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Lesson: The Gradient Vector Field

Worksheet • 14 Questions

Q1:

Find the gradient of 𝑓 ( π‘₯ , 𝑦 ) = π‘₯ + 𝑦 βˆ’ 1   .

  • A ⟨ 2 π‘₯ , 2 𝑦 ⟩
  • B ⟨ π‘₯ , 𝑦 ⟩
  • C ⟨ 2 𝑦 , 2 π‘₯ ⟩
  • D ⟨ 2 𝑦 βˆ’ 1 , 2 π‘₯ βˆ’ 1 ⟩
  • E ⟨ 2 π‘₯ βˆ’ 1 , 2 𝑦 βˆ’ 1 ⟩

Q2:

Compute the gradient for 𝑓 ( π‘₯ , 𝑦 , 𝑧 ) = π‘₯ 𝑦 𝑧 . s i n

  • A ( 𝑦 𝑧 π‘₯ 𝑦 𝑧 , π‘₯ 𝑧 π‘₯ 𝑦 𝑧 , π‘₯ 𝑦 π‘₯ 𝑦 𝑧 ) c o s c o s c o s
  • B ( 𝑦 𝑧 , π‘₯ 𝑧 , π‘₯ 𝑦 )
  • C ( π‘₯ 𝑧 π‘₯ 𝑦 𝑧 , 𝑦 𝑧 π‘₯ 𝑦 𝑧 , π‘₯ 𝑦 π‘₯ 𝑦 𝑧 ) c o s c o s c o s
  • D ( π‘₯ 𝑧 , 𝑦 𝑧 , π‘₯ 𝑦 )
  • E ( 𝑦 𝑧 π‘₯ 𝑦 𝑧 , π‘₯ 𝑦 π‘₯ 𝑦 𝑧 , π‘₯ 𝑧 π‘₯ 𝑦 𝑧 ) c o s c o s c o s

Q3:

Compute the gradient for 𝑓 ( π‘₯ , 𝑦 , 𝑧 ) = √ π‘₯ + 𝑦 + 𝑧 . 2 2 2

  • A ο€Ώ π‘₯ √ π‘₯ + 𝑦 + 𝑧 , 𝑦 √ π‘₯ + 𝑦 + 𝑧 , 𝑧 √ π‘₯ + 𝑦 + 𝑧  2 2 2 2 2 2 2 2 2
  • B ο€½ π‘₯ π‘₯ + 𝑦 + 𝑧 , 𝑦 π‘₯ + 𝑦 + 𝑧 , 𝑧 π‘₯ + 𝑦 + 𝑧  2 2 2 2 2 2 2 2 2
  • C ο€Ώ 𝑦 √ π‘₯ + 𝑦 + 𝑧 , π‘₯ √ π‘₯ + 𝑦 + 𝑧 , 𝑧 √ π‘₯ + 𝑦 + 𝑧  2 2 2 2 2 2 2 2 2
  • D ο€Ώ 1 √ π‘₯ + 𝑦 + 𝑧 , 1 √ π‘₯ + 𝑦 + 𝑧 , 1 √ π‘₯ + 𝑦 + 𝑧  2 2 2 2 2 2 2 2 2
  • E ο€Ώ π‘₯ √ π‘₯ + 𝑦 + 𝑧 , 𝑧 √ π‘₯ + 𝑦 + 𝑧 , 𝑦 √ π‘₯ + 𝑦 + 𝑧  2 2 2 2 2 2 2 2 2

Q4:

Find the gradient of 𝑓 ( π‘₯ , 𝑦 ) = 1 π‘₯ + 𝑦   .

  • A ο“’ βˆ’ 2 π‘₯ ( π‘₯ + 𝑦 ) , βˆ’ 2 𝑦 ( π‘₯ + 𝑦 ) ο““      
  • B ο“’ βˆ’ π‘₯ ( π‘₯ + 𝑦 ) , βˆ’ 𝑦 ( π‘₯ + 𝑦 ) ο““      
  • C ο“’ 2 π‘₯ ( π‘₯ + 𝑦 ) , βˆ’ 2 𝑦 ( π‘₯ + 𝑦 ) ο““      
  • D ο“’ 2 π‘₯ ( π‘₯ + 𝑦 ) , 2 𝑦 ( π‘₯ + 𝑦 ) ο““    
  • E ο“’ βˆ’ 2 π‘₯ ( π‘₯ + 𝑦 ) , 2 𝑦 ( π‘₯ + 𝑦 ) ο““    

Q5:

Compute the gradient for 𝑓 ( π‘₯ , 𝑦 , 𝑧 ) = π‘₯ 𝑒 .   

