习题1.4
第 1 题
求下列函数的偏导数:
(1) \(z = ax^2y + bxy^2\);
(3) \(z = \dfrac{x}{y} + \dfrac{y}{x}\);
(5) \(z = \ln(x + \sqrt{x^2 - y^2})\);
(7) \(z = \cos(1 + 2^{xy})\);
解
(1) \(\dfrac{\partial z}{\partial x} = 2axy + by^2\), \(\dfrac{\partial z}{\partial y} = ax^2 + 2bxy\).
(3) \(\dfrac{\partial z}{\partial x} = \dfrac{1}{y} -\dfrac{y}{x^2}\), \(\dfrac{\partial z}{\partial y} = \dfrac{1}{x} -\dfrac{x}{y^2}\).
(5) \(\dfrac{\partial z}{\partial x} = \dfrac{1}{x + \sqrt{x^2 - y^2}}\cdot(1 + \dfrac{2x}{2\sqrt{x^2 - y^2}}) = \dfrac{1}{\sqrt{x^2 - y^2}}\), \(\dfrac{\partial z}{\partial y} = \dfrac{1}{x + \sqrt{x^2 - y^2}}\cdot\dfrac{-2y}{2\sqrt{x^2 - y^2}} = \dfrac{y}{y^2 - x^2 - x\sqrt{x^2 - y^2}}\).
(7) \(\dfrac{\partial z}{\partial x} = -\sin(1 + 2^{xy})\cdot 2^{xy}\cdot \ln(2^y) = -2^{xy}y\sin(1 + 2^{xy})\ln 2\), \(\dfrac{\partial z}{\partial y} = -\sin(1 + 2^{xy})\cdot 2^{xy}\cdot \ln(2^x) = -2^{xy}x\sin(1 + 2^{xy})\ln 2\).
第 2 题
考察下列函数在坐标原点的可微性.
(1) \(f(x, y) = \sqrt{\left|x\right|}\cos y\);
(2) \(f(x, y) = \begin{cases}\dfrac{2xy}{\sqrt{x^2 + y^2}}, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0;\end{cases}\)
(3) \(f(x, y) = \begin{cases}\dfrac{x^2y^2}{\left(x^2 + y^2\right)^{\frac{3}{2}}}, &x^2 + y^2 \neq 0, \\ 0, &x^2 + y^2 = 0;\end{cases}\)
(4) \(f(x, y) = \left|x - y\right|\phi(x, y)\), 其中 \(\phi(x, y)\) 在原点的某个邻域内连续, 且 \(\phi(0, 0) = 0\).
解
(1) 由于 \(f'_x(0, 0) = \lim\limits_{\Delta x \to 0}\dfrac{f(\Delta x, 0) - f(0, 0)}{\Delta x} = \lim\limits_{\Delta x \to 0}\dfrac{\sqrt{\left|x\right|}}{\Delta x}\) 不存在, 所以该函数在原点不可微.
(2) 由于 \(f'_x(0, 0) = \lim\limits_{\Delta x \to 0}\dfrac{f(\Delta x, 0) - f(0, 0)}{\Delta x} = 0, f'_y(0, 0) = \lim\limits_{\Delta y \to 0}\dfrac{f(0, \Delta y) - f(0, 0)}{\Delta y} = 0\), 所以 \(\lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{f(\Delta x, \Delta y) - f'_x(0, 0)\Delta x - f'_y(0, 0)\Delta y - f(0, 0)}{\sqrt{\Delta x^2 + \Delta y^2}} = \lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{f(\Delta x, \Delta y) - 0 - 0 - 0}{\sqrt{\Delta x^2 + \Delta y^2}} = \lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{2\Delta x\Delta y}{\Delta x^2 + \Delta y^2}\). 由于沿着 \(\Delta y = k\Delta x\) 趋于 \((0, 0)\) 时极限值为 \(\dfrac{2k}{1 + k^2}\) 随着 \(k\) 的变化为变化, 所以该函数在原点不可微.
