EngComp e Prod 2018 1 Calculo II MCA502 P3 GABARITO
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EngComp e Prod 2018 1 Calculo II MCA502 P3 GABARITO


Disciplina<strong>cálculo</strong>36 materiais5 seguidores
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AVALIAÇÃO PRESENCIAL 
CADERNO DE PERGUNTAS 
 
curso: 
Engenharia de Computação/ 
Engenharia de Produção 
bimestre: 4º bimestre ano: 2018 | 2sem 
CÓDIGO DA PROVA 
P3 
 
\uf0b7 Preencha atentamente o cabeçalho de TODAS AS FOLHAS DE RESPOSTA que você utilizar. 
\uf0b7 Ao término da prova, entregue apenas a folha de resposta ao aplicador. Leve este caderno de 
perguntas consigo. 
Boa prova! 
 
disciplina: MCA502 \u2013 Cálculo II 
 
\uf0b7 É permitido o uso de formulário impresso. Não é permitido o uso de calculadora. 
 
Questão 1 (2 pontos) 
Sobre a função \ud835\udc53(\ud835\udc65, \ud835\udc66) = \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc65\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 definida em ]\u2212
\ud835\udf0b
2
,
\ud835\udf0b
2
[ \ud835\udc4b ]\u2212
\ud835\udf0b
2
,
\ud835\udf0b
2
[ é correto afirmar: 
 
a) (0,0) é seu único ponto crítico e é um ponto de máximo local. 
b) (0,0) é seu único ponto crítico e é um ponto de mínimo local. 
c) (0,0) é seu único ponto crítico e não é nem ponto de máximo nem de mínimo local. 
d) \ud835\udc53 não possui pontos críticos no domínio definido no enunciado. 
e) nenhuma das outras alternativas. 
 
Questão 2 (2 pontos) 
Calcule a massa da placa plana descrita por \ud835\udc652 + \ud835\udc662 \u2264 4, \ud835\udc66 \u2265 0, \ud835\udc65 \u2265 0, com densidade \ud835\udeff = \ud835\udeff(\ud835\udc65, \ud835\udc66) = \ud835\udc652 + \ud835\udc662. 
 
a) \ud835\udf0b 
b) 
2\ud835\udf0b
3
 
c) 
\ud835\udf0b
2
 
d) 2\ud835\udf0b 
e) Nenhuma das outras alternativas. 
 
 
Questão 3 (3 pontos) 
Calcule \u222b 3\ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66\ud835\udc51\ud835\udc65 \u2212 \ud835\udc523\ud835\udc65 \ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66\ud835\udc51\ud835\udc66 \ud835\udefe , sendo \ud835\udefe a curva esboçada 
abaixo, ligando o ponto (-3, 0 ) ao ponto (3, 0). 
 
 
 
 
 
Questão 4 (3 pontos) 
Calcule a massa do pedaço do cilindro \ud835\udc652 + \ud835\udc662 = 4 , \ud835\udc66 \u2265 0, acima do plano \ud835\udc67 = 0 e abaixo da superfície 
\ud835\udc67 = 8 \u2212 \ud835\udc65, com densidade \ud835\udeff = \ud835\udeff(\ud835\udc65, \ud835\udc66, \ud835\udc67) = \ud835\udc662. 
 
GABARITO 
curso: 
Engenharia de Computação/ 
Engenharia de Produção 
bimestre: 4º bimestre P3 
 
 
Questão 1 
Alternativa A. 
 
\ud835\udc53(\ud835\udc65, \ud835\udc66) = \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc65\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 \u21d2 
\ud835\udf15\ud835\udc53
\ud835\udf15\ud835\udc65
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc65\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 , 
\ud835\udf15\ud835\udc53
\ud835\udf15\ud835\udc66
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc65\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66 
 
\ud835\udf15\ud835\udc53
\ud835\udf15\ud835\udc65
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc65\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 = 0 \u21d2 \ud835\udc65 = 0 . 
 
\ud835\udf15\ud835\udc53
\ud835\udf15\ud835\udc66
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc65\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66 = 0 \u21d2 \ud835\udc66 = 0 . 
 
Logo, o único ponto crítico no domínio considerado é (0,0). 
 
 
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc652
(\ud835\udc65, \ud835\udc66) = 
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc662
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc65\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 \u21d2
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc652
(0,0) =
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc662
(0,0) = \u22121. 
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc65\ud835\udf15\ud835\udc66
(\ud835\udc65, \ud835\udc66) = \ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc65\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66 \u21d2
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc65\ud835\udf15\ud835\udc66
(0,0) = 0 
 
Segue que, \ud835\udc37 = |
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc652
(0,0)
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc65\ud835\udf15\ud835\udc66
(0,0)
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc65\ud835\udf15\ud835\udc66
(0,0)
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc662
(0,0)
| = |
\u22121 0
0 \u22121
| = 1 > 0. 
 
Como \ud835\udc37 = 1 > 0 e 
\ud835\udf15\ud835\udc52\ud835\udc53
\ud835\udf15\ud835\udc652
(0,0) = \u22121 < 0, segue que (0,0) é ponto de máximo local. 
 
