同济高数 第9章 第9-3-3题
📝 题目
3.求函数 $\displaystyle z=\frac{y}{x}$ 当 $x=2, y=1, \Delta x=0.1, \Delta y=-0.2$ 时的全增量和全微分.
💡 答案解析
[AI解答]
[AI解答]
第一步:计算全增量
全增量定义为
$$
\Delta z = f(x+\Delta x, y+\Delta y) - f(x, y)
$$
代入 $\displaystyle f(x,y)=\frac{y}{x}$,已知 $x=2, y=1, \Delta x=0.1, \Delta y=-0.2$,则
$$
x+\Delta x = 2+0.1 = 2.1, \quad y+\Delta y = 1 + (-0.2) = 0.8
$$
于是
$$
\Delta z = \frac{0.8}{2.1} - \frac{1}{2}
$$
先计算 $\displaystyle \frac{0.8}{2.1} = \frac{8}{21}$,$\displaystyle \frac{1}{2} = \frac{1}{2}$,通分:
$$
\frac{8}{21} - \frac{1}{2} = \frac{16}{42} - \frac{21}{42} = -\frac{5}{42}
$$
因此全增量为
$$
\boxed{\Delta z = -\frac{5}{42}}
$$
第二步:计算全微分
全微分公式为
$$
\mathrm{d}z = \frac{\partial z}{\partial x}\,\mathrm{d}x + \frac{\partial z}{\partial y}\,\mathrm{d}y
$$
先求偏导数:
$$
\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}\left(\frac{y}{x}\right) = -\frac{y}{x^2}, \quad \frac{\partial z}{\partial y} = \frac{1}{x}
$$
在点 $(x,y)=(2,1)$ 处:
$$
\left.\frac{\partial z}{\partial x}\right|_{(2,1)} = -\frac{1}{2^2} = -\frac{1}{4}, \quad \left.\frac{\partial z}{\partial y}\right|_{(2,1)} = \frac{1}{2}
$$
取 $\mathrm{d}x = \Delta x = 0.1$,$\mathrm{d}y = \Delta y = -0.2$,则
$$
\mathrm{d}z = \left(-\frac{1}{4}\right)\times 0.1 + \frac{1}{2}\times (-0.2) = -0.025 - 0.1 = -0.125
$$
即
$$
\boxed{\mathrm{d}z = -0.125}
$$
难度评级:★☆☆☆☆
(直接套用全增量与全微分定义,计算简单,无复杂技巧。)