GABARITO Lista2 Funções_limites e derivadas

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FACULDADE DE TECNOLOGIA DE CURITIBA 
LISTA DE EXERCÍCIOS - 2 
 
1. Funções de várias variáveis 
a) Seja 𝑓(𝑥, 𝑦, 𝑧) = 3𝑥 + 2𝑦 + 4𝑧 
i. Determine o 𝐷𝑓 
𝐷𝑓 = {(𝑥, 𝑦, 𝑧) ∈ ℝ
3} 
ii. Calcule 𝑓(1, −1,2) 
𝑓(1, −1,2) = 3(1) + 2(−1) + 4(2) 
𝑓(1, −1,2) = 9 
 
iii. Calcule 𝑓(ℎ, 𝑥, 𝑥) 
𝑓(ℎ, 𝑥, 𝑥)) = 3(ℎ) + 2(𝑥) + 4(𝑥) 
𝑓(ℎ, 𝑥, 𝑥)) = 3ℎ + 6𝑥 
 
b) Seja 𝑓(𝑥, 𝑦) =
𝑥+𝑦
𝑥−𝑦
 
i. Determine o 𝐷𝑓 
𝐷𝑓 = {(𝑥, 𝑦, 𝑧) ∈ ℝ
2 ; 𝑥 ≠ 𝑦} 
 
ii. Calcule 𝑓(2,3) 
𝑓(2,3) =
2 + 3
2 − 3
=
5
−1
= −5 
iii. Calcule 𝑓(𝑎 + 𝑏, 𝑎 − 𝑏) 
𝑓(𝑎 + 𝑏, 𝑎 − 𝑏) =
(𝑎 + 𝑏) + (𝑎 − 𝑏)
(𝑎 + 𝑏) − (𝑎 − 𝑏)
 
𝑓(𝑎 + 𝑏, 𝑎 − 𝑏) =
𝑎 + 𝑏 + 𝑎 − 𝑏
𝑎 + 𝑏 − 𝑎 + 𝑏
 
𝑓(𝑎 + 𝑏, 𝑎 − 𝑏) =
2𝑎
2𝑏
=
𝑎
𝑏
 
 
c) Seja 𝑓(𝑥, 𝑦) = √1 − 𝑥2 − 𝑦2 
i. Determine o 𝐷𝑓 
𝐷𝑓 = {(𝑥, 𝑦) ∈ ℝ
2; 𝑥2 + 𝑦2 ≤ 1 } 
ii.Represente graficamente o 
domínio da função
 
iii. Calcule 𝑓(0,0) 
𝑓(0,0) = √1 − 0 − 0 = 1 
 
 
 
 
 
 
Curso 
Disciplina Cálculo Diferencial e Integral II Data: Nota: 
Aluno 
Professor Paula Fernanda Vieira paulafernandagv@gmail.com 
2. Mostre que os limites não existem: 
a) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
5𝑦−𝑥
2𝑥−𝑦
 
𝑥 = 0 
𝑙𝑖𝑚
𝑥=0
𝑦→0
5𝑦 − 𝑥
2𝑥 − 𝑦
= 𝑙𝑖𝑚
 𝑥=0
 𝑦→0
5𝑦 − 0
2.0 − 𝑦
= 𝑙𝑖𝑚
 𝑥=0
 𝑦→0
5𝑦
−𝑦 
= −5 
𝑦 = 0 
𝑙𝑖𝑚
𝑦=0
𝑥→0
5𝑦 − 𝑥
2𝑥 − 𝑦
= 𝑙𝑖𝑚
 𝑥=0
 𝑦→0
5.0 − 𝑥
2. 𝑥 − 0
= 𝑙𝑖𝑚
 𝑥=0
 𝑦→0
−𝑥
2𝑥 
=
−1
2 
 
Como os resultados são diferentes o limite dado não existe. 
 
b) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥2−𝑦2
𝑥2+𝑦2
 
 
c) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥𝑦
𝑥2+𝑦2
 
 
 
d) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
2𝑥
√𝑥2+𝑦2
 
 
e) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥− 𝑦
2𝑥−𝑦
 
. 
f) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
3𝑥𝑦
4𝑥2+5𝑦2
 
 
 
 
g) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥2−4𝑦2
𝑥2+𝑦2
 
 
h) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥3
𝑥3+𝑦2
 
 
i) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑦4+3𝑥2𝑦2+2𝑦𝑥3
(𝑥2+𝑦2)
2 
 
 
 
j) 𝑙𝑖𝑚
(𝑥,𝑦)→(1,0)
(𝑥−1)2𝑦
(𝑥−1)4+ 𝑦2
 
 
3. Calcular os seguintes limites envolvendo indeterminações: 
a) 𝑙𝑖𝑚
(𝑥,𝑦)→(2,1)
𝑥3+𝑥2𝑦−2𝑥𝑦−2𝑥2−2𝑥+4
𝑥𝑦+𝑥−2𝑦−2
 
 
 
 
b) 𝑙𝑖𝑚
(𝑥,𝑦)→(2,3)
𝑥2𝑦−3𝑥2−4𝑥𝑦 +12𝑥+4𝑦−12
𝑥𝑦−3𝑥−2𝑦+6
 
 
 
 
4. Calcular os seguintes limites: 
a) 𝑙𝑖𝑚
(𝑥,𝑦)→(1,2)
𝑒𝑥𝑦 − 𝑒𝑦 + 1 
𝑙𝑖𝑚
(𝑥,𝑦)→(1,2)
𝑒1.2 − 𝑒2 + 1 
𝑙𝑖𝑚
(𝑥,𝑦)→(1,2)
1 = 1 
 
b) 𝑙𝑖𝑚
(𝑥,𝑦)→(1,2)
2𝑥𝑦 + 𝑥2 −
𝑥
𝑦
= 2.1.2 + 12 −
1
2
=
9
2
 
c) 𝑙𝑖𝑚
(𝑥,𝑦)→(2,1)
𝑥+𝑦−2
𝑥2+𝑦2
=
2+1−2
22+12
=
1
5
 
d) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,0)
𝑥√ 𝑥2 + 𝑦2 =0 
e) 𝑙𝑖𝑚
(𝑥,𝑦)→(−1,2)
𝑥3𝑦3 + 2𝑥𝑦2 + 𝑦 = −8 − 8 + 2 = −14 
f) 𝑙𝑖𝑚
(𝑥,𝑦)→(1,2)
3𝑥2𝑦 + 2𝑥𝑦2 − 2𝑥𝑦 = 6 + 8 − 4 = 10 
g) 𝑙𝑖𝑚
(𝑥,𝑦)→(2,1)
𝑥𝑦2−5𝑥+8
𝑥2+𝑦2+4𝑥𝑦
=
−2+10+8
4+1−8
= −
16
3
 
h) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,1)
𝑥2−3𝑥𝑦2
𝑥2+𝑦2
=
0
1
= 0 
i) 𝑙𝑖𝑚
(𝑥,𝑦)→(0,1)
𝑥2−𝑥𝑦
𝑥2−𝑦2
=
𝑥(𝑥−𝑦)
(𝑥−𝑦)(𝑥+𝑦)
= 
𝑥
(𝑥+𝑦)
=
𝑥
(𝑥+𝑦)
=
0
1
= 0 
 
 
5. Esboce os gráficos das funções: 
Obs.: Utilizar software : https://www.geogebra.org/3d?lang=pt 
 
a) 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑦² 
 
b) 𝑓(𝑥, 𝑦) = 3𝑦 + 2𝑥 
 
 
 
 
 
 
 
 
 
 
https://www.geogebra.org/3d?lang=pt
c) 𝑓(𝑥, 𝑦) = 10 − 𝑥² − 𝑦² 
 
d) 𝑓(𝑥, 𝑦) = √𝑥² + 𝑦² 
 
 
 