(3) 由于 \(f'_x(0, 0) = \lim\limits_{\Delta x \to 0}\dfrac{f(\Delta x, 0) - f(0, 0)}{\Delta x} = 0, f'_y(0, 0) = \lim\limits_{\Delta y \to 0}\dfrac{f(0, \Delta y) - f(0, 0)}{\Delta y} = 0\), 所以 \(\lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{f(\Delta x, \Delta y) - f'_x(0, 0)\Delta x - f'_y(0, 0)\Delta y - f(0, 0)}{\sqrt{\Delta x^2 + \Delta y^2}} = \lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{f(\Delta x, \Delta y) - 0 - 0 - 0}{\sqrt{\Delta x^2 + \Delta y^2}} = \lim\limits_{(\Delta x, \Delta y)\to (0, 0)}\dfrac{\Delta x^2\Delta y^2}{(\Delta x^2 + \Delta y^2)^2}\). 由于沿着 \(\Delta y = k\Delta x\) 趋于 \((0, 0)\) 时极限值为 \(\dfrac{k^2}{(1 + k^2)^2}\) 随着 \(k\) 的变化而变化, 所以该函数在原点不可微.
(4) 由于 \(f'_x(0, 0) = \lim\limits_{\Delta x \to 0}\dfrac{f(\Delta x, 0) - f(0, 0)}{\Delta x} = \lim\limits_{\Delta x \to 0}\dfrac{\left|\Delta x\right|\phi(\Delta x, 0)}{\Delta x} = 0\), \(f'_y(0, 0) = \lim\limits_{\Delta y \to 0}\dfrac{f(0, \Delta y) - f(0, 0)}{\Delta y} = \lim\limits_{\Delta y \to 0}\dfrac{\left|\Delta y\right|\phi(0, \Delta y)}{\Delta y} = 0\),
故
所以该函数在原点可微.
第 4 题
求下列函数的全微分:
(1) \(u = \sin\dfrac{1}{x^2 + y^2 + z^2}\), 在点 \(\left(\dfrac{\sqrt{2}}{2}, \dfrac{1}{2}, -\dfrac{1}{2}\right)\);
(3) \(z = (x + y)^2\);
(5) \(z = \dfrac{x - y}{x + y}\);
(7) \(u = \ln(1 + x^2 + y^2 + z^2)\);
解
(1) 记 \(\mathbf{r} = \left(\dfrac{\sqrt{2}}{2}, \dfrac{1}{2}, -\dfrac{1}{2}\right)\). \(\text{d}{u} = 2(x\text{d}{x} + y\text{d}{y} + z\text{d}{z})\cos\dfrac{1}{x^2 + y^2 + z^2}\). 代入得 \(\text{d}{u}\big\vert_{\mathbf{r}} = (\sqrt{2}\text{d}{x} + \text{d}{y} - \text{d}{z})\cos 1\).
(3) 由于 \(z = (x + y)^2 = x^2 + 2xy + y^2\), 故 \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 2x + 2y\). 所以 \(\text{d}{z} = 2(x + y){\text{d}{x} + \text{d}{y}}\).
(5) 由于 \(\dfrac{\partial z}{\partial x} = \dfrac{(x + y) - (x - y)}{(x + y)^2} = \dfrac{2y}{(x + y)^2}\), \(\dfrac{\partial z}{\partial y} = \dfrac{-(x + y) - (x - y)}{(x + y)^2} = -\dfrac{2x}{(x + y)^2}\), 故 \(\text{d}{z} = \dfrac{2(y\text{d}{x} - x\text{d}{y})}{(x + y)^2}\).
(7) \(\text{d}{u} = \dfrac{2x\text{d}{x} + 2y\text{d}{y} + 2z\text{d}{z}}{1 + x^2 + y^2 + z^2} = \dfrac{2(x\text{d}{x} + y\text{d}{y} + z\text{d}{z})}{1 + x^2 + y^2 + z^2}\).
第 7 题
设 \(\dfrac{\partial f}{\partial x}(x_0, y_0)\) 存在, \(\dfrac{\partial f}{\partial y}(x, y)\) 连续, 证明 \(f(x, y)\) 在点 \((x_0, y_0)\) 处可微.
证明
考虑函数 \(f\) 在点 \((x_0, y_0)\) 处的该变量 \(\Delta f = f(x_0 + h, y_0 + k) - f(x_0, y_0)\). 利用一元函数的中值定理得
其中 \(\beta = f_y(x_0 + h, y_0 + \mu k) - f_y(x_0, y_0)\).
由题知 \(f_y(x, y)\) 连续, 因此 \(\lim\limits_{(h, k)\to(0, 0)}\beta(h, k) = 0\). 由题知 \(\lim\limits_{(x, y)\to(x_0, y_0)}\dfrac{f(x_0 + h, y_0) - f(x_0, y_0)}{h} = \dfrac{\partial f}{\partial x}(x_0, y_0)\) 存在, 这说明 \(f(x_0 + h, y_0) - f(x_0, y_0)\) 是关于 \(h\) 的一阶无穷小. 不妨记 \(f(x_0 + h, y_0) - f(x_0, y_0) = \lambda h\). 故 \(\Delta f = \lambda h + f_y(x_0, y_0)k + \beta k = \lambda h + f_y(x_0, y_0)k + o(\rho)\). 此即 \(f\) 在点 \((x_0, y_0)\) 处可微. 命题得证.
第 8 题
设函数 \(f(x, y) = \sqrt[3]{xy}\), 证明: 函数 \(f\) 在原点处连续、偏导数存在, 但沿方向 \(l = (a, b)(ab \neq 0)\) 的方向导数不存在.
证明
由于 \(f\) 是初等函数的复合, 所以函数 \(f\) 在原点处连续. 因为 \(f'_x(0, 0) = \lim\limits_{\Delta x \to 0}\dfrac{f(\Delta x, 0) - f(0, 0)}{\Delta x} = 0, f'_y(0, 0) = \lim\limits_{\Delta y \to 0}\dfrac{f(0, \Delta y) - f(0, 0)}{\Delta y} = 0\), 所以 \(f\) 在原点处的偏导数均存在. 设 \(l\) 单位化得 \(l_0 = (\cos\theta, \sin\theta), \theta \neq \dfrac{k\pi}{2}, k \in \mathbb{N}\). 则沿着 \(l_0\) 的方向导数 \(\dfrac{\partial f}{\partial l_0} = \lim\limits_{d \to 0}\dfrac{f(d\cos \theta, d\sin\theta) - f(0, 0)}{d} = \sqrt[3]{\dfrac{\cos\theta\sin\theta}{d}}\). 由于 \(\theta \neq \dfrac{k\pi}{2}, k \in \mathbb{N}\), 故该极限不存在. 因此题中所表述的方向导数不存在. 命题得证.
第 11 题
求下列函数在点 \(P_0\) 处沿方向 \(l\) 的方向导数.
(1) \(z = \cos (x + y), P_0 = \left(0, \dfrac{\pi}{2}\right), l = (3, -4)\);
(3) \(z = \sum\limits_{i = 1}^n\sum\limits_{j = 1}^nx_ix_j, P(1, 1, \cdots, 1), l = (-1, -1, \cdots, -1)\);
解
(1) 由于 \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = -\sin(x + y)\), 且将 \(l\) 单位化后得到 \(r = \left(\dfrac{3}{5}, -\dfrac{4}{5}\right)\), 所以 \(\dfrac{\partial z}{\partial l} = -\dfrac{3}{5}\sin(0 + \dfrac{\pi}{2}) + \dfrac{4}{5}\sin(0 + \dfrac{\pi}{2}) = \dfrac{1}{5}\).
(3) 由于 \(\dfrac{\partial z}{\partial x_i} = 2\sum\limits_{j = 1}^nx_j\), 且将 \(l\) 单位化后得到 \(r = \left(-\dfrac{1}{\sqrt{n}}, -\dfrac{1}{\sqrt{n}}, \cdots, -\dfrac{1}{\sqrt{n}}\right)\), 所以将 \(P(1, 1, \cdots, 1)\) 代入得 \(\dfrac{\partial z}{\partial l} = 2\sum\limits_{j = 1}^nn\left(-\dfrac{1}{\sqrt{n}}\right) = -2n\sqrt{n}\).
第 12 题
求下列数量场的梯度.
(1) \(u(x, y) = \sqrt{x^2 + y^2}\);
(3) \(u(x_1, x_2, \cdots, x_n) = \sum\limits_{i = 1}^nx_i\);
解
(1) 由于 \(\dfrac{\partial u}{\partial x} = \dfrac{\partial u}{\partial y} = \dfrac{1}{\sqrt{x^2 + y^2}}\), 故 \(\nabla u = \left(\dfrac{1}{\sqrt{x^2 + y^2}}, \dfrac{1}{\sqrt{x^2 + y^2}}\right)\).
(3) 由于 \(\forall i \in [1, n], \dfrac{\partial u}{\partial x_i} = 1\), 故 \(\nabla u = (1, 1, \cdots, 1)\).
第 13 题
已知函数 \(u(x, y, z) = x^2 + y^2 + z^2 - xy - xz + yz\) 及点 \(P(1, 1, 1)\), 求 \(u\) 在 \(P\) 点的方向导数 \(\dfrac{\partial u}{\partial l}\) 的最值, 并指出取得最值时的方向, 以及哪个方向的方向导数为零.
