Atividade: Derivadas de Funções Vetoriais

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Nesta atividade, nós vamos praticar as derivadas do valor de funções vetoriais em uma variável tomando a derivada de cada componente.

Q1:

Calcule 𝑓 β€² ( 𝑠 ) , e determine a equação vetorial da reta tangente 𝐿 a 𝑓 ( 0 ) para 𝑓 ( 𝑠 ) = ( 2 𝑠 , 2 𝑠 , 𝑠 ) c o s s e n .

  • A 𝑓 β€² ( 𝑠 ) = ( βˆ’ 2 𝑠 , 2 𝑠 , 1 ) s e n c o s , 𝐿 ( 1 , 0 , 0 ) + 𝑑 ( 0 , 1 , 1 ) :
  • B 𝑓 β€² ( 𝑠 ) = ( βˆ’ 2 2 𝑠 , 2 2 𝑠 , 1 ) s e n c o s , 𝐿 ( 0 , 2 , 1 ) + 𝑑 ( 1 , 0 , 0 ) :
  • C 𝑓 β€² ( 𝑠 ) = ( βˆ’ 2 𝑠 , 2 𝑠 , 1 ) s e n c o s , 𝐿 ( 0 , 1 , 1 ) + 𝑑 ( 1 , 0 , 0 ) :
  • D 𝑓 β€² ( 𝑠 ) = ( βˆ’ 2 2 𝑠 , 2 2 𝑠 , 1 ) s e n c o s , 𝐿 ( 1 , 0 , 0 ) + 𝑑 ( 0 , 2 , 1 ) :
  • E 𝑓 β€² ( 𝑠 ) = ( 2 2 𝑠 , βˆ’ 2 2 𝑠 , 1 ) s e n c o s , 𝐿 ( 1 , 0 , 0 ) + 𝑑 ( 0 , βˆ’ 2 , 1 ) :

Q2:

Calcule 𝑓 β€² ( 𝑠 ) , e determine a equação vetorial da reta tangente a 𝑓 ( 0 ) para 𝑓 ( 𝑠 ) = ο€Ή 𝑠 + 1 , 𝑠 + 1 , 𝑠 + 1  2 3 .

  • A 𝑓 β€² ( 𝑠 ) = ο€Ή 2 , 2 𝑠 + 1 , 3 𝑠 + 1  2 , 𝐿 ( 2 , 1 , 1 ) + 𝑑 ( 2 , 0 , 0 ) :
  • B 𝑓 β€² ( 𝑠 ) = ο€Ή 1 , 2 𝑠 , 3 𝑠  2 , 𝐿 ( 1 , 0 , 0 ) + 𝑑 ( 1 , 1 , 1 ) :
  • C 𝑓 β€² ( 𝑠 ) = ο€Ή 2 , 2 𝑠 + 1 , 3 𝑠 + 1  2 , 𝐿 ( 2 , 0 , 0 ) + 𝑑 ( 2 , 1 , 1 ) :
  • D 𝑓 β€² ( 𝑠 ) = ο€Ή 1 , 2 𝑠 , 3 𝑠  2 , 𝐿 ( 1 , 1 , 1 ) + 𝑑 ( 1 , 0 , 0 ) :
  • E 𝑓 β€² ( 𝑠 ) = ( 1 , 2 𝑠 , 3 𝑠 ) , 𝐿 ( 1 , 1 , 1 ) + 𝑑 ( 1 , 0 , 0 ) :

Q3:

Calcule βƒ— 𝑓 β€² ( 𝑠 ) , e encontre a forma vetorial da equação da reta tangente em βƒ— 𝑓 ( 0 ) para βƒ— 𝑓 ( 𝑠 ) = ( 𝑒 + 1 , 𝑒 + 1 , 𝑒 + 1 ) 𝑠 2 𝑠 𝑠 2 .