 
Questão 2 
Alternativa D. 
Massa = \u222c \ud835\udeff(\ud835\udc65, \ud835\udc66)\ud835\udc51\ud835\udc34\ud835\udc37 = \u222c (\ud835\udc65
2 + \ud835\udc662)\ud835\udc51\ud835\udc34\ud835\udc37 . 
Vamos usar coordenadas polares no cálculo da integral dupla. 
Parametrização: 
\ud835\udc65 = \ud835\udc5f\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 0 \u2264 \ud835\udf03 \u2264
\ud835\udf0b
2
 , (\ud835\udc66 \u2265 0, \ud835\udc65 \u2265 0) 
\ud835\udc66 = \ud835\udc5f\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 0 \u2264 \ud835\udc5f \u2264 2 
Jacobiano = \ud835\udc5f. 
\u222c (\ud835\udc652 + \ud835\udc662 )\ud835\udc51\ud835\udc34\ud835\udc37 = \u222b \u222b \ud835\udc5f
2 \u2219 \ud835\udc5f\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 = \u222b
1
4
\ud835\udc5f4|
2
0
\ud835\udc51\ud835\udf03 =
\ud835\udf0b
2
0
2
0
\ud835\udf0b
2
0 
= \u222b 4
\ud835\udf0b
2
0 \ud835\udc51\ud835\udf03 = 4 \u2219
\ud835\udf0b
2
= 2\ud835\udf0b . 
 
 
disciplina: MCA502 \u2013 Cálculo II 
Questão 3 
Como a equação da curva não é fornecida, não conseguimos aplicar a definição de integral de linha. 
O campo é conservativo. Vamos obter sua função potencial: 
\ud835\udf15\ud835\udf11
\ud835\udf15\ud835\udc65
(\ud835\udc65, \ud835\udc66) = 3\ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 \u21d2 \ud835\udf11(\ud835\udc65, \ud835\udc66) = \ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 + \ud835\udc54(\ud835\udc66) 
\ud835\udf15\ud835\udf11
\ud835\udf15\ud835\udc66
(\ud835\udc65, \ud835\udc66) = \u2212\ud835\udc523\ud835\udc65 \ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66 \u21d2 \ud835\udf11(\ud835\udc65, \ud835\udc66) = \ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 + \u210e(\ud835\udc65) 
Comparando ambas as expressões, concluímos que \ud835\udf11(\ud835\udc65, \ud835\udc66) = \ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66 é uma função potencial. 
\u222b 3\ud835\udc523\ud835\udc65 \ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udc66\ud835\udc51\ud835\udc65 \u2212 \ud835\udc523\ud835\udc65 \ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udc66\ud835\udc51\ud835\udc66 \ud835\udefe = \ud835\udf11(3,0) \u2212 \ud835\udf11(\u22123,0) = \ud835\udc52
9\ud835\udc50\ud835\udc5c\ud835\udc600 \u2212 \ud835\udc52 \u22129\ud835\udc50\ud835\udc5c\ud835\udc600 = \ud835\udc529 \u2212 \ud835\udc52 \u22129 
 
 
Questão 4 
Massa = \u222c \ud835\udeff(\ud835\udc65, \ud835\udc66, \ud835\udc67)\ud835\udc51\ud835\udc34\ud835\udc37 = \u222c \ud835\udc66
2\ud835\udc51\ud835\udc34\ud835\udc37 . 
Parametrização: 
\ud835\udc65 = 2\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 0 \u2264 \ud835\udf03 \u2264 \ud835\udf0b , pois \ud835\udc66 \u2265 0 
\ud835\udc66 = 2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 0 \u2264 \ud835\udc67 \u2264 8 \u2212 2\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 
\ud835\udc67 = \ud835\udc67 \ud835\udc51\ud835\udc34 = 2\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 
\u222c \ud835\udc662\ud835\udc51\ud835\udc34\ud835\udc37 = \u222b \u222b 4\ud835\udc60\ud835\udc52\ud835\udc5b
2\ud835\udf032\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 = \u222b 8\ud835\udc60\ud835\udc52\ud835\udc5b 2\ud835\udf03 \u2219 \ud835\udc67|
8 \u2212 2\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03
0
\ud835\udc51\ud835\udf03 =
\ud835\udf0b
0
8\u22122\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03
0
\ud835\udf0b
0 
= \u222b (64\ud835\udc60\ud835\udc52\ud835\udc5b 2\ud835\udf03
\ud835\udf0b
0 \u2212 16\ud835\udc60\ud835\udc52\ud835\udc5b
2\ud835\udf03\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 )\ud835\udc51\ud835\udf03 = \u222b (
64(1\u2212\ud835\udc50\ud835\udc5c\ud835\udc602\ud835\udf03 )
2
\u2212
\ud835\udf0b
0 16\ud835\udc60\ud835\udc52\ud835\udc5b
2\ud835\udf03\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03)\ud835\udc51\ud835\udf03 = 
 
(32\ud835\udf03 \u2212 16\ud835\udc60\ud835\udc52\ud835\udc5b2\ud835\udf03 \u2212
16
3
\ud835\udc60\ud835\udc52\ud835\udc5b 3\ud835\udf03)) |
\ud835\udf0b
0
= 32\ud835\udf0b \u2212 0 + 0 = 32\ud835\udf0b