Question

вычислить угол между $grad u M_{1}$ и $grad u M_{2}$? Условие в рисунке

Answer 1

Ответ:

Угол между $\text{grad}\ u(M_{1})$ и $\text{grad}\ u(M_{2})$ равен $\arccos \bigg( -\dfrac{1}{3} \bigg)$

Пошаговое объяснение:

$u = \text{arctg} \ \dfrac{x}{y + z}; \ M_{1} (1;1;0), M_{2} (-1;0;1);$

$\dfrac{\partial u}{\partial x} = \dfrac{\partial}{\partial x} \bigg (\text{arctg} \ \dfrac{x}{y + z} \bigg) = \dfrac{\dfrac{\partial}{\partial x} \bigg(\dfrac{x}{y + z} \bigg)}{1 + \bigg(\dfrac{x}{y + z} \bigg)^{2} } =$

$=\dfrac{\dfrac{\dfrac{\partial u}{\partial x} \bigg(x \bigg)\bigg(y + z \bigg) - \dfrac{\partial u}{\partial x} \bigg(y + z \bigg)\bigg(x \bigg)}{(y + z)^{2}}}{1 + \dfrac{x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{1(y + z) - 0 \cdot x}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{(y + z)}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} }=$

$= \dfrac{(y + z)^{2}(y + z)}{(y + z)^{2}((y + z)^{2} + x^{2} )} = \dfrac{y + z}{(y + z)^{2} + x^{2}}$

$\dfrac{\partial u}{\partial y} = \dfrac{\partial}{\partial y} \bigg (\text{arctg} \ \dfrac{x}{y + z} \bigg) = \dfrac{\dfrac{\partial}{\partial y} \bigg(\dfrac{x}{y + z} \bigg)}{1 + \bigg(\dfrac{x}{y + z} \bigg)^{2} } =$

$=\dfrac{\dfrac{\dfrac{\partial u}{\partial y} \bigg(x \bigg)\bigg(y + z \bigg) - \dfrac{\partial u}{\partial y} \bigg(y + z \bigg)\bigg(x \bigg)}{(y + z)^{2}}}{1 + \dfrac{x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{0\cdot(y + z) - 1 \cdot x}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{-x}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} }=$

$= -\dfrac{x(y + z)^{2}}{(y + z)^{2}((y + z)^{2} + x^{2} )} = -\dfrac{x}{(y + z)^{2} + x^{2}}$

$\dfrac{\partial u}{\partial z} = \dfrac{\partial}{\partial z} \bigg (\text{arctg} \ \dfrac{x}{y + z} \bigg) = \dfrac{\dfrac{\partial}{\partial z} \bigg(\dfrac{x}{y + z} \bigg)}{1 + \bigg(\dfrac{x}{y + z} \bigg)^{2} } =$

$=\dfrac{\dfrac{\dfrac{\partial u}{\partial z} \bigg(x \bigg)\bigg(y + z \bigg) - \dfrac{\partial u}{\partial z} \bigg(y + z \bigg)\bigg(x \bigg)}{(y + z)^{2}}}{1 + \dfrac{x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{0 \cdot(y + z) - 1 \cdot x}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} } = \dfrac{\dfrac{-x}{(y + z)^{2}} }{\dfrac{(y + z)^{2} +x^{2}}{(y + z)^{2}} }=$

$= -\dfrac{x(y + z)^{2}}{(y + z)^{2}((y + z)^{2} + x^{2} )} = -\dfrac{x}{(y + z)^{2} + x^{2}}$

$\text{grad}\ u =\dfrac{\partial u}{\partial x} \ \overrightarrow{i} + \dfrac{\partial u}{\partial y} \ \overrightarrow{j} + \dfrac{\partial u}{\partial z} \ \overrightarrow{k} =$

