cos2 𝑡) = cos2 𝑡
𝑑
𝑑𝑡
sin3 𝑡 + sin3 𝑡
𝑑
𝑑𝑡
cos2 𝑡 
𝑑
𝑑𝑡
(2 sin3 𝑡 cos2 𝑡) = cos2 𝑡 (3 sin2 𝑡 cos 𝑡) + sin3 𝑡 (−2 cos 𝑡 sin 𝑡) 
𝑑
𝑑𝑡
(2 sin3 𝑡 cos2 𝑡) = 6 sin2 𝑡 cos3 𝑡 − 4 cos 𝑡 sin4 𝑡 
 
15.
𝑑
𝑑𝑥
(cot4 𝑡 − csc4 𝑡) 
𝑑
𝑑𝑥
(cot4 𝑡 − csc4 𝑡) =
𝑑
𝑑𝑥
cot4 𝑡 −
𝑑
𝑑𝑥
csc4 𝑡 
𝑑
𝑑𝑥
(cot4 𝑡 − csc4 𝑡) = 4 cot3 𝑡 (− csc2 𝑡) − 4 csc3 𝑡 (− csc 𝑡 cot 𝑡) 
𝑑
𝑑𝑥
(cot4 𝑡 − csc4 𝑡) = 4 csc4 𝑡 cot 𝑡 − 4 cot3 𝑡 csc2 𝑡 
 
16.
𝑑
𝑑𝑥
[(4𝑥2 + 7)2(2𝑥3 + 1)4] 
𝑑
𝑑𝑥
[(4𝑥2 + 7)2(2𝑥3 + 1)4] = (2𝑥3 + 1)4
𝑑
𝑑𝑥
 (4𝑥2 + 7)2 + (4𝑥2 + 7)2
𝑑
𝑑𝑥
(2𝑥3 + 1)4 
𝑑
𝑑𝑥
[(4𝑥2 + 7)2(2𝑥3 + 1)4] = 2(2𝑥3 + 1)4(4𝑥2 + 7)(8𝑥) + 4(4𝑥2 + 7)2(2𝑥3 + 1)3(6𝑥2) 
𝑑
𝑑𝑥
[(4𝑥2 + 7)2(2𝑥3 + 1)4] = 16𝑥(2𝑥3 + 1)4(4𝑥2 + 7) + 24𝑥2(4𝑥2 + 7)2(2𝑥3 + 1)3 
 
 
𝐸𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜𝑠 17 𝑎 24, 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑦 𝑎𝑝𝑜𝑦𝑒 𝑠𝑢 𝑟𝑒𝑠𝑝𝑢𝑒𝑠𝑡𝑎 𝑡𝑟𝑎𝑧𝑎𝑛𝑑𝑜 𝑙𝑎𝑠 𝑔𝑟á𝑓𝑖𝑐𝑎𝑠 
𝑑𝑒 𝑠𝑢 𝑟𝑒𝑠𝑝𝑢𝑒𝑠𝑡𝑎 𝑦 𝑑𝑒 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑛ú𝑚𝑒𝑟𝑖𝑐𝑎 𝑒𝑛 𝑥 𝑒𝑛 𝑒𝑙 𝑚𝑖𝑠𝑚𝑜 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑜 𝑑𝑒 𝑖𝑛𝑠𝑝𝑒𝑐𝑐𝑖ó𝑛 
17. 𝑓(𝑥) = (
𝑥 − 7
𝑥 + 2
)
2
 
𝑓′(𝑥) = 2 (
𝑥 − 7
𝑥 + 2
) (
𝑥 + 2 − 𝑥 + 7
(𝑥 + 2)2
) 
𝑓′(𝑥) = 18 (
𝑥 − 7
(𝑥 + 2)3
) 
 
 
 
18. 𝑓(𝑡) = (
3𝑡2 + 1
3𝑡3 + 1
)
2
 
𝑓′(𝑡) = 2 (
3𝑡2 + 1
3𝑡3 + 1
) (
6𝑡(3𝑡3 + 1) − 9𝑡2(3𝑡2 + 1)
(3𝑡3 + 1)2
) 
𝑓′(𝑡) = 2(−9𝑡4 − 9𝑡2 + 6𝑡) (
3𝑡2 + 1
(3𝑡3 + 1)3
) 
 
 
19. 𝑔(𝑡) = sin2(3𝑡2 − 1) 
𝑔′(𝑡) = 12𝑡 sin(3𝑡2 − 1) cos(3𝑡2 − 1) 
𝑔′(𝑡) = 6𝑡 sin 2(3𝑡2 − 1) 
 
 
 
 
20. 𝑔(𝑥) = tan2 𝑥2 
𝑔′(𝑥) = 4𝑥 tan 𝑥2 sec2 𝑥2 
 
21. 𝑓(𝑥) = (tan2 𝑥 − 𝑥2)3 
𝑓′(𝑥) = 3(2 tan 𝑥 sec2 𝑥 − 2𝑥)(tan2 𝑥 − 𝑥2)2 
 
22. 𝐺(𝑥) = (2 sin 𝑥 − 3 cos 𝑥)3 
𝐺′(𝑥) = 3(2 cos 𝑥 + 3 sin 𝑥)(2 sin 𝑥 − 3 cos 𝑥)2 
 
 
 
 
23. 𝐹(𝑥) = 4 cos(sin 3𝑥) 
𝐹′(𝑥) = −12 sin(sin 3𝑥) cos 3𝑥 
 
24. 𝑓(𝑥) = sin2(cos 2𝑥) 
𝑓′(𝑥) = 2 sin(cos 2𝑥) cos(cos 2𝑥) (− sin 2𝑥)2 
𝑓′(𝑥) = −2 sin 2(cos 2𝑥) sin 2𝑥 
 
 
 
 
𝐸𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜𝑠 25 𝑦 26, 𝑜𝑏𝑡𝑒𝑛𝑔𝑎 𝑢𝑛𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑟𝑒𝑐𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑎 𝑙𝑎 𝑐𝑢𝑟𝑣𝑎 𝑑𝑎𝑑𝑎 𝑒𝑛 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑖𝑛𝑑𝑖𝑐𝑎𝑑𝑜 
𝑎𝑝𝑜𝑦𝑒 𝑠𝑢𝑠 𝑟𝑒𝑠𝑝𝑢𝑒𝑠𝑡𝑎𝑠 𝑡𝑟𝑎𝑧𝑎𝑛𝑑𝑜 𝑙𝑎 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑑𝑒 𝑙𝑎 𝑐𝑢𝑟𝑣𝑎 𝑦 𝑙𝑎 𝑟𝑒𝑐𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒𝑛 𝑒𝑙 𝑚𝑖𝑠𝑚𝑜 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑜 𝑑𝑒 
𝑖𝑛𝑠𝑝𝑒𝑐𝑐𝑖ó𝑛 
25. 𝑦 = (𝑥2 − 1)2 𝑒𝑛 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 (2, 9) 
𝑦′ = 2(2𝑥)(𝑥2 − 1) 
 
 
𝑦′ = 4𝑥(𝑥2 − 1 
𝑅𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 ) 
𝑦′ = 4(2)(22 − 1) 
𝑦′ = 24 
𝐸𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑟𝑒𝑐𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 
𝑦 − 9 = 24(𝑥 − 2) 
 
26. 𝑦 = 2 tan 2𝑥 𝑒𝑛 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 (
1
8
𝜋, 4) 
𝑦′ = 4 sec2 2𝑥 
𝑅𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 ) 
𝑦′ = 4 sec2 45 
𝑦′ = 8 
𝐸𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑟𝑒𝑐𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 
𝑦 − 4 = 8 (𝑥 −
1
8
𝜋) 
 
𝐸𝑛 𝑙𝑜𝑠 𝑒𝑗𝑒𝑟𝑖𝑐𝑖𝑜𝑠 27 𝑎 30, 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛, 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒 𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑛𝑒𝑡𝑜 𝑑𝑒 𝑢𝑛 𝑐𝑢𝑒𝑟𝑝𝑜 𝑠𝑢𝑠𝑝𝑒𝑛𝑑𝑖𝑑𝑜 𝑑𝑒 𝑢𝑛 𝑟𝑒𝑠𝑜𝑟𝑡𝑒 
𝑞𝑢𝑒 𝑣𝑖𝑏𝑟𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑚𝑒𝑛𝑡𝑒 𝑑𝑜𝑛𝑑𝑒 𝑠 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑟𝑜𝑠 𝑒𝑠 𝑙𝑎 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑑𝑒𝑙 𝑐𝑢𝑒𝑟𝑝𝑜 𝑑𝑒𝑠𝑑𝑒 𝑠𝑢 𝑝𝑜𝑠𝑖𝑐𝑖ó𝑛 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 (𝑒𝑙 𝑜𝑟𝑖𝑔𝑒𝑛) 
𝑎 𝑙𝑜𝑠 𝑡 𝑠𝑒𝑛𝑔𝑢𝑛𝑑𝑜𝑠 𝑦 𝑒𝑙 𝑠𝑒𝑛𝑡𝑖𝑑𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑜 𝑒𝑠 ℎ𝑎𝑐𝑖𝑎 𝑎𝑟𝑟𝑖𝑏𝑎. 
𝑎)𝐶𝑎𝑙𝑐𝑢𝑙𝑒 𝑙𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑 𝑦 𝑙𝑎 𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑒𝑛𝑡𝑜 𝑝𝑎𝑟𝑎 𝑐𝑢𝑎𝑙𝑞𝑢𝑖𝑒𝑟 𝑡 
𝑏)𝑀𝑢𝑒𝑠𝑡𝑟𝑒 𝑞𝑢𝑒 𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑒𝑛𝑡𝑜 𝑒𝑠 𝑎𝑟𝑚ó𝑛𝑖𝑐𝑜 𝑠𝑖𝑚𝑝𝑙𝑒 
𝑐)𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑, 𝑒𝑙 𝑝𝑒𝑟𝑖𝑜𝑑𝑜 𝑦 𝑙𝑎 𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑒𝑛𝑡𝑜 
𝑑)𝑆𝑖𝑚𝑢𝑙𝑒 𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑒𝑛𝑡𝑜 ℎ𝑎𝑐𝑖𝑎 𝑎𝑟𝑟𝑖𝑏𝑎 𝑦 ℎ𝑎𝑐𝑖𝑎 𝑎𝑏𝑎𝑗𝑜 𝑑𝑒𝑙 𝑟𝑒𝑠𝑜𝑟𝑡𝑒 𝑒𝑛 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎𝑑𝑜𝑟𝑎 
𝑒)𝑇𝑟𝑎𝑐𝑒 𝑙𝑎 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑚𝑜𝑣𝑖𝑚𝑖𝑒𝑛𝑡𝑜 
27. 𝑠 = 6 sin (
1
4
𝜋𝑡) 
𝑎) 
𝑣 =
6
4
𝜋 cos (
1
4
𝜋𝑡) 
𝑎 = −
3
8
𝜋2 sin (
1
4
𝜋𝑡) 
 
