第九章 多元函数微分法及其应用

第九章

多元函数微分法及其应用

习题9-1

多元函数的基本概念

1. 判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并分别指出它们的聚点所成的点集（称为 $ \underset{\cdot}{导}\underset{\cdot}{集} $）和边界.

(1) $ \left| (x,y) \mid x \neq 0, y \neq 0 \right| $; (2) $ \left| (x,y) \mid 1 < x^{2} + y^{2} \leq 4 \right| $;

(3) $ \{(x,y)\mid y>x^{2}\} $;

$$ \left|\left(x,y\right)\mid x^{2}+\left(y-1\right)^{2}\geqslant1\right|\cap\left|\left(x,y\right)\mid x^{2}+\left(y-2\right)^{2}\leqslant4\right|. $$

解（1）集合是开集，无界集；导集为 $ \mathbb{R}^{2} $，边界为 $ \{(x,y)\mid x=0 \text{ 或 } y=0\} $

（2）集合既非开集，又非闭集，是有界集；导集为 $ \{(x,y)\mid1\leqslant x^{2}+y^{2}\leqslant4\} $，边界为 $ \{(x,y)\mid x^{2}+y^{2}=1\mid\cup\{(x,y)\mid x^{2}+y^{2}=4\}\} $

（3）集合是开集，区域，无界集；导集为 $ \{(x,y)\mid y\geqslant x^{2}\} $，边界为 $ \{(x,y)\mid y=x^{2}\} $

（4）集合是闭集，有界集；导集为集合本身，边界为 $ \{(x,y)\mid x^{2}+(y-1)^{2}=1\mid U\} $， $ \{(x,y)\mid x^{2}+(y-2)^{2}=4\} $.

1. 已知函数 $ f(x,y) = x^{2} + y^{2} - xyt\tan\frac{x}{y} $，试求 $ f(tx,ty) $.

解

$$ \begin{array}{l l}{\displaystyle f(t x,t y)}&{\displaystyle=\left(t x\right)^{2}+\left(t y\right)^{2}-\left(t x\right)\left(t y\right)\tan\frac{t x}{t y}}\\ {}&{}\\ {}&{\displaystyle=t^{2}\bigg(x^{2}+y^{2}-x y\tan\frac{x}{y}\bigg)}\\ {}&{}\\ {}&{\displaystyle=t^{2}f(x,y).}\\ \end{array} $$

1. 试证函数 $ F(x,y) = \ln x \cdot \ln y $ 满足关系式

$$ F\left(x y,u v\right)\;=\;F\left(x,u\right)\;+\;F\left(x,v\right)\;+\;F\left(y,u\right)\;+\;F\left(y,v\right). $$

证

$$ \begin{aligned}F\left(x y,u v\right)&=\ln(x y)\cdot\ln(u v)=(~\ln x+~\ln y)~(\ln~u+~\ln~v)\\&=~\ln~x\cdot\ln~u+~\ln~x\cdot\ln~v+~\ln~y\cdot\ln~u+~\ln~y\cdot\ln~v\\&=~F(x,u)~+~F(x,v)~+~F(y,u)~+~F(y,v).\\ \end{aligned} $$

1. 已知函数 $ f(u, v, w) = u^{w} + w^{u+v} $，试求 $ f(x + y, x - y, xy) $.

$$ f(x+y,x-y,xy)=(x+y)^{xy}+(xy)^{(x+y)+(x-y)}=(x+y)^{xy}+(xy)^{2x}. $$

1. 求下列各函数的定义域：

$$ (1)z=\ln(y^{2}-2x+1)；\quad(2)z=\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{x-y}}； $$

(3)

$$ z=\sqrt{x-\sqrt{y}}; $$

$$ z=\ln(y-x)+\frac{\sqrt{x}}{\sqrt{1-x^{2}-y^{2}}}; $$

(5) $ u = \sqrt{R^2 - x^2 - y^2 - z^2} + \frac{1}{\sqrt{x^2 + y^2 + z^2 - r^2}} $ ( $ R > r > 0 $);

(6) $ u = \arccos \frac{z}{\sqrt{x^2 + y^2}} $.

解 (1) $ \{(x,y)\mid y^{2}-2x+1>0\} $

(2) $ \{(x,y)\mid x+y>0,x-y>0\} $

(3) $ \{(x,y)\mid x\geqslant0,y\geqslant0,x^{2}\geqslant y\} $

(4) $ \left| (x,y) \mid y - x > 0, x \geqslant 0, x^{2} + y^{2} < 1 \right| $.

(5) $ \mid (x, y, z) \mid r^{2} < x^{2} + y^{2} + z^{2} \leqslant R^{2} \mid $

(6) $ \left| (x,y,z) \right| x^{2} + y^{2} - z^{2} \geqslant 0, x^{2} + y^{2} \neq 0 $.

注 本题是求多元函数的定义域，与求一元函数的定义域相类似，先写出构成该函数的各个简单函数的定义域，再求出表示这些定义域的集合的交集，即得所求函数的定义域.

1. 求下列各极限：

$$ \lim_{(x,y)\to(0,1)}\frac{1-xy}{x^{2}+y^{2}}; $$

$$ \lim_{(x,y)\to(1,0)}\frac{\ln(x+e^{y})}{\sqrt{x^{2}+y^{2}}}; $$

(3) $ \lim_{(x,y)\to(0,0)}\frac{2-\sqrt{xy+4}}{xy} $;

$$ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{2-e^{xy}}-1}; $$

(5) $ \lim_{(x,y)\to(2,0)}\frac{\tan(xy)}{y} $;

$$ \lim_{(x,y)\to(0,0)}\frac{1-\cos\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)\mathrm{e}^{x^{2}y^{2}}}. $$

解 (1) $ \lim_{(x,y)\to(0,1)}\frac{1-xy}{x^{2}+y^{2}}=\frac{1-0}{0+1}=1 $.

注 本题利用多元初等函数的连续性求极限，即极限值等于函数值。对于多元初等函数在点 $ P_{0} $ 处的极限，若 $ P_{0} $ 在该函数的定义区域内，均可利用此方法求极限。

(2) $ \lim_{(x,y)\to(1,0)}\frac{\ln(x+\mathrm{e}^{y})}{\sqrt{x^{2}+y^{2}}}=\frac{\ln(1+\mathrm{e}^{0})}{\sqrt{1+0}}=\ln2 $

(3) $ \lim_{(x,y)\to(0,0)}\frac{2-\sqrt{xy+4}}{xy}=\lim_{(x,y)\to(0,0)}\frac{4-(xy+4)}{xy(2+\sqrt{xy+4})} $

$ = \lim_{(x,y)\to(0,0)}\frac{-1}{2+\sqrt{xy+4}}=-\frac{1}{4} $

注 本题分母的极限为零，不能运用商的极限运算法则，而采用通过分母或分子有理化等方法，消去分母中趋于零的因子，再运用极限运算法则，这是求极限的基本方法之一.

$$ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{2-\mathrm{e}^{xy}}-1}=\lim_{(x,y)\to(0,0)}\frac{xy}{1-\mathrm{e}^{xy}}\cdot(\sqrt{2-\mathrm{e}^{xy}}+1)=-1\cdot2=-2. $$

注 本题利用 $ \mathrm{e}^{xy}-1\sim xy((x,y)\rightarrow(0,0)) $，相当于令 u=xy，当 $ (x,y)\rightarrow(0,0) $ 且 xy\neq0 时，有 u\rightarrow0 且 $ u\neq0 $，于是

$$ \lim_{(x,y)\to(0,0)}\frac{x y}{1-\mathrm{e}^{x y}}=\lim_{u\to0}\frac{u}{-u}=-1. $$

$$ \lim_{(x,y)\to(2,0)}\frac{\tan(xy)}{y}=\lim_{(x,y)\to(2,0)}\frac{\tan(xy)}{xy}\cdot x=1\cdot2=2. $$

注 本题利用 $ \tan(xy) \sim xy((x,y) \rightarrow (2,0)) $.

$$ \begin{aligned}\lim_{(x,y)\to(0,0)}\frac{1-\cos\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)\mathrm{e}^{x^{2}y^{2}}}&=\lim_{(x,y)\to(0,0)}\frac{1-\cos\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}\cdot\frac{x^{2}+y^{2}}{\mathrm{e}^{x^{2}y^{2}}}\\&=\frac{1}{2}\cdot0=0.\end{aligned} $$

注 本题利用 $ 1 - \cos(x^{2} + y^{2}) \sim \frac{1}{2}(x^{2} + y^{2})^{2}((x,y) \to (0,0)) $.

$ ^{*} $7. 证明下列极限不存在：

$$ \lim_{(x,y)\to(0,0)}\frac{x+y}{x-y}; $$

$$ \lim_{(x,y)\to(0,0)}\frac{x^{2}y^{2}}{x^{2}y^{2}+\left(x-y\right)^{2}}. $$

（1）当 $ (x,y) $沿直线y=kx趋于（0，0）时，有

$$ \lim_{\substack{(x,y)\to(0,0)\\ y=kx}}\frac{x+y}{x-y}=\lim_{x\to0}\frac{(1+k)x}{(1-k)x}=\frac{1+k}{1-k}(k\neq1). $$

显然它是随着 k 的值不同而改变的，故所求极限不存在.

（2）依次取 $ (x,y)\to(0,0) $的两种方式：y=x,y=-x，分别求极限：

$$ \lim_{_{(x,y)\to(0,0)\atop y=x}}\frac{x^{2}y^{2}}{x^{2}y^{2}+\left(x-y\right)^{2}}=\lim_{x\to0}\frac{x^{4}}{x^{4}}=1, $$

$$ \lim_{(x,y)\to(0,0)}\frac{x^{2}y^{2}}{x^{2}y^{2}+(x-y)^{2}}=\lim_{x\to0}\frac{x^{4}}{x^{4}+4x^{2}}=\lim_{x\to0}\frac{x^{2}}{x^{2}+4}=0. $$

两种方式求得的极限值不同，故所求极限不存在。

注 本题证明极限不存在所采用的方法是：找出两条不同的路径，使得点 P 沿这两条路径趋于 $ P_{0} $ 时， $ f(P) $ 的极限存在但不相等；或者找出一条特殊的路径，使得点 P 沿这条路径趋于 $ P_{0} $ 时， $ f(P) $ 的极限不存在。这是证明多元函数极限不存在常用的方法。

1. 函数 $ z = \frac{y^{2} + 2x}{y^{2} - 2x} $ 在何处是间断的？

解 这函数的定义域为 $ D = \{(x, y) \mid y^{2} - 2x \neq 0\} $，曲线 $ y^{2} - 2x = 0 $ 上各点均为 D 的聚点，且函数在这些点处没有定义，因此曲线 $ y^{2} - 2x = 0 $ 上各点均为函数的间断点.

$ ^{*} $9. 证明 $ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^{2}+y^{2}}}=0 $.

证 因为

$$ \left|\frac{xy}{\sqrt{x^{2}+y^{2}}}-0\right|\leqslant\frac{\frac{1}{2}(x^{2}+y^{2})}{\sqrt{x^{2}+y^{2}}}=\frac{1}{2}\sqrt{x^{2}+y^{2}}, $$

要使 $ \left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|<\varepsilon $，只要 $ \sqrt{x^2+y^2}<2\varepsilon $，所以 $ \forall\varepsilon>0 $，取 $ \delta=2\varepsilon $，则当 $ 0<\sqrt{x^2+y^2}<\delta $时，就有 $ \left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|<\varepsilon $成立，即 $ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0 $。

*10. 设 $ F(x,y)=f(x) $， $ f(x) $在 $ x_0 $处连续，证明：对任意 $ y_0\in\mathbb{R} $， $ F(x,y) $在 $ (x_0,y_0) $处连续。

证 设 $ P_0(x_0,y_0)\in\mathbb{R}^2 $，因为 $ f(x) $在 $ x_0 $处连续，所以 $ \forall\varepsilon>0 $， $ \exists\delta>0 $，当 $ |x-x_0|<\delta $时，有 $ |f(x)-f(x_0)|<\varepsilon $。从而，当 $ P(x,y)\in U(P_0,\delta) $时， $ |x-x_0|\leq\rho(P,P_0)<\delta $，因而有

$$ \left|F(x,y)-F(x_{0},y_{0})\right|=\left|f(x)-f(x_{0})\right|<\varepsilon, $$

即 $ F(x,y) $ 在 $ (x_{0},y_{0}) $ 处连续.

习题9-2

偏导数

1. 求下列函数的偏导数：

求下列函数的偏导数：

(1) $ z = x^3 y - y^3 x $; \quad (2) $ s = \frac{u^2 + v^2}{uv} $;

(3) $ z = \sqrt{\ln(xy)} $; \quad (4) $ z = \sin(xy) + \cos^2(xy) $

(5) $ z = \ln \tan \frac{x}{y} $; \quad (6) $ z = (1 + xy)^y $;

(7) $ u = x^{\frac{y}{i}} $; \quad (8) $ u = \arctan(x - y)^z $.

解 (1) $ \frac{\partial z}{\partial x} = 3x^{2}y - y^{3} $, $ \frac{\partial z}{\partial y} = x^{3} - 3y^{2}x $.

$$ \begin{array}{l}\displaystyle\frac{\partial s}{\partial u}=\frac{\displaystyle\frac{\partial}{\partial u}(u^{2}+v^{2})\cdot u v-(u^{2}+v^{2})\cdot\frac{\partial}{\partial u}(u v)}{(u v)^{2}}\\ =~\displaystyle\frac{2u^{2}v-(u^{2}+v^{2})v}{u^{2}v^{2}}\\ =~\displaystyle\frac{1}{v}-\frac{v}{u^{2}},\end{array} $$

$$ \begin{array}{l}\displaystyle\frac{\partial s}{\partial v}=\frac{\displaystyle\frac{\partial}{\partial v}(u^{2}+v^{2})\cdot u v-(u^{2}+v^{2})\cdot\frac{\partial}{\partial v}(u v)}{(u v)^{2}}\\ =~\displaystyle\frac{2u v^{2}-(u^{2}+v^{2})u}{u^{2}v^{2}}\\ =~\frac{1}{u}-\frac{u}{v^{2}}.\end{array} $$

(3)

$$ \frac{\partial z}{\partial x}=\frac{1}{2}\cdot\frac{1}{\sqrt{\ln\left(xy\right)}}\cdot\frac{1}{xy}\cdot y=\frac{1}{2x\sqrt{\ln\left(xy\right)}}, $$

$$ \frac{\partial z}{\partial y}=\frac{1}{2}\cdot\frac{1}{\sqrt{\ln\left(xy\right)}}\cdot\frac{1}{xy}\cdot x=\frac{1}{2y\sqrt{\ln\left(xy\right)}}. $$

(4)

$$ \begin{align*}\frac{\partial z}{\partial x}&=y\cos(xy)+2\cos(xy)\cdot\left[-\sin(xy)\right]\cdot y\\&=y\left[\cos(xy)-\sin(2xy)\right],\end{align*} $$

$$ \begin{align*}\frac{\partial z}{\partial y}&=x\cos(xy)+2\cos(xy)\cdot\left[-\sin(xy)\right]\cdot x\\&=x\left[\cos(xy)-\sin(2xy)\right].\end{align*} $$

(5)

$$ \frac{\partial z}{\partial x}\;=\;\cot\frac{x}{y}\cdot\sec^{2}\frac{x}{y}\cdot\frac{1}{y}\;=\;\frac{2}{y}\csc\frac{2x}{y}, $$

$$ \frac{\partial z}{\partial y}=\cot\frac{x}{y}\cdot\sec^{2}\frac{x}{y}\cdot\left(-\frac{x}{y^{2}}\right)=-\frac{2x}{y^{2}}\csc\frac{2x}{y}. $$

(6)

$$ \frac{\partial z}{\partial x}\;=\;y^{2}\left(1\;+\;x y\right)^{y-1}\;, $$

$$ \frac{\partial z}{\partial y}=\frac{\partial}{\partial y}\Big[\mathrm{e}^{y\ln\left(1+x y\right)}\Big]\;=\;\left(1+x y\right)^{y}\Bigg[\ln\left(1+x y\right)\;+\frac{x y}{1+x y}\Bigg]. $$

(7)

$$ \frac{\partial u}{\partial x}=\frac{y}{z}x^{\frac{y}{z}-1},\frac{\partial u}{\partial y}=\frac{1}{z}x^{\frac{y}{z}}\ln x,\frac{\partial u}{\partial z}=-\frac{y}{z^{2}}x^{\frac{y}{z}}\ln x. $$

(8)

$$ \frac{\partial u}{\partial x}=\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}}, $$

$$ \frac{\partial u}{\partial y}=-\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}}, $$

$$ \frac{\partial u}{\partial z}=\frac{\left(x-y\right)^{z}\ln\left(x-y\right)}{1+\left(x-y\right)^{2z}}. $$

1. 设 $ T = 2\pi\sqrt{\frac{l}{g}} $，求证 $ l\frac{\partial T}{\partial l} + g\frac{\partial T}{\partial g} = 0 $.

证 因为

$$ \frac{\partial T}{\partial l}\;=\;2\pi\cdot\frac{1}{2\sqrt{\frac{l}{g}}}\cdot\frac{1}{g}\;=\;\frac{\pi}{\sqrt{g l}}, $$

$$ \frac{\partial T}{\partial g}\;=\;2\pi\cdot\frac{1}{2\sqrt{\frac{l}{g}}}\cdot\Big(-\frac{l}{g^{2}}\Big)\;=\;-\frac{\pi}{g}\sqrt{\frac{l}{g}}, $$

所以

$$ l\frac{\partial T}{\partial l}+g\frac{\partial T}{\partial g}=\pi\sqrt{\frac{l}{g}}-\pi\sqrt{\frac{l}{g}}=0. $$

1. 设 $ z = e^{-(\frac{1}{x} + \frac{1}{y})} $，求证 $ x^{2} \frac{\partial z}{\partial x} + y^{2} \frac{\partial z}{\partial y} = 2z $.

证 因为

$$ \frac{\partial z}{\partial x}\;=\;\frac{1}{x^{2}}\mathrm{e}^{-(\frac{1}{x}+\frac{1}{y})}\;,\qquad\frac{\partial z}{\partial y}\;=\;\frac{1}{y^{2}}\mathrm{e}^{-(\frac{1}{x}+\frac{1}{y})}\;, $$

所以

$$ x^{2}\ \frac{\partial z}{\partial x}\ +\ y^{2}\ \frac{\partial z}{\partial y}\ =\ 2\mathrm{e}^{-\left(\frac{1}{x}+\frac{1}{y}\right)}\ \ =\ 2z. $$

1. 设 $ f(x, y) = x + (y - 1) \arcsin \sqrt{\frac{x}{y}} $，求 $ f_x(x, 1) $.