  • A ο‡° 2 π‘₯ 𝑒 , π‘₯ 𝑧 𝑒 , π‘₯ 𝑦 𝑒         
  • B ο‡° 2 𝑦 𝑒 , 𝑦 𝑧 𝑒 , 𝑦 𝑒         
  • C ο‡° π‘₯ 𝑧 𝑒 , 2 π‘₯ 𝑒 , π‘₯ 𝑦 𝑒         
  • D ο‡° π‘₯ 𝑦 𝑒 , π‘₯ 𝑧 𝑒 , 2 π‘₯ 𝑒         
  • E ο‡° π‘₯ 𝑧 𝑒 , π‘₯ 𝑦 𝑒 , 2 π‘₯ 𝑒         

Q6:

Suppose 𝑀 = 𝐹 ( πœ™ ( π‘₯ , 𝑦 ) ) with πœ™ = ( π‘₯ + 𝑦 , π‘₯ βˆ’ 𝑦 , π‘₯ 𝑦 )     . Express the gradient βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  (viewed as a 1 Γ— 2 matrix) in terms of the 1 Γ— 3 matrix βˆ‡ 𝐹 ( π‘ž ) , where π‘ž = πœ™ ο€Ό πœ‹ , βˆ’ 2 3  , and a matrix of partial derivatives of πœ™ .

  • A βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  = βˆ‡ 𝐹 ( π‘ž ) ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ 2 πœ‹ βˆ’ 4 3 2 πœ‹ 4 3 βˆ’ 2 3 πœ‹ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦
  • B βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  = βˆ‡ 𝐹 ( π‘ž ) ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ πœ‹ βˆ’ 2 3 πœ‹ βˆ’ 2 3 βˆ’ 2 3 πœ‹ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦
  • C βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  = βˆ‡ 𝐹 ( π‘ž ) ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ 2 πœ‹ 4 3 2 πœ‹ βˆ’ 4 3 βˆ’ 2 3 πœ‹ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦
  • D βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  = βˆ‡ 𝐹 ( π‘ž ) ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ 2 πœ‹ βˆ’ 4 3 2 πœ‹ βˆ’ 4 3 βˆ’ 2 3 πœ‹ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦
  • E βˆ‡ 𝑀 ο€Ό πœ‹ , βˆ’ 2 3  = βˆ‡ 𝐹 ( π‘ž ) ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ πœ‹ βˆ’ 2 3 πœ‹ 2 3 βˆ’ 2 3 πœ‹ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦

Q7:

For 𝑓 ( π‘₯ , 𝑦 , 𝑧 ) = 𝑧 π‘₯ + 𝑦 2 2 in Cartesian coordinates, find βˆ‡ 𝑓 in cylindrical coordinates.

  • A βˆ’ 2 𝑧 π‘Ÿ 𝑒 + 1 π‘Ÿ 𝑒 3 π‘Ÿ 2 𝑧
  • B 2 𝑧 π‘Ÿ 𝑒 βˆ’ 𝑧 π‘Ÿ 𝑒 3 π‘Ÿ 2 𝑧
  • C 2 𝑧 π‘Ÿ 𝑒 + 1 π‘Ÿ 𝑒 3 π‘Ÿ 2 𝑧
  • D 2 𝑧 π‘Ÿ 𝑒 + 𝑧 π‘Ÿ 𝑒 3 π‘Ÿ 2 𝑧
  • E βˆ’ 2 𝑧 π‘Ÿ 𝑒 + 𝑧 π‘Ÿ 𝑒 3 π‘Ÿ 2 𝑧

Q8:

Find the gradient of 𝑓 ( π‘₯ , 𝑦 ) = π‘₯ 𝑦 . l n

  • A ο€½ 1 π‘₯ , 1 𝑦 
  • B ( π‘₯ , 𝑦 ) l n l n
  • C ο€½ 1 𝑦 , 1 π‘₯ 
  • D ( π‘₯ , 𝑦 )
  • E ο€½ 1 π‘₯ 𝑦 , 1 π‘₯ 𝑦 

Q9:

Compute the gradient for 𝑓 ( π‘₯ , 𝑦 , 𝑧 ) = π‘₯ + 𝑦 + 𝑧 .   

