Задача 62132 Найти угол между градиентами функций u...
Условие
Решение
Находим
[m]\frac{ ∂u }{ ∂x }=(\frac{z^3}{xy^2})`_{x}=\frac{z^3}{y^2}\cdot( (x^{-1})`=\frac{z^3}{y^2} (-x^{-2})=-\frac{z^3}{x^2y^2}[/m]
[m]\frac{ ∂u }{ ∂y }=(\frac{z^3}{xy^2})`_{y}=\frac{z^3}{x}\cdot(y^{-2})`=\frac{z^3}{x}\cdot (-2y^{-3})=-\frac{2z^3}{xy^3}[/m]
[m]\frac{ ∂u }{ ∂z }=(\frac{z^3}{xy^2})`_{z}=\frac{1}{xy^2}\cdot (z^3)`=\frac{1}{xy^2}\cdot 3z^2=\frac{3z^2}{xy^2}[/m]
[m]\vec{grad u}|_{M}=\frac{ ∂u }{ ∂x }|_{M}\vec{i}+\frac{ ∂u }{ ∂y }|_{M}\vec{j}+\frac{ ∂u }{ ∂z }|_{M}\vec{k}[/m]
[m]\frac{ ∂u }{ ∂x }|_{M}=-\frac{(\sqrt{\frac{3}{2}})^3}{(\frac{1}{3})^2\cdot 2^2}=-\frac{27\sqrt{3}}{8\sqrt{2}}[/m]
[m]\frac{ ∂u }{ ∂y }|_{M}=-\frac{2(\sqrt{\frac{3}{2}})^3}{\frac{1}{3}\cdot 2^3}=\frac{9\sqrt{3}}{8\sqrt{2}}[/m]
[m]\frac{ ∂u }{ ∂z }|_{M}=\frac{3(\sqrt{\frac{3}{2}})^2}{\frac{1}{3}\cdot 2^2}=\frac{27}{8}[/m]
[red][m]\vec{grad u}|_{M}=-\frac{27\sqrt{3}}{8\sqrt{2}}\vec{i}+\frac{9\sqrt{3}}{8\sqrt{2}}\vec{j}+\frac{27}{8}\vec{k}[/m][/red]
Аналогично
[m]\vec{grad v}=\frac{ ∂v }{ ∂x }\vec{i}+\frac{ ∂v }{ ∂y }\vec{j}+\frac{ ∂v }{ ∂z }\vec{k}[/m]
Находим
[m]\frac{ ∂v }{ ∂x }=(9\sqrt{2}x^3-\frac{y^3}{2\sqrt{2}}-\frac{4z^3}{\sqrt{3}})`_{x}=9\sqrt{2}\cdot 3x^2=27\sqrt{2}x^2[/m]
[m]\frac{ ∂v }{ ∂y }=(9\sqrt{2}x^3-\frac{y^3}{2\sqrt{2}}-\frac{4z^3}{\sqrt{3}})`_{y}=-\frac{1}{2\sqrt{2}}\cdot 3y^2=-\frac{3y^2}{2\sqrt{2}}[/m]
[m]\frac{ ∂v }{ ∂z }=(9\sqrt{2}x^3-\frac{y^3}{2\sqrt{2}}-\frac{4z^3}{\sqrt{3}})`_{z}=-\frac{4}{\sqrt{3}}\cdot 3z^2=-\frac{12}{\sqrt{3}}z^2[/m]
[m]\vec{grad v}|_{M}=\frac{ ∂v }{ ∂x }|_{M}\vec{i}+\frac{ ∂v }{ ∂y }|_{M}\vec{j}+\frac{ ∂v }{ ∂z }|_{M}\vec{k}[/m]
[m]\frac{ ∂v }{ ∂x }|_{M}=27\sqrt{2}(\frac{1}{3})^2=3\sqrt{2}[/m]
[m]\frac{ ∂v }{ ∂y }|_{M}=-\frac{3}{2\sqrt{2}}\cdot 2^2=-3\sqrt{2}[/m]
[m]\frac{ ∂v }{ ∂z }|_{M}=-\frac{12}{\sqrt{3}}\cdot (\sqrt{\frac{3}{2}})^2=-6\sqrt{3}[/m]
[red][m]\vec{grad v}|_{M}=3\sqrt{2}\vec{i}-3\sqrt{2}\vec{j}-6\sqrt{3}\vec{k}[/m][/red]
Находим угол между векторами
[red][m]\vec{grad u}|_{M}=-\frac{27\sqrt{3}}{8\sqrt{2}}\vec{i}+\frac{9\sqrt{3}}{8\sqrt{2}}\vec{j}+\frac{27}{8}\vec{k}[/m][/red]
[red][m]\vec{grad v}|_{M}=3\sqrt{2}\vec{i}-3\sqrt{2}\vec{j}-6\sqrt{3}\vec{k}[/m][/red]
по формуле:
