Question

If
A=x^z^3i−2^x^2yzj+2y^z^4
, find
▽×A
at point (1,-1,1).

2j+3k
\
2i+j74k

i+3j+5k

3j+4k

Answer 1

Answer on Question #56129 – Math – Vector Calculus

If $A = x^{z^3}i - 2^{x^2}yzj + 2yz^4k$, find $\nabla \times A$ at point $(1, -1, 1)$.

Solution

Vector (cross) product $\nabla \times A$ can be rewritten in matrix form and computed:

$$\begin{array}{l} \nabla \times \boldsymbol{A} = \left| \begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{z^3} & -2^{x^2}yz & 2yz^4 \end{array} \right| = \left(\frac{\partial}{\partial y} 2yz^4 + \frac{\partial}{\partial z} 2^{x^2}yz\right)\boldsymbol{i} - \left(\frac{\partial}{\partial x} 2yz^4 - \frac{\partial}{\partial z} x^{z^3}\right)\boldsymbol{j} + \\ + \left(-\frac{\partial}{\partial x} 2^{x^2}yz - \frac{\partial}{\partial y} x^{z^3}\right)\boldsymbol{k} = \left(2z^4 y^{z^4-1} + 2^{x^2}y\right)\boldsymbol{i} + \left(3z^2 x^{z^3} \ln x\right)\boldsymbol{j} - \left(xyz2^{x^2+1} \ln 2\right)\boldsymbol{k}. \end{array}$$

Now substituting point $(1, -1, 1)$ into the expression for $\nabla \times \boldsymbol{A}$:

$$\nabla \times \boldsymbol{A}_{(1, -1, 1)} = (2 - 2)\boldsymbol{i} + (3\ln 1)\boldsymbol{j} - (-4\ln 2)\boldsymbol{k} = 4\ln 2\boldsymbol{k}.$$

Answer:

$$\nabla \times \boldsymbol{A}_{(1, -1, 1)} = 4\ln 2\boldsymbol{k}.$$

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