解
\(\dfrac{\partial u}{\partial x} = 2x - y - z, \dfrac{\partial u}{\partial y} = 2y - x + z, \dfrac{\partial u}{\partial z} = 2z - x + y\). 因此在 \(P\) 点处梯度的方向为 \((0, 2, 2)\), 单位化可得 \(l_0 = \left(0, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)\). 将 \(P(1, 1, 1)\) 以及 \(\pm l_0\) 代入可得 \(\dfrac{\partial u}{\partial l}\) 的最大值为 \(2\sqrt{2}\), 方向为 \(\left(0, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)\); 最小值为 \(-2\sqrt{2}\), 方向为 \(\left(0, -\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}\right)\). 对任意一个垂直于梯度的方向 \((k, 1, -1)(k \in \mathbb{R})\) 或 \((k, -1, 1)(k \in \mathbb{R})\),都有方向导数为 \(0\).
第 14 题
求下列函数的二阶偏导数 \(\dfrac{\partial^2 u}{\partial x^2}, \dfrac{\partial^2 u}{\partial y^2}, \dfrac{\partial^2 u}{\partial x\partial y}\).
(1) \(u = \cos^2(ax - by)\);
(3) \(u = xe^{-xy}\);
解
(1) 由于 \(\dfrac{\partial u}{\partial x} = 2\cos(ax - by)\cdot(-\sin(ax - by))\cdot a = -a\sin(2ax - 2by)\), \(\dfrac{\partial u}{\partial y} = 2\cos(ax - by)\cdot(-\sin(ax - by))\cdot (-b) = b\sin(2ax - 2by)\), 故 \(\dfrac{\partial ^2u}{\partial x^2} = -2a^2\cos(2ax - 2by), \dfrac{\partial ^2u}{\partial y^2} = -2b^2\cos(2ax - 2by), \dfrac{\partial ^2u}{\partial x\partial y} = 2ab\cos(2ax - 2by)\).
(3) 由于 \(\dfrac{\partial u}{\partial x} = e^{-xy} + -xye^{-xy} = (1 - xy)e^{-xy}\), \(\dfrac{\partial u}{\partial y} = xe^{-xy}\cdot(-x) = -x^2e^{-xy}\), 故 \(\dfrac{\partial ^2u}{\partial x^2} = ye^{-xy}(xy - 2), \dfrac{\partial ^2u}{\partial y^2} = x^3e^{-xy}, \dfrac{\partial ^2u}{\partial x\partial y} = xe^{-xy}(xy - 2)\).
第 15 题
证明下列函数满足相应的等式.
(1) \(u = 2\cos^2\left(x - \dfrac{y}{2}\right)\) 满足 \(2\dfrac{\partial^2u}{\partial y^2} + \dfrac{\partial^2u}{\partial x \partial y} = 0\);
(3) \(\begin{cases}u = e^x\cos y, \\ v = e^x\sin y\end{cases}\) 满足 Cauchy-Riemann 条件 \(\begin{cases}\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}, \\ \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}\end{cases}\), 且分别满足 Laplace 方程 \(\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\);
证明
(1) 由于 \(\dfrac{\partial u}{\partial y} = \sin(2x - y)\), 所以 \(\dfrac{\partial ^2u}{\partial y^2} = -\cos(2x - y), \dfrac{\partial ^2u}{\partial x\partial y} = 2\cos(2x - y)\). 代入有 \(2\dfrac{\partial^2u}{\partial y^2} + \dfrac{\partial^2u}{\partial x \partial y} = -2\cos(2x - y) + \cos(2x - y) = 0\). 命题得证.
(3) \(\dfrac{\partial u}{\partial x} = e^x\cos y, \dfrac{\partial u}{\partial y} = -e^x\sin y, \dfrac{\partial v}{\partial x} = e^x\sin y, \dfrac{\partial v}{\partial y} = e^x\cos y\). 代入可知其满足 Cauchy-Riemann 条件. \(\dfrac{\partial ^2u}{\partial x^2} = e^x\cos y, \dfrac{\partial ^2u}{\partial y^2} = -e^x\cos y, \dfrac{\partial ^2v}{\partial x^2} = e^x\sin y, \dfrac{\partial ^2v}{\partial y^2} = -e^x\sin y\). 代入可知其分别满足 Laplace 方程. 命题得证.