$=\dfrac{y + z}{(y + z)^{2} + x^{2}} \ \overrightarrow{i} -\dfrac{x}{(y + z)^{2} + x^{2}} \ \overrightarrow{j} -\dfrac{x}{(y + z)^{2} + x^{2}} \ \overrightarrow{k}$

$\text{grad}\ u(M_{1}) = \dfrac{1 + 0}{(1 + 0)^{2} + 1^{2}} \ \overrightarrow{i} -\dfrac{1}{(1 + 0)^{2} + 1^{2}} \ \overrightarrow{j} -\dfrac{1}{(1 + 0)^{2} + 1^{2}} \ \overrightarrow{k} =$

$= \dfrac{1}{2} \overrightarrow{i} - \dfrac{1}{2} \overrightarrow{j} - \dfrac{1}{2} \overrightarrow{k}$

$\text{grad}\ u(M_{2}) =\dfrac{0 + 1}{(0 + 1)^{2} + (-1)^{2}} \ \overrightarrow{i} -\dfrac{-1}{(0 + 1)^{2} + (-1)^{2}} \ \overrightarrow{j} -\dfrac{-1}{(0 + 1)^{2} + (-1)^{2}} \ \overrightarrow{k}=$

$= \dfrac{1}{2} \overrightarrow{i} + \dfrac{1}{2} \overrightarrow{j} + \dfrac{1}{2} \overrightarrow{k}$

Пусть:

$\displaystyle \left \{ {{\overrightarrow{a} = \dfrac{1}{2} \overrightarrow{i} - \dfrac{1}{2} \overrightarrow{j} - \dfrac{1}{2} \overrightarrow{k} } \atop {\overrightarrow{b} = \dfrac{1}{2} \overrightarrow{i} + \dfrac{1}{2} \overrightarrow{j} + \dfrac{1}{2} \overrightarrow{k} }} \right \Longleftrightarrow \left \{ {{\overrightarrow{a} \bigg( \dfrac{1}{2};-\dfrac{1}{2};-\dfrac{1}{2} \bigg)} \atop {\overrightarrow{b} \bigg( \dfrac{1}{2};\dfrac{1}{2};\dfrac{1}{2} \bigg)}} \right.$

$\cos \angle (\overrightarrow{a},\overrightarrow{b}) = \dfrac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| \cdot |\overrightarrow{b}|} = \dfrac{a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} }{\sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2} } \cdot \sqrt{b_{x}^{2} + b_{y}^{2} + b_{z}^{2} } } \Longrightarrow$

$\Longrightarrow \angle (\overrightarrow{a},\overrightarrow{b}) = \arccos \dfrac{a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} }{\sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2} } \cdot \sqrt{b_{x}^{2} + b_{y}^{2} + b_{z}^{2} } } =$

$= \arccos \dfrac{ \dfrac{1}{2} \cdot \dfrac{1}{2} - \dfrac{1}{2} \cdot \dfrac{1}{2} - \dfrac{1}{2} \cdot \dfrac{1}{2} }{\sqrt{ \bigg(\dfrac{1}{2}\bigg )^{2} + \bigg(\dfrac{1}{2}\bigg )^{2} + \bigg(\dfrac{1}{2}\bigg )^{2} } \cdot \sqrt{\bigg(\dfrac{1}{2}\bigg )^{2} + \bigg(-\dfrac{1}{2}\bigg )^{2}+ \bigg(-\dfrac{1}{2}\bigg )^{2} } } =$

$= \arccos \dfrac{-\dfrac{1}{4} }{\sqrt{\dfrac{3}{4} } \cdot \sqrt{\dfrac{3}{4} }} =\arccos \dfrac{-\dfrac{1}{4} }{\dfrac{3}{4} } = \arccos \bigg(-\dfrac{1 \cdot 4}{3 \cdot 4} \bigg) = \arccos \bigg( -\dfrac{1}{3} \bigg)$

Answer 2

Ответ:

≈109°

Пошаговое объяснение:

во вложении менее детализированная версия ответа.