．设函数u=x^3-xy^2-z,点M(1,1,0)则梯度gradu|M=

To find the gradient of a function at a point, we need to find the partial derivatives of the function with respect to each variable and evaluate them at the given point.

So,

∂u/∂x = 3x^2 - y^2
∂u/∂y = -2xy
∂u/∂z = -1

At point M(1,1,0), we have:

∂u/∂x|M = 3(1)^2 - (1)^2 = 2
∂u/∂y|M = -2(1)(1) = -2
∂u/∂z|M = -1

Therefore, the gradient of u at point M is:

grad u|M = (2, -2, -1)

相关问题

matlab绘制一标量场u=xy^2+yz^3的梯度箭头图，并求在点M（2，-1，1）处的梯度值

在MATLAB中，你可以使用`quiver3`函数来绘制三维空间中的标量场（如`u = xy^2 + yz^3`）的梯度箭头图。首先，你需要计算这个函数在各个网格点的梯度，然后确定箭头的方向和大小。以下是实现这一过程的步骤：

1. 定义标量场`u`：

[x, y, z] = meshgrid(-5:0.5:5, -5:0.5:5, -5:0.5:5); % 创建x, y, z的网格
u = x.*y.^2 + y.*z.^3; % 定义u(x, y, z)

2. 计算梯度向量`grad_u`：

[dx, dy, dz] = gradient(u, 'central'); % 使用中央差分法计算梯度

3. 绘制梯度箭头图：

figure;
quiver3(x(:), y(:), z(:), dx(:), dy(:), dz(:)); % 在三维空间显示箭头
hold on; % 保持当前图像以便添加更多的元素

4. 求解点M(2, -1, 1)处的梯度值：

[Mx, My, Mz] = deal(2, -1, 1); % 点M的坐标
[gradU_Mx, gradU_My, gradU_Mz] = gradient(u, Mx, My, Mz); % 在点M处的精确梯度值
disp(['在点M(2, -1, 1)处的梯度值为: ', num2str(gradU_Mx), ' i, ', num2str(gradU_My), ' j, ', num2str(gradU_Mz), ' k']);

完成以上代码后，你会看到标量场的梯度箭头图以及点M处的梯度数值。

证明 rot(gradu)=∇×∇u=0 div(rotA→)=∇⋅∇×A

1. 证明 $\operatorname{rot}(\operatorname{grad}u)=\nabla\times\nabla u=0$

根据向量分析中的恒等式，有：

$$
\begin{aligned}
\operatorname{rot}(\operatorname{grad}u) &= \nabla\times(\nabla u)\\
&=\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}\\
\end{vmatrix}\\
&=\left(\frac{\partial^2 u}{\partial y \partial z}-\frac{\partial^2 u}{\partial z \partial y}\right)\mathbf{i}+\left(\frac{\partial^2 u}{\partial z \partial x}-\frac{\partial^2 u}{\partial x \partial z}\right)\mathbf{j}+\left(\frac{\partial^2 u}{\partial x \partial y}-\frac{\partial^2 u}{\partial y \partial x}\right)\mathbf{k}\\
&=\mathbf{0}.
\end{aligned}
$$

因此，$\operatorname{rot}(\operatorname{grad}u)=\nabla\times\nabla u=0$。

2. 证明 $\operatorname{div}(\operatorname{rot}\mathbf{A})=\nabla\cdot(\nabla\times\mathbf{A})$

根据向量分析中的恒等式，有：

$$
\begin{aligned}
\operatorname{div}(\operatorname{rot}\mathbf{A}) &= \nabla\cdot(\nabla\times\mathbf{A})\\
&=\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
\frac{\partial}{\partial y} & \frac{\partial}{\partial z} & \frac{\partial}{\partial x}\\
A_x & A_y & A_z\\
\end{vmatrix}\\
&=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\\
&=\mathbf{0}.
\end{aligned}
$$

因此，$\operatorname{div}(\operatorname{rot}\mathbf{A})=\nabla\cdot(\nabla\times\mathbf{A})=0$。