习题1.6

第 2 题

下列方程中, 在哪些点附近可以确定一个函数 \(y = y(x)\) 或 \(z = z(x, y)\), 并求出相应的 \(\dfrac{\text{d}{y}}{\text{d}{x}}\) 或 \(\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}\).

(1) \((x^2 + y^2)^2 = a^2(y^2 - x^2)\);

(2) \(e^{-(x + y + z)} = x + y + z\);

解

(1) 设 \(F(x, y) = (x^2 + y^2)^2 - a^2(y^2 - x^2)\). 在满足 \(y_0 \neq 0, x_0^2 + y_0^2 \neq \dfrac{a^2}{2}\) 的点附近可以确定 \(y = y(x)\), 且 \(\dfrac{\text{d}{y}}{\text{d}{x}} = -\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}} = -\dfrac{2x^3 + (2y^2 + a^2)x}{2y^3 + (2x^2 - a^2)y}\).

(2) 设 \(F(x, y, z) = e^{-(x + y + z)} - (x + y + z)\). 在 \(\mathbb{R}^2\) 上的所有点附近都可以确定 \(z = z(x, y)\), 且 \(\dfrac{\partial z}{\partial x} = -\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial z}} = -\dfrac{-e^{-(x + y + z)} - 1}{-e^{-(x + y + z)} - 1} = -1, \dfrac{\partial z}{\partial y} = -\dfrac{\dfrac{\partial F}{\partial y}}{\dfrac{\partial F}{\partial z}} = -\dfrac{-e^{-(x + y + z)} - 1}{-e^{-(x + y + z)} - 1} = -1\).

第 3 题

下列方程均确定了函数 \(z = z(x, y)\), 分别求解下列各表达式的值.

(1) \(f(ax - cz, ay - bz) = 0\), \(f\) 可微, 计算: \(c\dfrac{\partial z}{\partial x} + b\dfrac{\partial z}{\partial y}\);

(3) \(f(x, x + y, x + y + z) = 0\), \(f\) 二阶可微, 计算: \(\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}, \dfrac{\partial ^2z}{\partial x^2}\).

解

(1) \(\dfrac{\partial z}{\partial x} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial z}} = -\dfrac{f_1'\cdot a}{f_1'\cdot (-c) + f_2'\cdot (-b)} = \dfrac{f_1'\cdot a}{f_1'\cdot c + f_2'\cdot b}\). 同理 \(\dfrac{\partial z}{\partial y} = \dfrac{f_2'\cdot a}{f_1'\cdot c + f_2'\cdot b}\). 故 \(c\dfrac{\partial z}{\partial x} + b\dfrac{\partial z}{\partial y} = \dfrac{f_1'\cdot ac}{f_1'\cdot c + f_2'\cdot b} + \dfrac{f_2'\cdot ab}{f_1'\cdot c + f_2'\cdot b} = a\).

(3) \(\dfrac{\partial z}{\partial x} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial z}} = -\dfrac{f_1' + f_2' + f_3'}{f_3'}\), \(\dfrac{\partial z}{\partial y} = -\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial f}{\partial z}} = -\dfrac{f_2' + f_3'}{f_3'}\). 故

\[ \begin{aligned} \dfrac{\partial ^2z}{\partial x^2} &= \dfrac{\partial }{\partial x}\left(\dfrac{\partial z}{\partial x}\right) \\ & = -\dfrac{[(f_{11}'' + f_{12}'' + f_{13}''\cdot\dfrac{\partial z}{\partial x}) + (f_{21}'' + f_{22}'' + f_{23}''\cdot\dfrac{\partial z}{\partial x}) + (f_{31}'' + f_{32}'' + f_{33}''\cdot\dfrac{\partial z}{\partial x})]\cdot f_3'}{(f_3')^2} \\ & - \dfrac{(f_1' + f_2' + f_3')(f_{31}'' + f_{32}'' + f_{33}''\dfrac{\partial z}{\partial x})}{(f_3')^2} \\ & = - \dfrac{(f_{11}'' + 2f_{12}'' + f_{22}'')}{f_3'} + \dfrac{2(f_1' + f_2')(f_{13}'' + f_{23}'')}{(f_3')^2} - \dfrac{(f_1' + f_2')^2f_{33}''}{(f_3')^3} \end{aligned} \]