解

$$ f_{x}\left(x,y\right)=1+\frac{y-1}{\sqrt{1-\frac{x}{y}}}\cdot\frac{1}{2\sqrt{\frac{x}{y}}}\cdot\frac{1}{y}, $$

$$ f_{x}(x,1)=1. $$

1. 曲线 $ \{\begin{aligned}z&=\frac{x^{2}+y^{2}}{4},\\ y&=4\end{aligned}. $ 在点 $ (2,4,5) $ 处的切线对于 x 轴的倾角是多少？

解 设 $ z = f(x, y) $. 按偏导数的几何意义, $ f_{x}(2, 4) $ 就是曲线在点 $ (2, 4, 5) $ 处的切线对于 x 轴的斜率, 而 $ f_{x}(2, 4) = \frac{1}{2} x \bigg|_{x=2} = 1 $, 即 $ k = \tan \alpha = 1 $, 于是倾角 $ \alpha = \frac{\pi}{4} $.

1. 求下列函数的 $ \frac{\partial^{2}z}{\partial x^{2}}, \frac{\partial^{2}z}{\partial y^{2}} $ 和 $ \frac{\partial^{2}z}{\partial x \partial y} $.

$$ (1)z=x^{4}+y^{4}-4x^{2}y^{2};\quad(2)z=\arctan\frac{y}{x}; $$

(3) z = $ y^{x} $

解（1）

$$ \frac{\partial z}{\partial x}\;=\;4x^{3}\;-\;8xy^{2}\;,\qquad\frac{\partial^{2}z}{\partial x^{2}}\;=\;12x^{2}\;-\;8y^{2}\;, $$

$$ \frac{\partial z}{\partial y}\;=\;4y^{3}\;-\;8x^{2}y\;,\qquad\frac{\partial^{2}z}{\partial y^{2}}\;=\;12y^{2}\;-\;8x^{2}\;, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial}{\partial y}(4x^{3}-8xy^{2})={}-16xy. $$

(2)

$$ \frac{\partial z}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^{2}}\cdot\left(-\frac{y}{x^{2}}\right)=-\frac{y}{x^{2}+y^{2}},\quad\frac{\partial^{2}z}{\partial x^{2}}=\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}. $$

$$ \frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{y}{x}\right)^{2}}\cdot\frac{1}{x}=\frac{x}{x^{2}+y^{2}},\quad\frac{\partial^{2}z}{\partial y^{2}}=-\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial}{\partial y}\bigg(-\frac{y}{x^{2}+y^{2}}\bigg)=-\frac{\left(x^{2}+y^{2}\right)-y\cdot2y}{\left(x^{2}+y^{2}\right)^{2}}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}. $$

(3)

$$ \frac{\partial z}{\partial x}\;=\;y^{x}\ln y,\quad\frac{\partial^{2}z}{\partial x^{2}}\;=\;y^{x}\ln^{2}y, $$

$$ \frac{\partial z}{\partial y}\;=\;x y^{x-1}\;,\quad\frac{\partial^{2}z}{\partial y^{2}}\;=\;x(x-1)y^{x-2}\;, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial}{\partial y}(y^{x}\ln y)=y^{x-1}\left(1+x\ln y\right). $$

1. 设 $ f(x,y,z) = xy^{2} + yz^{2} + zx^{2} $，求 $ f_{xx}(0,0,1) $， $ f_{xz}(1,0,2) $， $ f_{yz}(0,-1,0) $ 及 $ f_{zzx}(2,0,1) $.

解 因为

$$ f_{x}=y^{2}+2x z,\quad f_{x x}=2z,\quad f_{x z}=2x, $$

$$ f_{y}=2x y+z^{2},\quad f_{y z}=2z, $$

$$ f_{z}=2y z+x^{2},\quad f_{z z}=2y,\quad f_{z z x}=0, $$

所以 $ f_{xx}(0,0,1)=2 $, $ f_{xz}(1,0,2)=2 $, $ f_{yz}(0,-1,0)=0 $, $ f_{zzx}(2,0,1)=0 $.

1. 设 $ z = x \ln(xy) $，求 $ \frac{\partial^{3}z}{\partial x^{2}\partial y} $ 及 $ \frac{\partial^{3}z}{\partial x\partial y^{2}} $.

解

$$ \frac{\partial z}{\partial x}\;=\;\ln(x y)\;+\;x\cdot\frac{y}{x y}\;=\;\ln(x y)\;+\;1, $$

$$ \frac{\partial^{2}z}{\partial x^{2}}\;=\;\frac{y}{x y}\;=\;\frac{1}{x}\;,\quad\frac{\partial^{3}z}{\partial x^{2}\partial y}\;=\;0\;, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}\;=\;\frac{x}{x y}\;=\;\frac{1}{y}\;,\quad\frac{\partial^{3}z}{\partial x\partial y^{2}}\;=\;-\frac{1}{y^{2}}. $$

1. 验证：

(1) $ y = e^{-kn^2 t} \sin nx $ 满足 $ \frac{\partial y}{\partial t} = k \frac{\partial^2 y}{\partial x^2} $;

(2) $ r=\sqrt{x^{2}+y^{2}+z^{2}} $ 满足 $ \frac{\partial^{2}r}{\partial x^{2}}+\frac{\partial^{2}r}{\partial y^{2}}+\frac{\partial^{2}r}{\partial z^{2}}=\frac{2}{r} $

证 （1）因为 $ \frac{\partial y}{\partial t} = -kn^{2}e^{-kn^{2}t}\sin nx $， $ \frac{\partial y}{\partial x} = n e^{-kn^{2}t}\cos nx $，

$$ \frac{\partial^{2}y}{\partial x^{2}}\;=\;\frac{\partial}{\partial x}(n\mathrm{e}^{-k n^{2}t}\cos n x)\;=\;-n^{2}\mathrm{e}^{-k n^{2}t}\sin n x, $$

所以

$$ \frac{\partial y}{\partial t}=k(-\ n^{2}\mathrm{e}^{-k n^{2}t}\sin n x)=k\frac{\partial^{2}y}{\partial x^{2}}. $$

