同济高数 第9章 第9-4-7题

[AI解答]

[AI解答]

已知 $ z = \arctan \frac{x}{y} $，且 $ x = u+v $，$ y = u-v $。

首先，由链式法则： $$ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} $$ $$ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} $$

计算偏导数： $$ \frac{\partial z}{\partial x} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \frac{1}{y} = \frac{y}{x^2 + y^2} $$ $$ \frac{\partial z}{\partial y} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \left(-\frac{x}{y^2}\right) = -\frac{x}{x^2 + y^2} $$

又： $$ \frac{\partial x}{\partial u} = 1,\quad \frac{\partial x}{\partial v} = 1 $$ $$ \frac{\partial y}{\partial u} = 1,\quad \frac{\partial y}{\partial v} = -1 $$

于是： $$ \frac{\partial z}{\partial u} = \frac{y}{x^2+y^2} \cdot 1 + \left(-\frac{x}{x^2+y^2}\right) \cdot 1 = \frac{y - x}{x^2+y^2} $$ $$ \frac{\partial z}{\partial v} = \frac{y}{x^2+y^2} \cdot 1 + \left(-\frac{x}{x^2+y^2}\right) \cdot (-1) = \frac{y + x}{x^2+y^2} $$

相加： $$ \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v} = \frac{(y-x)+(y+x)}{x^2+y^2} = \frac{2y}{x^2+y^2} $$

代入 $ x = u+v $，$ y = u-v $： $$ x^2+y^2 = (u+v)^2 + (u-v)^2 = (u^2+2uv+v^2)+(u^2-2uv+v^2) = 2(u^2+v^2) $$ $$ 2y = 2(u-v) $$

因此： $$ \frac{\partial z}{\partial u}+\frac{\partial z}{\partial v} = \frac{2(u-v)}{2(u^2+v^2)} = \frac{u-v}{u^2+v^2} $$

验证成立。