题目
[题目]函数 =ln (x+sqrt ({y)^2+(z)^2}) 在点A(1,0,1)处沿点A-|||-指向点 B(3,-2,2) 方向的方向导数为 __
题目解答
答案
解析
步骤 1：计算偏导数
首先，我们需要计算函数 $u=\ln (x+\sqrt {{y}^{2}+{z}^{2}})$ 在点A(1,0,1)处的偏导数。根据链式法则，我们有：
$$\dfrac {\partial u}{\partial x}=\dfrac {1}{x+\sqrt {{y}^{2}+{z}^{2}}}$$
$$\dfrac {\partial u}{\partial y}=\dfrac {y}{(x+\sqrt {{y}^{2}+{z}^{2}})\sqrt {{y}^{2}+{z}^{2}}}$$
$$\dfrac {\partial u}{\partial z}=\dfrac {z}{(x+\sqrt {{y}^{2}+{z}^{2}})\sqrt {{y}^{2}+{z}^{2}}}$$
步骤 2：计算偏导数在点A(1,0,1)处的值
将点A(1,0,1)的坐标代入上述偏导数表达式中，得到：
$$\dfrac {\partial u}{\partial x}|_{(1,0,1)}=\dfrac {1}{1+\sqrt {{0}^{2}+{1}^{2}}}=\dfrac {1}{2}$$
$$\dfrac {\partial u}{\partial y}|_{(1,0,1)}=\dfrac {0}{(1+\sqrt {{0}^{2}+{1}^{2}})\sqrt {{0}^{2}+{1}^{2}}}=0$$
$$\dfrac {\partial u}{\partial z}|_{(1,0,1)}=\dfrac {1}{(1+\sqrt {{0}^{2}+{1}^{2}})\sqrt {{0}^{2}+{1}^{2}}}=\dfrac {1}{2}$$
步骤 3：计算方向导数
方向导数的计算公式为：
$$\dfrac {\partial u}{\partial n}=\dfrac {\partial u}{\partial x}\cos \alpha +\dfrac {\partial u}{\partial y}\cos \beta +\dfrac {\partial u}{\partial z}\cos \gamma$$
其中，$\overrightarrow {n}=\{ \cos \alpha ,\cos \beta ,\cos \gamma \} $ 是从点A指向点B的单位向量。根据点A(1,0,1)和点B(3,-2,2)的坐标，我们得到向量$\overrightarrow {AB}=(2,-2,1)$。单位向量为：
$$\overrightarrow {n}=\dfrac {\overrightarrow {AB}}{|\overrightarrow {AB}|}=\dfrac {(2,-2,1)}{\sqrt {{2}^{2}+{(-2)}^{2}+{1}^{2}}}=\dfrac {1}{3}(2,-2,1)$$
因此，$\cos \alpha =\dfrac {2}{3}$，$\cos \beta =-\dfrac {2}{3}$，$\cos \gamma =\dfrac {1}{3}$。将这些值代入方向导数的计算公式中，得到：
$$\dfrac {\partial u}{\partial n}|_{(1,0,1)}=\dfrac {1}{2}\times \dfrac {2}{3}+0\times (-\dfrac {2}{3})+\dfrac {1}{2}\times \dfrac {1}{3}=\dfrac {1}{2}$$
首先，我们需要计算函数 $u=\ln (x+\sqrt {{y}^{2}+{z}^{2}})$ 在点A(1,0,1)处的偏导数。根据链式法则，我们有：
$$\dfrac {\partial u}{\partial x}=\dfrac {1}{x+\sqrt {{y}^{2}+{z}^{2}}}$$
$$\dfrac {\partial u}{\partial y}=\dfrac {y}{(x+\sqrt {{y}^{2}+{z}^{2}})\sqrt {{y}^{2}+{z}^{2}}}$$
$$\dfrac {\partial u}{\partial z}=\dfrac {z}{(x+\sqrt {{y}^{2}+{z}^{2}})\sqrt {{y}^{2}+{z}^{2}}}$$
步骤 2：计算偏导数在点A(1,0,1)处的值
将点A(1,0,1)的坐标代入上述偏导数表达式中，得到：
$$\dfrac {\partial u}{\partial x}|_{(1,0,1)}=\dfrac {1}{1+\sqrt {{0}^{2}+{1}^{2}}}=\dfrac {1}{2}$$
$$\dfrac {\partial u}{\partial y}|_{(1,0,1)}=\dfrac {0}{(1+\sqrt {{0}^{2}+{1}^{2}})\sqrt {{0}^{2}+{1}^{2}}}=0$$
$$\dfrac {\partial u}{\partial z}|_{(1,0,1)}=\dfrac {1}{(1+\sqrt {{0}^{2}+{1}^{2}})\sqrt {{0}^{2}+{1}^{2}}}=\dfrac {1}{2}$$
步骤 3：计算方向导数
方向导数的计算公式为：
$$\dfrac {\partial u}{\partial n}=\dfrac {\partial u}{\partial x}\cos \alpha +\dfrac {\partial u}{\partial y}\cos \beta +\dfrac {\partial u}{\partial z}\cos \gamma$$
其中，$\overrightarrow {n}=\{ \cos \alpha ,\cos \beta ,\cos \gamma \} $ 是从点A指向点B的单位向量。根据点A(1,0,1)和点B(3,-2,2)的坐标，我们得到向量$\overrightarrow {AB}=(2,-2,1)$。单位向量为：
$$\overrightarrow {n}=\dfrac {\overrightarrow {AB}}{|\overrightarrow {AB}|}=\dfrac {(2,-2,1)}{\sqrt {{2}^{2}+{(-2)}^{2}+{1}^{2}}}=\dfrac {1}{3}(2,-2,1)$$
因此，$\cos \alpha =\dfrac {2}{3}$，$\cos \beta =-\dfrac {2}{3}$，$\cos \gamma =\dfrac {1}{3}$。将这些值代入方向导数的计算公式中，得到：
$$\dfrac {\partial u}{\partial n}|_{(1,0,1)}=\dfrac {1}{2}\times \dfrac {2}{3}+0\times (-\dfrac {2}{3})+\dfrac {1}{2}\times \dfrac {1}{3}=\dfrac {1}{2}$$