（2）因为

$$ \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{r}, $$

$$ \frac{\partial^{2}r}{\partial x^{2}}\;=\;\frac{\partial}{\partial x}\bigg(\frac{x}{r}\bigg)\;=\;\frac{1}{r}\;-\;\frac{x}{r^{2}}\;\cdot\;\frac{x}{r}\;=\;\frac{r^{2}\;-\;x^{2}}{r^{3}}, $$

由函数关于自变量的对称性，得

$$ \frac{\partial^{2}r}{\partial y^{2}}\;=\;\frac{r^{2}\;-\;y^{2}}{r^{3}},\quad\frac{\partial^{2}r}{\partial z^{2}}\;=\;\frac{r^{2}\;-\;z^{2}}{r^{3}}, $$

所以

$$ \frac{\partial^{2}r}{\partial x^{2}}+\frac{\partial^{2}r}{\partial y^{2}}+\frac{\partial^{2}r}{\partial z^{2}}=\frac{r^{2}-x^{2}}{r^{3}}+\frac{r^{2}-y^{2}}{r^{3}}+\frac{r^{2}-z^{2}}{r^{3}}=\frac{2}{r}. $$

习题9-3

全微分

1. 求下列函数的全微分：

(1)

$$ z=x y+\frac{x}{y}; $$

(2)

$$ z=e^{\frac{1}{r}}; $$

(3)

$$ z=\frac{y}{\sqrt{x^{2}+y^{2}}}; $$

(4)

$$ u=x^{yz}. $$

解（1）因为

$$ \frac{\partial z}{\partial x}\;=\;y\;+\;\frac{1}{y}\;,\quad\frac{\partial z}{\partial y}\;=\;x\;-\;\frac{x}{y^{2}}\;, $$

所以

$$ \mathrm{d}z\;=\;\frac{\partial z}{\partial x}\mathrm{d}x\;+\;\frac{\partial z}{\partial y}\mathrm{d}y\;=\;\left(\;y\;+\;\frac{1}{y}\right)\mathrm{d}x\;+\;\left(\;x\;-\;\frac{x}{y^{2}}\right)\mathrm{d}y. $$

（2）因为

$$ \frac{\partial z}{\partial x}\;=\;-\;\frac{y}{x^{2}}\mathrm{e}^{\frac{y}{x}}\;,\quad\frac{\partial z}{\partial y}\;=\;\frac{1}{x}\mathrm{e}^{\frac{y}{x}}\;, $$

所以

$$ \mathrm{d}z\;=\;\frac{\partial z}{\partial x}\mathrm{d}x\;+\;\frac{\partial z}{\partial y}\mathrm{d}y\;=\;-\;\frac{1}{x^{2}}\mathrm{e}^{\frac{x}{x}}\left(y\mathrm{d}x\;-\;x\mathrm{d}y\right). $$

（3）因为

$$ \frac{\partial z}{\partial x}=\frac{-y}{x^{2}+y^{2}}\cdot\frac{x}{\sqrt{x^{2}+y^{2}}}=\frac{-xy}{\left(x^{2}+y^{2}\right)^{3/2}}, $$

$$ \frac{\partial z}{\partial y}=\frac{\sqrt{x^{2}+y^{2}}-y\cdot\frac{y}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{x^{2}}{\left(x^{2}+y^{2}\right)^{3/2}}, $$

所以

$$ \mathrm{d}z\;=\;\frac{\partial z}{\partial x}\mathrm{d}x\;+\;\frac{\partial z}{\partial y}\mathrm{d}y\;=\;-\;\frac{x}{\left(x^{2}\;+\;y^{2}\right)^{3/2}}\left(y\mathrm{d}x\;-\;x\mathrm{d}y\right). $$

（4）因为

$$ \frac{\partial u}{\partial x}\;=\;y z x^{y z-1}\;,\quad\frac{\partial u}{\partial y}\;=\;z x^{y z}\mathrm{l n}\;x,\quad\frac{\partial u}{\partial z}\;=\;y x^{y z}\mathrm{l n}\;x, $$

所以

$$ \mathrm{d}u~=~\frac{\partial u}{\partial x}\mathrm{d}x~+~\frac{\partial u}{\partial y}\mathrm{d}y~+~\frac{\partial u}{\partial z}\mathrm{d}z~=~y z x^{y z-1}\mathrm{d}x~+~z x^{y z}\mathrm{l n}~x\mathrm{d}y~+~y x^{y z}\mathrm{l n}~x\mathrm{d}z. $$

1. 求函数 $ z = \ln(1 + x^{2} + y^{2}) $ 当 x = 1, y = 2 时的全微分.

解 因为

$$ \frac{\partial z}{\partial x}\;=\;\frac{2x}{1\;+\;x^{2}\;+\;y^{2}},\qquad\frac{\partial z}{\partial y}\;=\;\frac{2y}{1\;+\;x^{2}\;+\;y^{2}}, $$

$$ \frac{\partial z}{\partial x}\bigg|_{\substack{x=1\\ y=2}}=\frac{1}{3},\quad\frac{\partial z}{\partial y}\bigg|_{\substack{x=1\\ y=2}}=\frac{2}{3}, $$

所以

$$ \mathrm{d}z\left|\begin{array}{c}\mathrm{x}=1\\\mathrm{y}=2\end{array}\right.=\left.\frac{1}{3}\mathrm{d}x\right.+\left.\frac{2}{3}\mathrm{d}y.\right. $$

1. 求函数 $ 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.

解

$$ \Delta z~=~\frac{y~+~\Delta y}{x~+~\Delta x}-\frac{y}{x},\quad\mathrm{d}z~=~-\frac{y}{x^{2}}\Delta x~+~\frac{1}{x}\Delta y. $$

当x=2,y=1, $ \Delta x=0.1,\Delta y=-0.2 $时，全增量

$$ \Delta z=\frac{1+(-0.2)}{2+0.1}-\frac{1}{2}=-0.119, $$

全微分

$$ \mathrm{d}z=-\frac{1}{4}\cdot0.1+\frac{1}{2}\cdot(-0.2)=-0.125. $$

1. 求函数 $ z = e^{xy} $ 当 x = 1, y = 1, $ \Delta x = 0.15 $, $ \Delta y = 0.1 $ 时的全微分.