𝑏)La aceleración es proporcional al desplazamiento y de signo contrario; 
𝑎(𝑡) = −
3
8
𝜋2 sin (
1
4
𝜋𝑡) → −
1
16
𝜋2 (6 sin (
1
4
𝜋𝑡)) = −
1
16
𝜋2 
𝑐) 
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑 = 6 
𝑃𝑒𝑟𝑖𝑜𝑑𝑜 
1
4
𝜋𝑝 = 2𝜋 → 𝑃 = 8 
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 
1
𝑃
=
1
8
→ 𝑂𝑠𝑐𝑖𝑙𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑜𝑟 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 
 
28. 𝑠 = (3 cos
1
6
𝜋𝑡) 
𝑎) 
𝑣 = −
1
2
𝜋 sin (
1
6
𝜋𝑡) 
𝑎 = −
1
12
𝜋2 cos (
1
6
𝜋𝑡) 
𝑏)La aceleración es proporcional al desplazamiento y de signo contrario; 
𝑎(𝑡) = −
1
12
𝜋2 cos (
1
6
𝜋𝑡) → −
1
36
𝜋2 (3 cos (
1
6
𝜋𝑡)) = −
1
36
𝜋2 
𝑐) 
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑 = 3 
𝑃𝑒𝑟𝑖𝑜𝑑𝑜 
1
6
𝜋𝑝 = 2𝜋 → 𝑃 = 12 
 
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 
1
𝑃
=
1
12
→ 𝑂𝑠𝑐𝑖𝑙𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑜𝑟 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 
 
29. 𝑠 = 4 cos π (2𝑡 −
1
3
) → 𝑠 = 4 cos (2πt −
π
3
) 
𝑎) 
𝑣 = −4(2𝜋) sin (2πt −
π
3
) 
𝑎 = −4(2𝜋)2 cos (2πt −
π
3
) 
𝑏)La aceleración es proporcional al desplazamiento y de signo contrario; 
𝑎(𝑡) = −4(2𝜋)2 cos (2πt −
π
3
) → −(2𝜋)2 (4 cos (2πt −
π
3
)) = −(2𝜋)2 
𝑐) 
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑 = 4 
𝑃𝑒𝑟𝑖𝑜𝑑𝑜 
2πP −
π
3
= 2𝜋 → 6𝜋𝑝 − 𝜋 = 6𝜋 → 𝑝 =
7
6
 
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 
1
𝑃
=
1
7
6
=
6
7
→ 𝑂𝑠𝑐𝑖𝑙𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑜𝑟 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 
 
 
30. 𝑠 = 8 sin 𝜋 (3𝑡 +
1
2
) 
𝑎) 
𝑣 = 8(3𝜋) cos (𝜋 (3𝑡 +
1
2
)) 
𝑎 = −8(3𝜋)2 sin (𝜋 (3𝑡 +
1
2
)) 
𝑏)La aceleración es proporcional al desplazamiento y de signo contrario; 
𝑎(𝑡) = −8(3𝜋)2 sin (𝜋 (3𝑡 +
1
2
)) → −(3𝜋)2 (8 sin (𝜋 (3𝑡 +
1
2
))) = −(3𝜋)2 
𝑐) 
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑 = 8 
𝑃𝑒𝑟𝑖𝑜𝑑𝑜 
3𝜋𝑃 +
𝜋
2
= 2𝜋 → 6𝜋𝑃 + 𝜋 = 4𝜋 → 𝑃 =
1
2
 
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 
𝑓 =
1
𝑃
→
1
1
2
→ 2