求函数 u=xy+yz+xz 在点P(1,2,3)处沿P点的向径方向的方向导数.

步骤 1：计算偏导数
首先，我们需要计算函数 u=xy+yz+xz 在点 P(1,2,3) 处的偏导数。
- $\dfrac {\partial u}{\partial x} = y + z$
- $\dfrac {\partial u}{\partial y} = x + z$
- $\dfrac {\partial u}{\partial z} = x + y$

步骤 2：代入点 P(1,2,3)
将点 P(1,2,3) 的坐标代入偏导数中，得到：
- $\dfrac {\partial u}{\partial x}{|}_{p} = 2 + 3 = 5$
- $\dfrac {\partial u}{\partial y}{|}_{p} = 1 + 3 = 4$
- $\dfrac {\partial u}{\partial z}{|}_{p} = 1 + 2 = 3$

步骤 3：计算向径方向
向径方向 $\overrightarrow {i}=\overrightarrow {OP}=\{ 1,2,3\} $，需要将其单位化，得到单位向量 $\overrightarrow {i} = \left\langle \dfrac {1}{\sqrt {{1}^{2}+{2}^{2}+{3}^{2}}},\dfrac {2}{\sqrt {14}},\dfrac {3}{\sqrt {14}}\right\rangle = \left\langle \dfrac {1}{\sqrt {14}},\dfrac {2}{\sqrt {14}},\dfrac {3}{\sqrt {14}}\right\rangle$

步骤 4：计算方向导数
方向导数为梯度向量与单位向量的点积，即：
$\dfrac {\partial u}{\partial u}||=\langle 5,4,3\rangle \cdot \langle \dfrac {1}{\sqrt {14}},\dfrac {2}{\sqrt {14}},\dfrac {3}{\sqrt {14}}\rangle = 5 \cdot \dfrac {1}{\sqrt {14}} + 4 \cdot \dfrac {2}{\sqrt {14}} + 3 \cdot \dfrac {3}{\sqrt {14}} = \dfrac {5}{\sqrt {14}} + \dfrac {8}{\sqrt {14}} + \dfrac {9}{\sqrt {14}} = \dfrac {22}{\sqrt {14}}$