解

$$ \mathrm{d}z~=~\frac{\partial z}{\partial x}\Delta x~+~\frac{\partial z}{\partial y}\Delta y~=~y\mathrm{e}^{x y}\Delta x~+~x\mathrm{e}^{x y}\Delta y. $$

当 x=1, y=1, $ \Delta x=0.15,\Delta y=0.1 $ 时，全微分

$$ \mathrm{d}z=\mathrm{e}\cdot0.15+\mathrm{e}\cdot0.1=0.25\mathrm{e}. $$

1. 考虑二元函数 $ f(x,y) $ 的下面四条性质：

(1) $ f(x,y) $ 在点 $ (x_{0},y_{0}) $ 连续；

(2) $ f_{x}(x,y) $， $ f_{y}(x,y) $在点 $ (x_{0},y_{0}) $连续；

(3) $ f(x,y) $在点 $ (x_{0},y_{0}) $可微分；

(4) $ f_{x}(x_{0},y_{0}),f_{y}(x_{0},y_{0}) $存在.

若用“ $ P \Rightarrow Q $”表示可由性质P推出性质Q，则下列四个选项中正确的是（）.

(A) $ (2)\Rightarrow(3)\Rightarrow(1) $ (B) $ (3)\Rightarrow(2)\Rightarrow(1) $

(C) $ (3)\Rightarrow(4)\Rightarrow(1) $ (D) $ (3)\Rightarrow(1)\Rightarrow(4) $

解 由于二元函数偏导数存在且连续是二元函数可微分的充分条件，二元函数可微分必定可（偏）导，二元函数可微分必定连续，因此选项(A)正确.

选项(B)中(3) $ \not\ni(2) $，选项(C)中(4) $ \not\ni(1) $，选项(D)中(1) $ \not\ni(4) $。

$ ^{*} $6. 计算 $ \sqrt{(1.02)^{3}}+(1.97)^{3} $的近似值.

解 设 $ z = \sqrt{x^{3} + y^{3}} $，则

$$ \begin{aligned}\sqrt{\left(x+\Delta x\right)^{3}+\left(y+\Delta y\right)^{3}}&=\sqrt{x^{3}+y^{3}}+\Delta z\approx\sqrt{x^{3}+y^{3}}+\mathrm{d}z\\&=\sqrt{x^{3}+y^{3}}+\frac{3x^{2}\Delta x+3y^{2}\Delta y}{2\sqrt{x^{3}+y^{3}}}.\end{aligned} $$

取x=1,y=2, $ \Delta x=0.02,\Delta y=-0.03 $，可得

$$ \sqrt{\left(1.02\right)^{3}+\left(1.97\right)^{3}}\approx\sqrt{1+2^{3}}+\frac{3\cdot1\cdot0.02+3\cdot2^{2}\cdot\left(-0.03\right)}{2\sqrt{1+2^{3}}}=2.95. $$

1. 计算(1.97) $ ^{1.05} $的近似值( $ \ln2=0.693 $).

解 设 $ z = x^{y} $，则

$$ \left(x+\Delta x\right)^{y+\Delta y}=x^{y}+\Delta z\approx x^{y}+\mathrm{d}z=x^{y}+y x^{y-1}\Delta x+x^{y}\ln x\cdot\Delta y. $$

取x=2,y=1, $ \Delta x=-0.03,\Delta y=0.05 $,可得

$$ (1.97)^{1.05}\approx2-0.03+2\ln2\cdot0.05=1.97+0.0693\approx2.039. $$

$ ^{*} $8. 已知边长为 x = 6m 与 y = 8m 的矩形，如果 x 边增加 5cm 而 y 边减少 10cm，问这个矩形的对角线的近似变化怎样？

解 矩形的对角线的长为 $ z = \sqrt{x^{2} + y^{2}} $

$$ \Delta z\;\approx\;\mathrm{d}z\;=\;\frac{\partial z}{\partial x}\Delta x\;+\;\frac{\partial z}{\partial y}\Delta y\;=\;\frac{1}{\sqrt{x^{2}+y^{2}}}\!\left(x\Delta x\;+\;y\Delta y\right). $$

当x=6,y=8, $ \Delta x=0.05,\Delta y=-0.1 $时，

$$ \Delta z\approx\frac{1}{\sqrt{6^{2}+8^{2}}}(6\cdot0.05-8\cdot0.1)=-0.05, $$

即，这个矩形的对角线的长减少大约5 cm.

$ ^{*} $9. 设有一无盖圆柱形容器，容器的壁与底的厚度均为0.1 cm，内高为20 cm，内半径为4 cm。求容器外壳体积的近似值。

解 圆柱体的体积公式为 $ V = \pi R^{2} H $，圆柱形容器的外壳体积就是圆柱体体积 V 的增量 $ \Delta V $，而

$$ \Delta V\simeq\mathrm{d}V=\frac{\partial V}{\partial R}\Delta R+\frac{\partial V}{\partial H}\Delta H=2\pi R H\Delta R+\pi R^{2}\Delta H. $$

当 R=4, H=20, $ \Delta R=\Delta H=0.1 $ 时，

$$ \Delta V\approx2\cdot3.14\cdot4\cdot20\cdot0.1+3.14\cdot4^{2}\cdot0.1\approx55.3. $$

即容器外壳的体积大约是55.3 cm $ ^{3} $.

$ ^{*} $10. 设有直角三角形，测得其两直角边的长分别为 $ (7 \pm 0.1) $ cm 和 $ (24 \pm 0.1) $ cm. 试求利用上述两值来计算斜边长度时的绝对误差.

解 设两直角边长度分别为 x 和 y，则斜边长度为 $ z = \sqrt{x^{2} + y^{2}} $

$$ \begin{align*}|\Delta z|&\approx|\mathrm{d}z|=\left|\frac{\partial z}{\partial x}\Delta x+\frac{\partial z}{\partial y}\Delta y\right|\leqslant\left|\frac{\partial z}{\partial x}\right|\left|\Delta x\right|+\left|\frac{\partial z}{\partial y}\right|\left|\Delta y\right|\\&=\frac{1}{\sqrt{x^{2}+y^{2}}}(x\mid\Delta x\mid+y\mid\Delta y\mid)\leqslant\frac{1}{\sqrt{x^{2}+y^{2}}}(x\delta_{x}+y\delta_{y}),\end{align*} $$

便得

$$ \delta_{z}=\frac{1}{\sqrt{x^{2}+y^{2}}}(x\delta_{x}+y\delta_{y}). $$

当x=7,y=24, $ \delta_{x}=0.1,\delta_{y}=0.1 $时，

$$ \delta_{z}=\frac{1}{\sqrt{7^{2}+24^{2}}}(7\cdot0.1+24\cdot0.1)=0.124. $$

即计算斜边长度 z 的绝对误差约为 0.124 cm.

