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.