6. Calcular as derivadas parciais de 1a ordem, usando a definição: 
 
a) 𝑓(𝑥, 𝑦) = 2𝑥2𝑦 + 3𝑥𝑦2 − 4 
𝜕𝑧
𝜕𝑥
= lim
∆𝑥→0
(𝑓(𝑥+∆𝑥,𝑦)−𝑓(𝑥,𝑦)
∆𝑥
 
lim
∆𝑥→0
2(𝑥 + ∆𝑥)2𝑦 + 3(𝑥 + ∆𝑥)𝑦2 − 4 − (2𝑥2𝑦 + 3𝑥𝑦2 − 4)
∆𝑥
 
lim
∆𝑥→0
2(𝑥2 + 2𝑥∆𝑥 + ∆𝑥2)𝑦 + 3(𝑥 + ∆𝑥)𝑦2 − 4 − (2𝑥2𝑦 + 3𝑥𝑦2 − 4)
∆𝑥
 
lim
∆𝑥→0
2𝑥2𝑦 + 4𝑥∆𝑥𝑦 + 2∆𝑥2𝑦 + 3𝑥𝑦2 + 3∆𝑥𝑦2 − 4 − 2𝑥2𝑦 − 3𝑥𝑦2 + 4)
∆𝑥
 
lim
∆𝑥→0
2𝑥2𝑦 + 4𝑥∆𝑥𝑦 + 2∆𝑥2𝑦 + 3𝑥𝑦2 + 3∆𝑥𝑦2 − 4 − 2𝑥2𝑦 − 3𝑥𝑦2 + 4)
∆𝑥
 
lim
∆𝑥→0
4𝑥∆𝑥𝑦 + 2∆𝑥2𝑦 + 3∆𝑥𝑦2 
∆𝑥
= lim
∆𝑥→0
∆𝑥( 4𝑥𝑦 + 2∆𝑥𝑦 + 3𝑦2 )
∆𝑥
 
lim
∆𝑥→0
 4𝑥𝑦 + 2∆𝑥𝑦 + 3𝑦2 = 4𝑥𝑦 + 3𝑦2 
𝜕𝑧
𝜕𝑥
= 4𝑥𝑦 + 3𝑦2 
 
 
𝜕𝑧
𝜕𝑥
= lim
∆𝑦→0
(𝑓(𝑥,𝑦+∆𝑦)−𝑓(𝑥,𝑦)
∆𝑦
 
lim
∆𝑦→0
2𝑥2(𝑦 + ∆𝑦) + 3𝑥(𝑦 + ∆𝑦)2 − 4 − (2𝑥2𝑦 + 3𝑥𝑦2 − 4)
∆𝑦
 
lim
∆𝑦→0
2𝑥2𝑦 + 2𝑥2∆𝑦 + 3𝑥(𝑦2 + 2𝑦∆𝑦 + ∆𝑦2 ) − 4 − (2𝑥2𝑦 + 3𝑥𝑦2 − 4)
∆𝑦
 
lim
∆𝑦→0
2𝑥2𝑦 + 2𝑥2∆𝑦 + 3𝑥𝑦2 + 6𝑥𝑦∆𝑦 + 3𝑥∆𝑦2 − 4 − 2𝑥2𝑦 − 3𝑥𝑦2 + 4)
∆𝑦
 
lim
∆𝑦→0
2𝑥2∆𝑦 +6𝑥𝑦∆𝑦+3𝑥∆𝑦2 
∆𝑦
= lim
∆𝑦→0
 
∆𝑦(2𝑥2 +6𝑥𝑦+3𝑥∆𝑦) 
∆𝑦
 
lim
∆𝑦→0
 2𝑥2 + 6𝑥𝑦 + 3𝑥∆𝑦 = 2𝑥2 + 6𝑥𝑦 
𝜕𝑧
𝜕𝑦
= 2𝑥2 + 6𝑥𝑦 
 
 
 
 
 
 
 
 
b) 𝑓(𝑥, 𝑦) = 5𝑥𝑦 − 𝑥2 
 
 
7. Calcular as derivadas parciais de 1a ordem: 
 
a) 𝑓(𝑥, 𝑦) = 8𝑥3𝑦 + 2𝑦2 + 4𝑥2 
𝜕𝑧
𝜕𝑥
= 24𝑥2 𝑦 + 8𝑥 
𝜕𝑧
𝜕𝑦
= 8𝑥3 + 4𝑦 
b) 𝑓(𝑥, 𝑦) = 2𝑥5𝑦 + 9𝑥𝑦3 − 90 
𝜕𝑧
𝜕𝑥
= 10𝑥4 𝑦 + 9𝑦3 
𝜕𝑧
𝜕𝑦
= 2𝑥5 + 27𝑥𝑦2 
 
c) 𝑓(𝑥, 𝑦) = 20𝑥𝑦 − 𝑥4 
𝜕𝑧
𝜕𝑥
= 20𝑦 − 4𝑥3 
𝜕𝑧
𝜕𝑦
= 20𝑥 
 
d) 𝑓(𝑥, 𝑦) =
𝑥2𝑦
2𝑦2+ 𝑥2
 (usar regra do quociente) 
 
e) 𝑓(𝑥, 𝑦) = 𝑒𝑥
2+𝑦2+4 
 
 
 
 
f) 𝑓(𝑥, 𝑦, 𝑧) =
𝑥𝑦𝑧
√𝑥2+𝑦2+𝑧2
 
 
g) ℎ(𝑢, 𝑣, 𝑤, 𝑡) = 𝑢2 + 𝑣2 − ln (𝑤𝑡)