  • A βƒ— 𝑓 β€² ( 𝑠 ) = ο€Ί 𝑒 + 1 , 2 𝑒 + 1 , 2 𝑠 β‹… 𝑒 + 1  𝑠 2 𝑠 𝑠 2 , 𝐿 ( 2 , 2 , 2 ) + 𝑑 ( 2 , 3 , 1 ) :
  • B βƒ— 𝑓 β€² ( 𝑠 ) = ο€Ί 𝑒 , 2 𝑒 , 2 𝑠 β‹… 𝑒  𝑠 2 𝑠 𝑠 2 , 𝐿 ( 1 , 2 , 0 ) + 𝑑 ( 2 , 2 , 2 ) :
  • C βƒ— 𝑓 β€² ( 𝑠 ) = ο€Ί 𝑒 + 1 , 2 𝑒 + 1 , 2 𝑠 β‹… 𝑒 + 1  𝑠 2 𝑠 𝑠 2 , 𝐿 ( 2 , 3 , 1 ) + 𝑑 ( 2 , 2 , 2 ) :
  • D βƒ— 𝑓 β€² ( 𝑠 ) = ο€Ί 𝑒 , 2 𝑒 , 2 𝑠 β‹… 𝑒  𝑠 2 𝑠 𝑠 2 , 𝐿 ( 2 , 2 , 2 ) + 𝑑 ( 1 , 2 , 0 ) :
  • E βƒ— 𝑓 β€² ( 𝑠 ) = ο€Ί 𝑒 , 𝑒 , 𝑒  𝑠 2 𝑠 𝑠 2 , 𝐿 ( 2 , 2 , 2 ) + 𝑑 ( 1 , 1 , 1 ) :

Q4:

Considere a curva βƒ— π‘Ÿ ( 𝑠 ) = ( 2 𝑠 , 2 𝑠 , 2 𝑠 ) s e n s e n c o s 2 . Determine βƒ— π‘Ÿ β€² ( 𝑠 ) e encontre a tangente 𝐿 para a curva quando 𝑠 = 0 .

  • A βƒ— π‘Ÿ β€² ( 𝑠 ) = ( βˆ’ 2 2 𝑠 , 2 𝑠 𝑠 , 2 𝑠 ) c o s s e n c o s s e n , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 2 , 0 ) :
  • B βƒ— π‘Ÿ β€² ( 𝑠 ) = ( 2 2 𝑠 , 2 2 𝑠 , βˆ’ 2 𝑠 ) c o s s e n s e n , 𝐿 ( 2 , 0 , 0 ) + 𝑑 ( 0 , 0 , 2 ) :
  • C βƒ— π‘Ÿ β€² ( 𝑠 ) = ( 2 2 𝑠 , 4 𝑠 , βˆ’ 2 𝑠 ) c o s c o s s e n 2 , 𝐿 ( 2 , 2 , 0 ) + 𝑑 ( 0 , 0 , 2 ) :
  • D βƒ— π‘Ÿ β€² ( 𝑠 ) = ( 2 2 𝑠 , 4 𝑠 𝑠 , βˆ’ 2 𝑠 ) c o s s e n c o s s e n , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 0 , 0 ) :
  • E βƒ— π‘Ÿ β€² ( 𝑠 ) = ( βˆ’ 2 𝑠 , 2 𝑠 , 2 𝑠 ) c o s c o s s e n 2 , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 1 , 0 ) :

Q5:

Dado que π‘Ÿ ( 𝑑 ) = π‘Ž 𝑑 βƒ— 𝚀 + 𝑑 𝑒 βƒ— πš₯ + 𝑐 𝑑 βƒ— π‘˜ s e n c o s 2 𝑏 𝑑 2 , onde π‘Ž e 𝑏 sΓ£o constantes, encontre π‘Ÿ β€² ( 𝑑 ) .

  • A βˆ’ 2 π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 βƒ— 𝚀 + 𝑒 ( 1 + 𝑏 𝑑 ) βƒ— πš₯ + 2 𝑐 𝑐 𝑑 𝑐 𝑑 βƒ— π‘˜ s e n c o s c o s s e n 𝑏 𝑑
  • B π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 βƒ— 𝚀 + 𝑒 ( 1 + 𝑏 𝑑 ) βƒ— πš₯ βˆ’ 𝑐 𝑐 𝑑 𝑐 𝑑 βƒ— π‘˜ s e n c o s c o s s e n 𝑏 𝑑
  • C 2 π‘Ž 𝑑 π‘Ž 𝑑 βƒ— 𝚀 + 𝑒 ( 1 + 𝑑 ) βƒ— πš₯ βˆ’ 2 𝑐 𝑑 𝑐 𝑑 βƒ— π‘˜ s e n c o s c o s s e n 𝑏 𝑑
  • D 2 π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 βƒ— 𝚀 + 𝑒 ( 1 + 𝑏 𝑑 ) βƒ— πš₯ βˆ’ 2 𝑐 𝑐 𝑑 𝑐 𝑑 βƒ— π‘˜ s e n c o s c o s s e n 𝑏 𝑑
  • E βˆ’ π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 βƒ— 𝚀 + 𝑒 ( 1 + 𝑑 ) βƒ— πš₯ + 𝑐 𝑐 𝑑 𝑐 𝑑 βƒ— π‘˜ s e n c o s c o s s e n 𝑏 𝑑

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