  • A ⟨ 2 π‘₯ , 2 𝑦 , 2 𝑧 ⟩
  • B ⟨ π‘₯ , 𝑦 , 𝑧 ⟩
  • C ⟨ 2 𝑦 , 2 π‘₯ , 2 𝑧 ⟩
  • D ⟨ 2 , 2 , 2 ⟩
  • E ⟨ 2 π‘₯ , 2 𝑧 , 2 𝑦 ⟩

Q10:

Suppose 𝑀 = 𝐹 ( πœ™ ( π‘₯ , 𝑦 ) ) with πœ™ = ( πœ™ , πœ™ , πœ™ ) 1 2 3 and that π‘ž = πœ™ ( 𝑝 ) for a point 𝑝 ∈ ℝ 2 . Express the gradient βˆ‡ 𝑀 ( 𝑝 ) (viewed as a 1 Γ— 2 matrix) in terms of the 1 Γ— 3 matrix βˆ‡ 𝐹 ( π‘ž ) and a matrix of partial derivatives of πœ™ .

  • A βˆ‡ 𝑀 ( 𝑝 ) = βˆ‡ 𝐹 ( π‘ž ) β‹… ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦ 1 1 2 2 3 3
  • B βˆ‡ 𝑀 ( 𝑝 ) = βˆ‡ 𝐹 ( π‘ž ) β‹… ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦ 1 3 2 2 3 1
  • C βˆ‡ 𝑀 ( 𝑝 ) = βˆ‡ 𝐹 ( π‘ž ) β‹… ⎑ ⎒ ⎒ ⎒ ⎣ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• 𝑦 ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ 1 2 3 1 2 3
  • D βˆ‡ 𝑀 ( 𝑝 ) = βˆ‡ 𝐹 ( π‘ž ) β‹… ⎑ ⎒ ⎒ ⎒ ⎣ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• π‘₯ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ 1 2 3 1 2 3
  • E βˆ‡ 𝑀 ( 𝑝 ) = βˆ‡ 𝐹 ( π‘ž ) β‹… ⎑ ⎒ ⎒ ⎒ ⎒ ⎒ ⎒ ⎣ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ πœ• πœ™ πœ• 𝑦 πœ• πœ™ πœ• π‘₯ ⎀ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ βŽ₯ ⎦ 1 1 2 2 3 3

Q11:

Compute the gradient of the function 𝑓 ( π‘₯ , 𝑦 ) = π‘₯ 𝑒 2 𝑦 .

  • A ο€Ή 2 π‘₯ 𝑒 , π‘₯ 𝑒  𝑦 2 𝑦
  • B ( π‘₯ 𝑒 , 2 𝑒 ) 𝑦 𝑦
  • C ο€Ή π‘₯ 𝑒 , 2 π‘₯ 𝑒  2 𝑦 𝑦
  • D ( 2 𝑒 , π‘₯ 𝑒 ) 𝑦 𝑦
  • E ο€Ή 2 π‘₯ 𝑒 , 𝑦 π‘₯ 𝑒  𝑦 2 𝑦

Q12:

Find a function so that the vector field is a gradient field.

  • A
  • B
  • C
  • D
  • E

Q13:

Compute the gradient of the function 𝑓 ( π‘₯ , 𝑦 ) = 2 π‘₯ + 5 𝑦 .

  • A ⟨ 2 , 5 ⟩
  • B ⟨ 2 π‘₯ , 5 𝑦 ⟩
  • C ⟨ 5 , 2 ⟩
  • D  1 5 , 1 2 ο‡·
  • E  1 2 , 1 5 ο‡·

Q14:

We explore an example where a vector field F = ⟨ 𝐹 , 𝐹 ⟩ 1 2 satisfies πœ• 𝐹 πœ• 𝑦 βˆ’ πœ• 𝐹 πœ• π‘₯ = 0 1 2 but does not come from a potential. On the plane with the origin removed, consider the vector field F ( π‘₯ , 𝑦 ) = ο“’ βˆ’ 𝑦 π‘₯ + 𝑦 , π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2 .

