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Show that the vector field F(x,y,z)=⟨−ycos(9x),9xsin(−y),0⟩ is not a gradient vector field by computing its curl. How does this show what you intended? curl(F)=∇×F=⟨0,_,_)

User Dady
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Answer:

In this case, the computed curl of F = ⟨0, 0, 9ysin(9x) + cos(9x)⟩ shows that F is not a gradient vector field, as intended.

Step-by-step explanation:

To show that the vector field F(x, y, z) = ⟨-ycos(9x), 9xsin(-y), 0⟩ is not a gradient vector field, we can compute its curl. The curl of a vector field measures the rotational behavior of the field.

To compute the curl of F, we use the formula:

curl(F) = ∇ × F = (dFz/dy - dFy/dz) i + (dFx/dz - dFz/dx) j + (dFy/dx - dFx/dy) k

where ∇ is the del operator and d/dx, d/dy, and d/dz are partial derivatives with respect to x, y, and z, respectively.

Let's calculate the partial derivatives:

dFz/dy = 0

dFy/dz = 0

dFx/dz = 0

dFz/dx = (d/dx)(-ycos(9x)) = ysin(9x) * 9 = 9ysin(9x)

dFy/dx = (d/dx)(9xsin(-y)) = 9sin(-y)

dFx/dy = (d/dy)(-ycos(9x)) = -cos(9x)

Now, we can substitute these partial derivatives into the curl formula:

curl(F) = (∇ × F) = (0 - 0) i + (0 - 0) j + (9ysin(9x) - (-cos(9x))) k

= 9ysin(9x) + cos(9x) k

From the curl formula, we can see that the z-component of the curl is non-zero (cos(9x) ≠ 0), which indicates that the vector field F is not a gradient vector field.

Therefore, the computed curl of F = ⟨0, 0, 9ysin(9x) + cos(9x)⟩ shows that F is not a gradient vector field, as intended.

User Arif Dewi
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