Trabajo 10_Asqui B_8076 - Parvs BJ

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ESCUELA SUPERIOR POLITÉCNICA DE CHIMBORAZO 
 
 
 
 
 
 
 
 
 
 
 
 
PRIMER SEMESTRE 
PARALELO ¨B¨ 
ANÁLISIS MATEMÁTICO I 
 
BORIS JOSUE ASQUI VACA 
2020-2021
 
 
 
𝐷𝑒𝑟𝑖𝑣𝑎 𝑙𝑎𝑠 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒𝑠 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐𝑎𝑠. 
8. {
𝑥 = cos
𝜃
2
 
𝑦 = sin 2𝜃
; −
𝜋
2
≤ 𝜃 <
𝜋
2
 
𝑑𝑥
𝑑𝜃
= −
1
2
sin
𝜃
2
; 
𝑑𝑦
𝑑𝜃
= 2 cos 2𝜃 
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
2 cos 2𝜃
−
1
2
sin 𝜃
 
𝑑𝑦
𝑑𝑥
=
4 cos 2𝜃
− sin 𝜃
 
 
9. {
𝑥 = 5𝑡2 
𝑦 =
4
𝑡2
; −3 ≤ 𝑡 < 3 
𝑑𝑥
𝑑𝑡
= 10𝑡; 
𝑑𝑦
𝑑𝑡
=
−8
𝑡3
 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
−8
𝑡3
10𝑡
 
𝑑𝑦
𝑑𝑥
=
−4
5𝑡4
 
 
10. {
𝑥 = 3 + 2 tan 𝜃
𝑦 = 4 + csc 𝜃
; 0 ≤ 𝜃 < 2𝜋 
𝑑𝑥
𝑑𝜃
= 2 sec2 𝜃 ; 
𝑑𝑦
𝑑𝜃
= − csc 𝜃 cot 𝜃 
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
− csc 𝜃 cot 𝜃
2 sec2 𝜃
 
𝑑𝑦
𝑑𝑥
− csc 𝜃 cot 𝜃
2 sec2 𝜃
 
 
 
 
 
 
 
 
𝐸𝑛 𝑙𝑎𝑠 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑎𝑟𝑎𝑚é𝑡𝑟𝑖𝑐𝑎𝑠 𝑜𝑏𝑡é𝑛 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑙𝑎 𝑝𝑒𝑛𝑑𝑖𝑒𝑛𝑡𝑒 𝑒𝑛 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑖𝑛𝑑𝑖𝑐𝑎𝑑𝑜: 
11. {
𝑥 = 1 + sin 𝑡
𝑦 = 1 − cos 𝑡
; 0 ≤ 𝑡 ≤ 2𝜋; 𝑡 =
𝜋
3
 
𝑑𝑥
𝑑𝑡
= cos 𝑡 ; 
𝑑𝑦
𝑑𝑡
= sin 𝑡 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
sin 𝑡
cos 𝑡
 
𝑑𝑦
𝑑𝑥
=
sin 𝑡
cos 𝑡
→ 𝑚 = tan 𝑡 
𝑚 = tan
𝜋
3
→ √3 
 
12. {
𝑥 = 𝑚𝑡2 + 𝑏
𝑦 = 𝑛𝑡 + 𝑎
; 𝑚 ≤ 𝑡 ≤ 2𝑛; 𝑡 = 2𝑚𝑛 
𝑑𝑥
𝑑𝑡
= 2𝑚𝑡; 
𝑑𝑦
𝑑𝑡
= 𝑛 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
𝑛
2𝑚𝑡
 
𝑑𝑦
𝑑𝑥
=
𝑛
2𝑚𝑡
→
𝑑𝑦
𝑑𝑥
=
𝑛
2𝑚𝑡
 
𝑑𝑦
𝑑𝑥
=
𝑛
2𝑚2𝑛
=
1
2𝑚2
 
 
13. {
𝑥 = 𝑏(2 − 3𝑡)
𝑦 = 𝑎𝑡2
; 3𝑎 ≤ 𝑡 ≤ 2𝑏; 𝑡 =
𝑏
2𝑎
 
𝑑𝑥
𝑑𝑡
= −3𝑏; 
𝑑𝑦
𝑑𝑡
= 2𝑎𝑡 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
2𝑎𝑡
−3𝑏
 
𝑑𝑦
𝑑𝑥
=
2𝑎𝑡
−3𝑏
→ 𝑚 =
2𝑎𝑡
−3𝑏
 
𝑚 =
2𝑎 (
𝑏
2𝑎
)
−3𝑏
→ −
1
3
 
 
 
 
 
14. {
𝑥 = 𝑏(2 − 3𝑡)
𝑦 = 𝑎𝑡2
; 3𝑎 ≤ 𝑡 ≤ 2𝑏; 𝑡 =
𝑏
2𝑎
 
𝑑𝑥
𝑑𝑡
= −3𝑏; 
𝑑𝑦
𝑑𝑡
= 2𝑎𝑡 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
2𝑎𝑡
−3𝑏
 
𝑑𝑦
𝑑𝑥
=
2𝑎𝑡
−3𝑏
→ 𝑚 =
2𝑎𝑡
−3𝑏
 
𝑚 =
2𝑎 (
𝑏
2𝑎
)
−3𝑏
→ −
1
3
 
 
15. {
𝑥 = 2 cot2 𝜃
𝑦 = 4 cot 𝜃
; 0 ≤ 𝜃 ≤ 2𝜋; 𝜃 =
𝜋
4
 
𝑑𝑥
𝑑𝜃
= −4 cot 𝜃 csc2 𝜃 
𝑑𝑦
𝑑𝜃
= −4 csc2 𝜃 
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
−4 csc2 𝜃
−4 cot 𝜃 csc2 𝜃
; 𝑚 =
1
cot 𝜃
 
𝑚 =
1
cot
𝜋
4
= 1 
 
16. {
𝑥 = 5𝑡2
𝑦 = 𝑡 − 𝑡2
; −3 < 𝑡 < 3; 𝑡 = 1 
𝑑𝑥
𝑑𝑡
= 10𝑡 ; 
𝑑𝑦
𝑑𝑡
= 1 − 2𝑡 
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
1 − 2𝑡
10𝑡
 
𝑑𝑦
𝑑𝑥
=
1 − 2𝑡
10𝑡
→ 𝑚 =
1 − 2𝑡
10𝑡
 
𝑚 =
1 − 2(1)
10(1)
→ −
1
10