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Question 27.6: The vector field v is derivable from the potential Φ = 2xy +......

The vector field v is derivable from the potential Φ = 2xy + zx. Find v.

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If v is derivable from the potential Φ, then v = ∇Φ and so

v=∇ϕ=(2y+z)i+2xj+xk\displaystyle\mathbf{v}=\nabla\phi=(2y+z)\mathbf{i}+2x\mathbf{j}+x\mathbf{k}

This vector field is conservative as is easily verified by finding curl v. In fact,

∇×v=∣ijk∂∂x∂∂y∂∂z2y+z2xx∣\nabla \times v = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y + z & 2x & x\end{vmatrix}

= 0i−(1−1)j+(2−2)k

= 0

Indeed, recall from Example 26.10 that curl (grad Φ) is identically zero for any Φ.

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