Question

For the vectors □(□(→┬a )) =xyz□(→┬i )-2xz^2 □(→┬j )+xz□(→┬k ) and □(→┬b )=2z□(→┬i )+y□(→┬j )-x□(→┬k)
Find ∂^2/∂x∂y (□(→┬a )×□(→┬b )) at (2,0,-1)

Answer 1

Answer on Question #46340 – Math – Vector Calculus

Problem.

For the vectors $\square (\square (\rightarrow_{\mathsf{T}}\mathsf{a})) = xyz\square (\rightarrow_{\mathsf{T}}\mathsf{i}) - 2xz^2\square (\rightarrow_{\mathsf{T}}\mathsf{j}) + xz\square (\rightarrow_{\mathsf{T}}\mathsf{k})$ and $\square (\rightarrow_{\mathsf{T}}\mathsf{b}) = 2z\square (\rightarrow_{\mathsf{T}}\mathsf{i}) + y\square (\rightarrow_{\mathsf{T}}\mathsf{j}) - x\square (\rightarrow_{\mathsf{T}}\mathsf{k})$

Find $\partial^2 \frac{\partial x \partial y}{\partial (\rightarrow_{\mathsf{T}} \mathsf{a}) \times (\rightarrow_{\mathsf{T}} \mathsf{b})}$ at $(2,0,-1)$

Remark: I suppose that the statement is incorrectly formatted. The correct statement is:

"For the vectors $\vec{a} = xyz\vec{i} - 2xz^2\vec{j} + xz\vec{k}$ and $\vec{b} = 2z\vec{i} + y\vec{j} - x\vec{k}$

Find $\frac{\partial^2}{\partial x \partial y} (\vec{a} \times \vec{b})$ at $(2,0,-1)$"

Solution:

$$\begin{array}{l} \vec{a} \times \vec{b} = \det \left[ \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ xyz & -2xz^2 & xz \\ 2z & y & -x \end{array} \right] = (2x^2z^2 - xyz)\vec{i} + (x^2yz + 2z^2x)\vec{j} + (y^2xz + 4z^3x)\vec{k} \\ \frac{\partial^2}{\partial x \partial y} (\vec{a} \times \vec{b}) = -z\vec{i} + 2xz\vec{j} + 2yz\vec{k} \end{array}$$

For $(x,y,z) = (2,0,-1)$ we have $\frac{\partial^2}{\partial x\partial y} (\vec{a}\times \vec{b})(2,0, - 1) = \vec{i} -4\vec{j} +0\vec{k} = (1, - 4,0).$

Answer: $\vec{i} - 4\vec{j} + 0\vec{k} = (1, -4,0)$

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