📝 题目
14.证明函数 $u=\frac{1}{r}$ 满足拉普拉斯方程:
$$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}=0 $$
其中 $r=\sqrt{x^{2}+y^{2}+z^{2}}$ . * 15 .求 $\sqrt{1.02^{3}+1.97^{3}}$ 的近似值.
💡 答案与解析
[AI解答]
[AI解答]
**14. 证明函数 $ u = \frac{1}{r} $ 满足拉普拉斯方程**
已知 $ r = \sqrt{x^2 + y^2 + z^2} $,则 $$ u = \frac{1}{r} = (x^2 + y^2 + z^2)^{-1/2}. $$
先求一阶偏导: $$ \frac{\partial u}{\partial x} = -\frac{1}{2}(x^2 + y^2 + z^2)^{-3/2} \cdot 2x = -\frac{x}{r^3}. $$ 同理: $$ \frac{\partial u}{\partial y} = -\frac{y}{r^3},\quad \frac{\partial u}{\partial z} = -\frac{z}{r^3}. $$
再求二阶偏导。先对 $ x $: $$ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left( -\frac{x}{r^3} \right). $$ 使用商法则或乘积法则,将 $ -\frac{x}{r^3} $ 看作 $ -x \cdot r^{-3} $: $$ \frac{\partial}{\partial x}(-x) = -1,\quad \frac{\partial}{\partial x}(r^{-3}) = -3 r^{-4} \cdot \frac{x}{r} = -\frac{3x}{r^5}. $$ 所以: $$ \frac{\partial^2 u}{\partial x^2} = -1 \cdot r^{-3} + (-x) \cdot \left( -\frac{3x}{r^5} \right) = -\frac{1}{r^3} + \frac{3x^2}{r^5}. $$
同理: $$ \frac{\partial^2 u}{\partial y^2} = -\frac{1}{r^3} + \frac{3y^2}{r^5},\quad \frac{\partial^2 u}{\partial z^2} = -\frac{1}{r^3} + \frac{3z^2}{r^5}. $$
相加得: $$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = -\frac{3}{r^3} + \frac{3(x^2 + y^2 + z^2)}{r^5}. $$ 由于 $ x^2 + y^2 + z^2 = r^2 $,所以: $$ = -\frac{3}{r^3} + \frac{3r^2}{r^5} = -\frac{3}{r^3} + \frac{3}{r^3} = 0. $$ 因此 $ u = \frac{1}{r} $ 满足拉普拉斯方程。
**难度:★★☆☆☆**
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**15. 求 $\sqrt{1.02^{3} + 1.97^{3}}$ 的近似值**
考虑函数 $$ f(x,y) = \sqrt{x^3 + y^3}. $$ 取 $ x_0 = 1,\ y_0 = 2 $,则 $$ f(1,2) = \sqrt{1 + 8} = \sqrt{9} = 3. $$
计算偏导数: $$ f_x = \frac{3x^2}{2\sqrt{x^3 + y^3}},\quad f_y = \frac{3y^2}{2\sqrt{x^3 + y^3}}. $$ 在 $(1,2)$ 处: $$ f_x(1,2) = \frac{3 \cdot 1^2}{2 \cdot 3} = \frac{3}{6} = \frac{1}{2}, \quad f_y(1,2) = \frac{3 \cdot 4}{2 \cdot 3} = \frac{12}{6} = 2. $$
取增量 $\Delta x = 0.02,\ \Delta y = -0.03$,则近似值: $$ f(1.02, 1.97) \approx f(1,2) + f_x(1,2)\Delta x + f_y(1,2)\Delta y = 3 + \frac{1}{2} \cdot 0.02 + 2 \cdot (-0.03). $$ 计算: $$ \frac{1}{2} \cdot 0.02 = 0.01,\quad 2 \cdot (-0.03) = -0.06, $$ 所以: $$ 3 + 0.01 - 0.06 = 2.95. $$
因此近似值为 $ 2.95 $。
**难度:★★☆☆☆**