EngComp e Prod 2018 1 Calculo II MCA502 P4 GABARITO
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EngComp e Prod 2018 1 Calculo II MCA502 P4 GABARITO


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1 
 
AVALIAÇÃO PRESENCIAL 
CADERNO DE PERGUNTAS 
 
curso: Engenharia de Computação/ Engenharia de Produção bimestre: 4º bimestre ano: 2018 | 2sem 
CÓDIGO DA PROVA 
P4 
 
\u2022 Preencha atentamente o cabeçalho de TODAS AS FOLHAS DE RESPOSTA que você utilizar. 
\u2022 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 
 
\u2022 É permitido o uso de formulário impresso. Não é permitido o uso de calculadora. 
 
Questão 1 (2 pontos) 
Calcule a massa do arco de hélice \u210e(\ud835\udc61\ud835\udc61) = (\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, \ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, \ud835\udc61\ud835\udc61), 0 \u2264 \ud835\udc61\ud835\udc61 \u2264 2 \ud835\udf0b\ud835\udf0b com densidade \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67) = 8\ud835\udc67\ud835\udc67. 
 
a) 4\u221a2\ud835\udf0b\ud835\udf0b2 
b) 16\u221a2\ud835\udf0b\ud835\udf0b2 
c) 8\u221a2\ud835\udf0b\ud835\udf0b2 
d) 16\ud835\udf0b\ud835\udf0b2 
e) Nenhuma das outras alternativas. 
 
Questão 2 (2 pontos) 
A temperatura de uma placa plana é dada por \ud835\udc47\ud835\udc47 = \ud835\udc47\ud835\udc47(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66), em que x e y são as coordenadas de um ponto 
genérico da placa. Um besouro anda sobre a placa seguindo uma trajetória \ud835\udc65\ud835\udc65 = \ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61) e \ud835\udc66\ud835\udc66 = \ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61), em que t é o 
tempo em segundos. Sabe-se que quando \ud835\udc61\ud835\udc61 = 3 o besouro está na posição (5,4) e que \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(3) = \u22127, \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(3) =3, \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(5,4) = 3 e \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(5,4) = 6. Podemos afirmar: 
 
a) No instante t = 3 a temperatura no besouro está aumentando 3 unidades por segundo. 
b) No instante t = 3 a temperatura no besouro está diminuindo 2 unidades por segundo. 
c) No instante t = 3 a temperatura no besouro está aumentando 39 unidades por segundo. 
d) No instante t = 3 a temperatura no besouro está diminuindo 1 unidade por segundo. 
e) Nenhuma das outras alternativas. 
 
 
Questão 3 (3 pontos) 
Calcule \u222b (1 \u2212 \ud835\udc66\ud835\udc66\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66))\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + \ud835\udc65\ud835\udc65\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udefe\ud835\udefe , sendo \ud835\udefe\ud835\udefe a curva 
esboçada ao lado, ligando o ponto (-3, \ud835\udf0b\ud835\udf0b ) ao ponto (1, 0). 
 
 
 
Questão 4 (3 pontos) 
Calcule \u222c (\ud835\udc67\ud835\udc672 \u2212 2\ud835\udc65\ud835\udc65)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \u2227 \ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 + (6\ud835\udc66\ud835\udc66 + \ud835\udc60\ud835\udc603\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 \u2227 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + (2\ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc67 \u2212 4\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 \u2227 \ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udc46\ud835\udc46 através da superfície S que é o bordo 
do paralelepípedo cujos vértices são os oito pontos (0,0,0) , (2,0,0), (2,3,0), (0,3,0), (0,0,1), (2,0,1), (2,3,1) e 
(0,3,1) orientada com a normal que aponta para dentro. 
2 
 
GABARITO 
curso: Engenharia de Computação/ Engenharia de Produção bimestre: 4º bimestre P4 
 