$ ^{*} $11. 测得一块三角形土地的两边边长分别为 $ (63 \pm 0.1) $ m 和 $ (78 \pm 0.1) $ m，这两边的夹角为 $ 60^{\circ} \pm 1^{\circ} $. 试求三角形面积的近似值，并求其绝对误差和相对误差.

解 设三角形的两边长分别为 a 和 b，它们的夹角为 $ \theta $，则三角形的面积为

$$ S\;=\;{\frac{1}{2}}a b\sin\;\theta, $$

$$ \begin{align*}|\Delta S|&\approx|\mathrm{d}S|=\left|\frac{\partial S}{\partial a}\Delta a+\frac{\partial S}{\partial b}\Delta b+\frac{\partial S}{\partial\theta}\Delta\theta\right|\\&\leq\left|\frac{\partial S}{\partial a}\right|\left|\Delta a\right|+\left|\frac{\partial S}{\partial b}\right|\left|\Delta b\right|+\left|\frac{\partial S}{\partial\theta}\right|\left|\Delta\theta\right|\\&=\frac{1}{2}b\sin\theta\left|\Delta a\right|+\frac{1}{2}a\sin\theta\left|\Delta b\right|+\frac{1}{2}a b\cos\theta\left|\Delta\theta\right|\\&\leq\frac{1}{2}b\sin\theta\delta_{a}+\frac{1}{2}a\sin\theta\delta_{b}+\frac{1}{2}a b\cos\theta\delta_{\theta},\end{align*} $$

便得

$$ \delta_{_\mathrm{S}}\;=\;\frac{1}{2}b\sin\;\theta\delta_{a}\;+\;\frac{1}{2}a\sin\;\theta\delta_{b}\;+\;\frac{1}{2}a b\cos\;\theta\delta_{\theta}. $$

当 a = 63, b = 78, $ \theta = \frac{\pi}{3} $, $ \delta_{a} = 0.1 $, $ \delta_{b} = 0.1 $, $ \delta_{\theta} = \frac{\pi}{180} $ 时，三角形面积的近似值为

$$ S=\frac{1}{2}\cdot63\cdot78\cdot\sin\frac{\pi}{3}=2127.8(m^{2})， $$

绝对误差为

$$ \begin{array}{l}\displaystyle \delta_{S}=\frac{1}{2} \cdot 78 \cdot \frac{\sqrt{3}}{2} \cdot 0.1 + \frac{1}{2} \cdot 63 \cdot \frac{\sqrt{3}}{2} \cdot 0.1 + \frac{1}{2} \cdot 63 \cdot 78 \cdot \frac{1}{2} \cdot \frac{\pi}{180}\\ \qquad = 27.6(\mathrm{~m}^2)\end{array} $$

相对误差为

$$ \frac{\delta_{S}}{S}=\frac{27.6}{2\;127.8}=1.30\%. $$

1. 利用全微分证明：两数之和的绝对误差等于它们各自的绝对误差之和.

证 设 u = x + y，则

$$ \begin{align*}|\Delta u|&\approx|\mathrm{d}u|=\left|\frac{\partial u}{\partial x}\Delta x+\frac{\partial u}{\partial y}\Delta y\right|\\&=\left|\Delta x+\Delta y\right|\leqslant\left|\Delta x\right|+\left|\Delta y\right|\leqslant\delta_{x}+\delta_{y},\end{align*} $$

便得

$$ \delta_{u}=\delta_{x}+\delta_{y}, $$

即两数之和的绝对误差等于它们各自的绝对误差之和.

$ ^{*} $13. 利用全微分证明：乘积的相对误差等于各因子的相对误差之和，商的相对误差等于被除数及除数的相对误差之和.

证 设 $ u = xy, v = \frac{x}{y} $，则

$$ \left|\Delta u\right|\approx\left|\mathrm{d}u\right|=\left|y\Delta x+x\Delta y\right|\leq\left|y\right|\left|\Delta x\right|+\left|x\right|\left|\Delta y\right|\leq\left|y\right|\delta_{x}+\left|x\right|\delta_{y}, $$

$$ \left|\Delta v\right|\approx\left|\mathrm{d}v\right|=\left|\frac{y\Delta x-x\Delta y}{y^{2}}\right|\leqslant\frac{\left|y\right|\left|\Delta x\right|+\left|x\right|\left|\Delta y\right|}{\left|y\right|^{2}}\leqslant\frac{\left|y\right|\delta_{x}+\left|x\right|\delta_{y}}{\left|y\right|^{2}}, $$

便得

$$ \delta_{u}=\left|y\right|\delta_{x}+\left|x\right|\delta_{y},\quad\delta_{v}=\frac{\left|y\right|\delta_{x}+\left|x\right|\delta_{y}}{\left|y\right|^{2}}, $$

$$ \frac{\delta_{u}}{\mid u\mid}=\frac{\mid y\mid\delta_{x}+\mid x\mid\delta_{y}}{\mid x y\mid}=\frac{\delta_{x}}{\mid x\mid}+\frac{\delta_{y}}{\mid y\mid}, $$

$$ \frac{\delta_{v}}{\left|v\right|}=\frac{1}{\left|\frac{x}{y}\right|}\cdot\frac{\left|y\right|\delta_{x}+\left|x\right|\delta_{y}}{\left|y\right|^{2}}=\frac{\delta_{x}}{\left|x\right|}+\frac{\delta_{y}}{\left|y\right|}. $$

即乘积的相对误差等于各因子的相对误差之和，商的相对误差等于被除数及除数的相对误差之和.

习题9-4

多元复合函数的求导法则

1. 设 $ z = u^{2} + v^{2} $，而 u = x + y， $ v = x - y $，求 $ \frac{\partial z}{\partial x} $， $ \frac{\partial z}{\partial y} $.

解

$$ \frac{\partial z}{\partial x}\;=\;\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}\;+\;\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}\;=\;2u\cdot1\;+\;2v\cdot1\;=\;2\left(u\;+\;v\right)\;=\;4x, $$

$$ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y}=2u\cdot1+2v\cdot(-1)=2(u-v)=4y. $$

1. 设 $ z = u^{2} \ln v $，而 $ u = \frac{x}{y} $， $ v = 3x - 2y $，求 $ \frac{\partial z}{\partial x} $， $ \frac{\partial z}{\partial y} $.

