If \(\vec{A}= xyi – 8xy^2z^2j + 2xyzk \). Find \( \vec{∇}\vec{A} \ at \ point (1,-2, 4)\)

Answer: \(\nabla\cdot\vec A = 506\) at (1, −2, 4).

Here \(A_x = xy\), \(A_y = -8xy^{2}z^{2}\), \(A_z = 2xyz\). Taking \(\nabla\vec A\) as the divergence \(\nabla\cdot\vec A\):

\[\nabla\cdot\vec A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}\]

\[= y - 16xyz^{2} + 2xy\]

At \((1, -2, 4)\):

• \(y = -2\)
• \(-16xyz^{2} = -16(1)(-2)(16) = 512\)
• \(2xy = 2(1)(-2) = -4\)

\[\nabla\cdot\vec A = -2 + 512 - 4 = 506\]

If the curl is required instead:

\[\nabla\times\vec A = (2xz + 16xy^{2}z)\hat i - 2yz\,\hat j - (8y^{2}z^{2} + x)\hat k\]

At \((1, -2, 4)\): \(\nabla\times\vec A = 264\hat i + 16\hat j - 513\hat k\).