Questão 1 
Alternativa B. 
Massa = \u222b \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc50\ud835\udc50 = \u222b \u2016\u210e\u2032(\ud835\udc61\ud835\udc61)\u20162\ud835\udf0b\ud835\udf0b0\u210e \ud835\udeff\ud835\udeff\ufffd\u210e(\ud835\udc61\ud835\udc61)\ufffd\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 
\u210e(\ud835\udc61\ud835\udc61) = (\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, \ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, \ud835\udc61\ud835\udc61)) \u21d2\u210e\u2032(\ud835\udc61\ud835\udc61) = (\u2212\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, 1). 
\u2016\u210e\u2032(\ud835\udc61\ud835\udc61)\u2016 = \u221a\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc602\ud835\udc61\ud835\udc61 + \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc502\ud835\udc61\ud835\udc61 + 1 = \u221a2 e também \ud835\udeff\ud835\udeff\ufffd\u210e(\ud835\udc61\ud835\udc61)\ufffd = 8\ud835\udc67\ud835\udc67 = 8\ud835\udc61\ud835\udc61 
Massa = \u222b \u221a2 \u22192\ud835\udf0b\ud835\udf0b0 8\ud835\udc61\ud835\udc61\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 = 4\u221a2\ud835\udc61\ud835\udc612|2\ud835\udf0b\ud835\udf0b0 = 16\u221a2\ud835\udf0b\ud835\udf0b2. 
 
Questão 2 
Alternativa E. 
\ud835\udc51\ud835\udc51\ud835\udf15\ud835\udf15
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61) = \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61),\ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61)) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61) + \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61),\ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61)) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61)\u21d2 
 \ud835\udc51\ud835\udc51\ud835\udf15\ud835\udf15
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(3) = \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(5,4) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(3) + \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(5,4) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(3) = 3 \u2219 (\u22127) + 6 \u2219 3 = \u22123 
Assim, no instante t = 3 a temperatura no besouro está diminuindo 3 unidades por segundo. 
 
Questão 3 \u2013 ANULADA (redistribuir os pontos entre as outras questões) 
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\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = 1 + \ud835\udc66\ud835\udc66\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66) \u21d2 \ud835\udf11\ud835\udf11(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65 \u2212 cos (\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66) + \ud835\udc54\ud835\udc54(\ud835\udc66\ud835\udc66) 
\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66) \u21d2 \ud835\udf11\ud835\udf11(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) = \u2212cos (\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66) + \u210e(\ud835\udc65\ud835\udc65) 
Comparando ambas as expressões, concluímos que \ud835\udf11\ud835\udf11(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65 \u2212 cos (\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66) é uma função potencial. 
\ufffd 2\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc652\ud835\udc66\ud835\udc66)\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + \ud835\udc65\ud835\udc652\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60(\ud835\udc65\ud835\udc652\ud835\udc66\ud835\udc66)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 
\ud835\udefe\ud835\udefe
= \ud835\udf11\ud835\udf11(1,0) \u2212 \ud835\udf11\ud835\udf11(\u22123,\ud835\udf0b\ud835\udf0b) = 1 \u2212 \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc500 \u2212 (\u22123 \u2212 cos(\u22123\ud835\udf0b\ud835\udf0b)) = 1 \u2212 1 \u2212 (\u22123 + 1) 
 = 2 
 
Questão 4 
Vamos aplicar o Teorema de Gauss. Lembre-se que neste teorema a normal à superfície deve apontar para 
fora. Como no enunciado a normal está para dentro vai aparecer um sinal de \u201c-\u201c. 
 \ud835\udc3c\ud835\udc3c = \u222c (\ud835\udc67\ud835\udc672 \u2212 2\ud835\udc65\ud835\udc65)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \u2227 \ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 + (6\ud835\udc66\ud835\udc66 + \ud835\udc60\ud835\udc603\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 \u2227 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + (2\ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc67 \u2212 4\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 \u2227 \ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udc46\ud835\udc46 = \u2212 \u222d (\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ufffd\u20d7\ufffd\ud835\udc39\ud835\udc49\ud835\udc49 )\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 = 
\u2212\u222d (\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
+ \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
+ \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc67\ud835\udc67\ud835\udc49\ud835\udc49
)\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 = \u2212\u222d (\u22122 + 6 + 2\ud835\udc65\ud835\udc65 \u2212 4)\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 = \u2212\u222b \u222b \u222b 2\ud835\udc65\ud835\udc65\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 = \u2212\u222b \u222b \ud835\udc65\ud835\udc652|20\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 =3010203010\ud835\udc49\ud835\udc49 = \u2212\ufffd \ufffd 4\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 = \u22124 \u2219 1 \u2219 3 = \u2212123
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disciplina: MCA502 \u2013 Cálculo II