Алгебра: только быстро надо с решением

только быстро надо с решением

Ответы

Празеодим 17.04.2021 09:47

$u=x^2yz^3 \\ \\ \frac{\partial u}{\partial x} =(x^2yz^3)'_x=2xyz^3 \\ \\ \frac{\partial u}{\partial y} =(x^2yz^3)'_y=x^2z^3 \\ \\ \frac{\partial u}{\partial z} =(x^2yz^3)'_z=3x^2yz^2 \\ \\ M(2, \frac{1}{3}, \sqrt{\frac{3}{2}})$

$(\frac{\partial u}{\partial x} )_M=2\cdot2\cdot\frac{1}{3}\cdot (\sqrt{\frac{3}{2}})^3=\frac{4}{3}\cdot \frac{3\sqrt{3}}{2\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6} \\ \\ (\frac{\partial u}{\partial y} )_M=2^2\cdot (\sqrt{\frac{3}{2}})^3=4\cdot \frac{3\sqrt{3}}{2\sqrt{2}}=\frac{6\sqrt{3}}{\sqrt{2}}=3\sqrt{6} \\ \\ (\frac{\partial u}{\partial z} )_M=3\cdot 2^2\cdot \frac{1}{3}\cdot (\sqrt{\frac{3}{2}})^2=4\cdot \frac{3}{2}=6$

$grad \, u=(\sqrt{6};3\sqrt{6};6)$

$\upsilon=\frac{4\sqrt{6}}{x} -\frac{\sqrt{6}}{9y}+\frac{3}{z}\\ \\ \frac{\partial \upsilon}{\partial x} =(\frac{4\sqrt{6}}{x} - \frac{\sqrt{6}}{9y}+\frac{3}{z})'_x=-\frac{4\sqrt{6}}{x^2} \\ \\ \frac{\partial \upsilon}{\partial y} =(\frac{4\sqrt{6}}{x} -\frac{\sqrt{6}}{9y}+\frac{3}{z})'_y=-(-\frac{\sqrt{6}}{9y^2})=\frac{\sqrt{6}}{9y^2} \\ \\ \frac{\partial \upsilon}{\partial z} =(\frac{4\sqrt{6}}{x} -\frac{\sqrt{6}}{9y}+\frac{3}{z})'_z=-\frac{3}{z^2} \\ \\ M(2, \frac{1}{3}, \sqrt{\frac{3}{2}})$

$(\frac{\partial \upsilon}{\partial x})_M=-\frac{4\sqrt{6}}{2^2}=-\sqrt{6} \\\\(\frac{\partial \upsilon}{\partial y})_M=\frac{\sqrt{6}}{9\cdot (\frac{1}{3})^2}=\sqrt{6} \\ \\ (\frac{\partial \upsilon}{\partial z})_M=-\frac{3}{(\sqrt{\frac{3}{2}})^2}=-2$

$grad \, \upsilon=(-\sqrt{6};\sqrt{6};-2)$

$\cos{\varphi}=\frac{\overrightarrow{a} \cdot \overrightarrow{b} }{|\overrightarrow{a} |\cdot |\overrightarrow{b} |} \\ \\ \cos{\varphi}=\frac{a_x\cdot b_x+a_y\cdot b_y+a_z\cdot b_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot \sqrt{b_x^2+b_y^2+b_z^2}} \\ \\ \cos{\varphi}=\frac{\sqrt{6}\cdot( -\sqrt{6})+3\sqrt{6}\cdot \sqrt{6}+6 \cdot (-2)}{\sqrt{(\sqrt{6})^2+(3\sqrt{6})^2+6^2}\cdot \sqrt{(-\sqrt{6})^2+(\sqrt{6})^2+(-2)^2}}=$