第 4 题

设方程 \(f(u^2 - x^2, u^2 - y^2, u^2 - z^2) = 0\) 确定了函数 \(u = u(x, y, z)\), 其中 \(f\) 可微, 证明:

\[ \dfrac{1}{x}\dfrac{\partial u}{\partial x} + \dfrac{1}{y}\dfrac{\partial u}{\partial y} + \dfrac{1}{z}\dfrac{\partial u}{\partial z} = \dfrac{1}{u}. \]

证明

\[ \dfrac{\partial u}{\partial x} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial u}} = -\dfrac{f_x'\cdot(-2x)}{f_x'\cdot (2u) + f_y'\cdot (2u) + f_z'\cdot (2u)} = \dfrac{x}{u}\cdot\dfrac{f_x'}{f_x' + f_y' + f_z'} \]

同理有 \(\dfrac{\partial u}{\partial y} = \dfrac{y}{u}\cdot\dfrac{f_y'}{f_x' + f_y' + f_z'}, \dfrac{\partial u}{\partial z} = \dfrac{z}{u}\cdot\dfrac{f_z'}{f_x' + f_y' + f_z'}\). 故 \(\dfrac{1}{x}\dfrac{\partial u}{\partial x} + \dfrac{1}{y}\dfrac{\partial u}{\partial y} + \dfrac{1}{z}\dfrac{\partial u}{\partial z} = \dfrac{1}{u}\cdot\dfrac{f_x' + f_y' + f_z'}{f_x' + f_y' + f_z'} = \dfrac{1}{u}\). 命题得证.

第 5 题

方程组 \(\begin{cases}x = u + v, \\ y = u - v, \\ z = u^2v^2\end{cases}\) 能否确定 \(z\) 是 \(x, y\) 的函数? 如果能, 求 \(\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}\); 如果不能, 说明理由.

解

由方程组可解得

\[ \begin{cases} u = \dfrac{x + y}{2}, \\ v = \dfrac{x - y}{2}. \end{cases} \]

因此 \(z = u^2v^2 = \dfrac{1}{16}(x^4 - 2x^2y^2 + y^4)\) 是 \(x, y\) 的函数. \(\dfrac{\partial z}{\partial x} = \dfrac{1}{4}(x^3 - xy^2)\), \(\dfrac{\partial z}{\partial y} = \dfrac{1}{4}(y^3 - x^2y)\).

第 6 题

方程组 \(\begin{cases}x + y + z + z^2 = 0, \\ x + y^2 + z + z^3 = 0\end{cases}\) 在点 \(P(-1, 1, 0)\) 附近能否确定向量值函数 \(\begin{pmatrix}y\\z\end{pmatrix}=\mathbf{f}(x)\), 如果能确定, 求出 \(y'(-1), z'(-1)\).

解

记 \(F(x, y, z) = x + y + z + z^2, G(x, y, z) = x + y^2 + z + z^3\). 计算得 \(\dfrac{\partial(F, G)}{\partial(y, z)} = \begin{pmatrix}1 & 1 + 2z \\ 2y & z + 3z^2\end{pmatrix}\). 由于 \(\dfrac{\partial(F, G)}{\partial(y, z)}\bigg\vert_{(1, 0)} = \begin{pmatrix}1 & 1 \\ 2 & 0\end{pmatrix}\) 可逆, 故能确定向量值函数 \(\mathbf{f}(x)\). 且有 \(y(-1) = 1, z(-1) = 0\). 对 \(\begin{cases}F(x, y, z) = 0, \\ G(x, y, z) = 0\end{cases}\) 求导得 \(\begin{cases}1 + y'(x) + z'(x) + 2z(x)z'(x) = 0, \\ 1 + 2y(x)y'(x) + z'(x) + 3z^2(x)z'(x) = 0.\end{cases}\) 将 \(x = 1\) 代入有 \(\begin{cases}1 + y'(-1) + z'(-1) = 0, \\1 + 2y'(-1) + z'(-1) = 0.\end{cases}\) 解得 \(y'(-1) = 0, z'(-1) = -1\).

第 9 题

求下列向量值函数的逆映射的 Jacobi 矩阵以及 Jacobi 行列式.

(1) \(\begin{cases}u = x^2 - y^2, \\ v = 2xy;\end{cases}\)

(3) \(\begin{cases}u = x^3 - y^3, \\ v = xy^2;\end{cases}\)

解

(1) \(J = \begin{pmatrix}\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}\end{pmatrix}^{-1} = \begin{pmatrix}2x & -2y \\ 2y & 2x\end{pmatrix}^{-1} = \dfrac{1}{2(x^2 + y^2)}\begin{pmatrix}x & y \\ -y & x\end{pmatrix}\). \(\left|J\right| = \dfrac{1}{4(x^2 + y^2)}\).