On the (open) half-plane π‘₯ > 0 , we can define the angle function πœƒ ( π‘₯ , 𝑦 ) = ο€» 𝑦 π‘₯  a r c t a n . This is well defined and gives a value between βˆ’ πœ‹ 2 and πœ‹ 2 . What is the gradient βˆ‡ πœƒ ?

  • A ο“’ βˆ’ 𝑦 π‘₯ + 𝑦 , π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • B ο“’ βˆ’ π‘₯ π‘₯ + 𝑦 , 𝑦 π‘₯ + 𝑦 ο““ 2 2 2 2
  • C ο“’ 𝑦 π‘₯ + 𝑦 , βˆ’ π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • D ο“’ βˆ’ 𝑦 π‘₯ + 𝑦 , βˆ’ π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • E ο“’ 𝑦 π‘₯ + 𝑦 , π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2

Using the figure shown, use πœƒ above to define the angle function πœƒ ( π‘₯ , 𝑦 ) 1 on the region 𝑦 > 0 by a suitable composition with a πœ‹ 2 rotation.

  • A πœƒ ( π‘₯ , 𝑦 ) = πœƒ ( 𝑦 , βˆ’ π‘₯ ) + πœ‹ 2 1
  • B πœƒ ( π‘₯ , 𝑦 ) = πœƒ ( βˆ’ 𝑦 , βˆ’ π‘₯ ) + πœ‹ 2 1
  • C πœƒ ( π‘₯ , 𝑦 ) = πœƒ ( 𝑦 , βˆ’ π‘₯ ) 1
  • D πœƒ ( π‘₯ , 𝑦 ) = πœƒ ( βˆ’ 𝑦 , π‘₯ ) 1
  • E πœƒ ( π‘₯ , 𝑦 ) = πœƒ ( βˆ’ 𝑦 , π‘₯ ) + πœ‹ 2 1

What is βˆ‡ πœƒ ( π‘₯ , 𝑦 ) 1 ?

  • A ο“’ βˆ’ 𝑦 π‘₯ + 𝑦 , π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • B ο“’ βˆ’ π‘₯ π‘₯ + 𝑦 , 𝑦 π‘₯ + 𝑦 ο““ 2 2 2 2
  • C ο“’ 𝑦 π‘₯ + 𝑦 , βˆ’ π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • D ο“’ 𝑦 π‘₯ + 𝑦 , π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2
  • E ο“’ βˆ’ 𝑦 π‘₯ + 𝑦 , βˆ’ π‘₯ π‘₯ + 𝑦 ο““ 2 2 2 2

Since πœƒ and πœƒ 1 agree on the quadrant π‘₯ > 0 , 𝑦 > 0 , we can define the angle 𝑇 ( π‘₯ , 𝑦 ) at each point of the union with values between βˆ’ πœ‹ 2 and 3 πœ‹ 2 . Using this, what is ο„Έ β‹… 𝐢 F r d , where 𝐢 is the arc of the unit circle from ο€Ώ 1 2 , βˆ’ √ 3 2  to ο€Ώ βˆ’ 1 √ 2 , 1 √ 2  ? Answer in terms of πœ‹ .

  • A 1 3 πœ‹ 1 2
  • B 1 2 πœ‹ 1 3
  • C 3 πœ‹ 5
  • D πœ‹ 1 2
  • E πœ‹ 5

In the same way, we can define πœƒ 2 on the half-plane π‘₯ < 0 and πœƒ 3 on 𝑦 < 0 . Hence, evaluate the line integral ο„Έ β‹… 𝐢 F r d around the circle of radius √ 2 , starting from 𝑃 ( 1 , βˆ’ 1 ) and going counterclockwise back to 𝑃 , stating your answer in terms of πœ‹ .

  • A 2 πœ‹
  • B πœ‹ √ 2
  • C πœ‹
  • D √ 2 πœ‹
  • E πœ‹ 2
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