Conservative Vector Field: Finding the Value of 'a

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SUMMARY

The vector field F(x,y,z) = 2xz i + ay^3 j + (x^2 + y^4) k is conservative when the scalar 'a' equals 4. The solution was derived using the 3D curl test, confirming that the necessary conditions for conservativeness are satisfied. Specifically, the calculations showed that the partial derivatives lead to the conclusion that a must be 4 for the vector field to be conservative.

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  • Understanding of vector fields and their components
  • Knowledge of the 3D curl test for determining conservativeness
  • Familiarity with partial derivatives and their applications
  • Basic concepts of scalar fields in multivariable calculus
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Homework Statement



For what value(s) of the scalar 'a' is the vector field
F(x,y,z)= 2xz i + ay^3 j + (x^2 + y^4) k conservative



The Attempt at a Solution



F1=2xz
F2=ay^3z
F3=(x^2 + y^4)

I used 3D curl test??

1)(partial F2)/(partial dx) - (partial F1)/ (partial dy)= 0-0 = 0
2)(partial F3)/(partial dy) - (partial F2)/ (partial dz)= 4y^3 - ay^3=0 so a=4
3)(partial F1)/(partial dz) - (partial F3)/ (partial dx)= 2x-2x = 0

my answer would be a=4
 
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If your Fy is supposed to be ay^3z then you're right. Your initial statement of the problem has Fy = ay^3.
 
yes it should be ay^3z j, Thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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