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Lesson: Differentiating Vector Valued Functions

In this lesson, we will learn how to determine the derivatives of vector-valued functions in one variable by taking the derivative of each component.

Worksheet • 5 Questions

Q1:

Given that π‘Ÿ ( 𝑑 ) = π‘Ž 𝑑 + 𝑑 𝑒 + 𝑐 𝑑 s i n c o s 2 𝑏 𝑑 2 i j k , where π‘Ž and 𝑏 are constants, find π‘Ÿ β€² ( 𝑑 ) .

  • A 2 π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 + 𝑒 ( 1 + 𝑏 𝑑 ) βˆ’ 2 𝑐 𝑐 𝑑 𝑐 𝑑 s i n c o s c o s s i n i j k 𝑏 𝑑
  • B βˆ’ π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 + 𝑒 ( 1 + 𝑑 ) + 𝑐 𝑐 𝑑 𝑐 𝑑 s i n c o s c o s s i n i j k 𝑏 𝑑
  • C π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 + 𝑒 ( 1 + 𝑏 𝑑 ) βˆ’ 𝑐 𝑐 𝑑 𝑐 𝑑 s i n c o s c o s s i n i j k 𝑏 𝑑
  • D 2 π‘Ž 𝑑 π‘Ž 𝑑 + 𝑒 ( 1 + 𝑑 ) βˆ’ 2 𝑐 𝑑 𝑐 𝑑 s i n c o s c o s s i n i j k 𝑏 𝑑
  • E βˆ’ 2 π‘Ž π‘Ž 𝑑 π‘Ž 𝑑 + 𝑒 ( 1 + 𝑏 𝑑 ) + 2 𝑐 𝑐 𝑑 𝑐 𝑑 s i n c o s c o s s i n i j k 𝑏 𝑑

Q2:

Consider the curve r ( 𝑠 ) = ( 2 𝑠 , 2 𝑠 , 2 𝑠 ) s i n s i n c o s 2 . Determine r β€² ( 𝑠 ) and find the tangent 𝐿 to the curve when 𝑠 = 0 .

  • A r β€² ( 𝑠 ) = ( 2 2 𝑠 , 4 𝑠 𝑠 , βˆ’ 2 𝑠 ) c o s s i n c o s s i n , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 0 , 0 ) :
  • B r β€² ( 𝑠 ) = ( βˆ’ 2 𝑠 , 2 𝑠 , 2 𝑠 ) c o s c o s s i n 2 , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 1 , 0 ) :
  • C r β€² ( 𝑠 ) = ( 2 2 𝑠 , 2 2 𝑠 , βˆ’ 2 𝑠 ) c o s s i n s i n , 𝐿 ( 2 , 0 , 0 ) + 𝑑 ( 0 , 0 , 2 ) :
  • D r β€² ( 𝑠 ) = ( 2 2 𝑠 , 4 𝑠 , βˆ’ 2 𝑠 ) c o s c o s s i n 2 , 𝐿 ( 2 , 2 , 0 ) + 𝑑 ( 0 , 0 , 2 ) :
  • E r β€² ( 𝑠 ) = ( βˆ’ 2 2 𝑠 , 2 𝑠 𝑠 , 2 𝑠 ) c o s s i n c o s s i n , 𝐿 ( 0 , 0 , 2 ) + 𝑑 ( 2 , 2 , 0 ) :

Q3:

Calculate 𝑓 β€² ( 𝑠 ) , and find the vector form of the equation of the tangent line at 𝑓 ( 0 ) for 𝑓 ( 𝑠 ) = ο€Ή 𝑠 + 1 , 𝑠 + 1 , 𝑠 + 1  2 3 .

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