(3) \(J = \begin{pmatrix}\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}\end{pmatrix}^{-1} = \begin{pmatrix}3x^2 & -3y^2 \\ y^2 & 2xy\end{pmatrix}^{-1} = \dfrac{1}{6x^3y + 3y^4}\begin{pmatrix}2xy & 3y^2 \\ -y^2 & 3x^2\end{pmatrix}\). \(\left|J\right| = \dfrac{1}{6x^3y + 3y^4}\).

第 10 题

下列由可微向量值函数 \(\mathbf{g}:(\xi, \eta)\mapsto(u, v)\) 和 \(\mathbf{f}:(x, y)\mapsto(\xi, \eta)\) 复合而成的复合向量值函数 \(\mathbf{g}\circ\mathbf{f}\) 在 \((x_0, y_0)\) 的邻域内能否确定可微的逆向量值函数 \((\mathbf{g}\circ\mathbf{f})^{-1}\)?

(1) \(\begin{cases}u = \xi^2 - \eta^2, \\ v = 2\xi\eta,\end{cases}\begin{cases}\xi = e^x\cos y,\\ \eta = e^x\sin y,\end{cases}(x_0, y_0) = (1, 0)\);

(3) \(\begin{cases}u = \xi^5 + \eta, \\ v = \eta^5 - \xi,\end{cases}\begin{cases}\xi = x^3-y^3,\\ \eta = x^2+2y^2,\end{cases}(x_0, y_0) = (1, 0)\).

解

(1) \(\mathbf{g}\) 的逆映射的 Jacobi 矩阵为 \(J_g = \begin{pmatrix}\dfrac{\partial u}{\partial \xi} & \dfrac{\partial u}{\partial \eta} \\ \dfrac{\partial v}{\partial \xi} & \dfrac{\partial v}{\partial \eta}\end{pmatrix}^{-1} = \begin{pmatrix}2\xi & -2\eta \\ 2\eta & 2\xi\end{pmatrix}^{-1} = \dfrac{1}{2(\xi^2 + \eta^2)}\begin{pmatrix}\xi & \eta \\ -\eta & \xi\end{pmatrix}\). \(\mathbf{f}\) 的逆映射的 Jacobi 矩阵为 \(J_f = \begin{pmatrix}\dfrac{\partial \xi}{\partial x} & \dfrac{\partial \xi}{\partial y} \\ \dfrac{\partial \eta}{\partial x} & \dfrac{\partial \eta}{\partial y}\end{pmatrix}^{-1} = \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}^{-1} = \dfrac{1}{e^{2x}}\begin{pmatrix}e^x\cos y & e^x\sin y \\ -e^x\sin y & e^x\cos y\end{pmatrix}\). 当 \(x = x_0 = 1, y = y_0 = 0\) 时, \(\left|J_f\right| \neq 0, \left|J_g\right| \neq 0\), 故能确定. 且 \((\mathbf{g}\circ\mathbf{f})^{-1}\) 的 Jacobi 矩阵为 \(J_fJ_g\).

(3) \(\mathbf{g}\) 的逆映射的 Jacobi 矩阵为 \(J_g = \begin{pmatrix}\dfrac{\partial u}{\partial \xi} & \dfrac{\partial u}{\partial \eta} \\ \dfrac{\partial v}{\partial \xi} & \dfrac{\partial v}{\partial \eta}\end{pmatrix}^{-1} = \begin{pmatrix}5\xi^4 & 1 \\ -1 & 5\eta^4\end{pmatrix}^{-1} = \dfrac{1}{25\xi^4\eta^4 + 1}\begin{pmatrix}5\eta^4 & -1 \\ 1 & 5\xi^4\end{pmatrix}\). \(\mathbf{f}\) 的逆映射的 Jacobi 矩阵为 \(J_f = \begin{pmatrix}\dfrac{\partial \xi}{\partial x} & \dfrac{\partial \xi}{\partial y} \\ \dfrac{\partial \eta}{\partial x} & \dfrac{\partial \eta}{\partial y}\end{pmatrix}^{-1} = \begin{pmatrix}3x^2 & -3y^2 \\ 2x & 4y\end{pmatrix}^{-1} = \dfrac{1}{12x^2y + 6xy^2}\begin{pmatrix}4y & 3y^2 \\ -2x & 3x^2\end{pmatrix}\). 当 \(x = x_0 = 1, y = y_0 = 0\) 时, \(J_f\) 前的系数分母为 \(0\), 故 \(J_f\) 不存在, 不能确定.