题目
(2018)设 (x,y,z)=xyi-yzj+zxk, 则 rotF(1,1,0)= __ .

题目解答
答案

解析
步骤 1:计算 rotF 的表达式
rotF 是向量场 F 的旋度,其计算公式为:
rotF = $\nabla \times F$ = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy & -yz & zx \end{vmatrix}$
步骤 2:计算行列式
根据行列式的计算方法,我们有:
rotF = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy & -yz & zx \end{vmatrix}$ = $\left( \frac{\partial (zx)}{\partial y} - \frac{\partial (-yz)}{\partial z} \right)i - \left( \frac{\partial (zx)}{\partial x} - \frac{\partial (xy)}{\partial z} \right)j + \left( \frac{\partial (-yz)}{\partial x} - \frac{\partial (xy)}{\partial y} \right)k$
步骤 3:计算偏导数
计算上述偏导数,我们得到:
rotF = $\left( 0 - (-y) \right)i - \left( z - 0 \right)j + \left( 0 - x \right)k$ = $yi - zj - xk$
步骤 4:代入点 (1,1,0)
将点 (1,1,0) 代入 rotF 的表达式中,我们得到:
rotF(1,1,0) = $1i - 0j - 1k$ = $i - k$
rotF 是向量场 F 的旋度,其计算公式为:
rotF = $\nabla \times F$ = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy & -yz & zx \end{vmatrix}$
步骤 2:计算行列式
根据行列式的计算方法,我们有:
rotF = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy & -yz & zx \end{vmatrix}$ = $\left( \frac{\partial (zx)}{\partial y} - \frac{\partial (-yz)}{\partial z} \right)i - \left( \frac{\partial (zx)}{\partial x} - \frac{\partial (xy)}{\partial z} \right)j + \left( \frac{\partial (-yz)}{\partial x} - \frac{\partial (xy)}{\partial y} \right)k$
步骤 3:计算偏导数
计算上述偏导数,我们得到:
rotF = $\left( 0 - (-y) \right)i - \left( z - 0 \right)j + \left( 0 - x \right)k$ = $yi - zj - xk$
步骤 4:代入点 (1,1,0)
将点 (1,1,0) 代入 rotF 的表达式中,我们得到:
rotF(1,1,0) = $1i - 0j - 1k$ = $i - k$