解

$$ \begin{align*}\frac{\partial z}{\partial x}&=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}=2u\ln v\cdot\frac{1}{y}+\frac{u^{2}}{v}\cdot3\\&=\frac{2x}{y^{2}}\ln(3x-2y)+\frac{3x^{2}}{(3x-2y)y^{2}},\end{align*} $$

$$ \begin{align*}\frac{\partial z}{\partial y}&=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y}=2u\ln v\cdot\left(-\frac{x}{y^{2}}\right)+\frac{u^{2}}{v}\cdot(-2)\\&=-\frac{2x^{2}}{y^{3}}\ln(3x-2y)-\frac{2x^{2}}{(3x-2y)y^{2}}.\end{align*} $$

1. 设 $ z = e^{x-2y} $，而 $ x = \sin t $， $ y = t^{3} $，求 $ \frac{dz}{dt} $.

解

$$ \begin{align*}\frac{\mathrm{d}z}{\mathrm{d}t}&=\frac{\partial z}{\partial x}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial z}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{e}^{x-2y}\cdot\cos t+\mathrm{e}^{x-2y}\cdot(-2)\cdot3t^{2}\\&=\mathrm{e}^{x-2y}(\cos t-6t^{2})=\mathrm{e}^{\sin t-2t^{3}}(\cos t-6t^{2}).\end{align*} $$

1. 设 $ z = \arcsin(x - y) $，而 x = 3t, $ y = 4t^{3} $，求 $ \frac{dz}{dt} $.

解

$$ \begin{array}{l}\frac{\mathrm{d}z}{\mathrm{d}t}\quad=\quad\frac{\partial z}{\partial x}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial z}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}t}\\ \quad=\quad\frac{1}{\sqrt{1-\left(x-y\right)^{2}}}\cdot3+\frac{\left(-1\right)}{\sqrt{1-\left(x-y\right)^{2}}}\cdot12t^{2}\\ \quad=\quad\frac{3\left(1-4t^{2}\right)}{\sqrt{1-\left(3t-4t^{3}\right)^{2}}}.\end{array} $$

1. 设 $ z = \arctan(xy) $，而 $ y = e^{x} $，求 $ \frac{dz}{dx} $.

解

$$ \begin{aligned}\frac{\mathrm{d}z}{\mathrm{d}x}&=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y}{1+x^{2}y^{2}}+\frac{x}{1+x^{2}y^{2}}\cdot\mathrm{e}^{x}\\&=\frac{\left(1+x\right)\mathrm{e}^{x}}{1+x^{2}\mathrm{e}^{2x}}.\end{aligned} $$

1. 设 $ u = \frac{\mathrm{e}^{ax}(y - z)}{a^2 + 1} $，而 $ y = a \sin x $, $ z = \cos x $，求 $ \frac{du}{dx} $.

解

$$ \begin{aligned}\frac{\mathrm{d}u}{\mathrm{d}x}&=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial u}{\partial z}\cdot\frac{\mathrm{d}z}{\mathrm{d}x}\\&=\frac{a\mathrm{e}^{ax}\left(y-z\right)}{a^{2}+1}+\frac{\mathrm{e}^{ax}}{a^{2}+1}\cdot a\cos x+\frac{\mathrm{e}^{ax}}{a^{2}+1}\cdot\left(-1\right)\cdot\left(-\sin x\right)\\&=\frac{\mathrm{e}^{ax}}{a^{2}+1}\left(a^{2}\sin x-a\cos x+a\cos x+\sin x\right)\\&=\mathrm{e}^{ax}\sin x.\end{aligned} $$

1. 设 $ z = \arctan \frac{x}{y} $，而 $ x = u + v, y = u - v $，验证

$$ \frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}=\frac{u-v}{u^{2}+v^{2}}. $$

证

$$ \begin{aligned}\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}&=\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial u}\right)+\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial v}\right)\\&=\frac{\frac{1}{y}}{1+\left(\frac{x}{y}\right)^{2}}\cdot1+\frac{-\frac{x}{y^{2}}}{1+\left(\frac{x}{y}\right)^{2}}\cdot1+\frac{\frac{1}{y}}{1+\left(\frac{x}{y}\right)^{2}}\cdot1+\\&\quad\frac{-\frac{x}{y^{2}}}{1+\left(\frac{x}{y}\right)^{2}}\cdot(-1)\\&=\frac{2y}{x^{2}+y^{2}}=\frac{u-v}{u^{2}+v^{2}},\end{aligned} $$

故等式成立.

1. 求下列函数的一阶偏导数（其中 f 具有一阶连续偏导数）：

(1)

$$ u=f(x^{2}-y^{2},e^{xy}); $$

(2)

$$ u=f\biggl(\frac{x}{y},\frac{y}{z}\biggr); $$

(3) u = f(x,xy,xyz).

解（1）将中间变量 $ x^{2}-y^{2} $， $ e^{xy} $依次编为1,2号，则

$$ \frac{\partial u}{\partial x}\;=\;f_{1}^{\prime}\cdot\frac{\partial}{\partial x}(x^{2}\;-\;y^{2})\;+\;f_{2}^{\prime}\cdot\frac{\partial}{\partial x}(e^{x y})\;=\;2x f_{1}^{\prime}\;+\;y e^{x y}f_{2}^{\prime}, $$

$$ \frac{\partial u}{\partial y}=f_{1}^{\prime}\cdot\frac{\partial}{\partial y}(x^{2}-y^{2})+f_{2}^{\prime}\cdot\frac{\partial}{\partial y}(e^{xy})=-\ 2yf_{1}^{\prime}+x e^{xy}f_{2}^{\prime}. $$

（2）令 $ s = \frac{x}{y} $, $ t = \frac{y}{z} $，则 $ u = f(s, t) $

$$ \frac{\partial u}{\partial x}\;=\;\frac{\partial f}{\partial s}\cdot\frac{\partial s}{\partial x}\;=\;\frac{1}{y}f_{s}\;, $$

$$ \frac{\partial u}{\partial y}=\frac{\partial f}{\partial s}\cdot\frac{\partial s}{\partial y}+\frac{\partial f}{\partial t}\cdot\frac{\partial t}{\partial y}=-\frac{x}{y^{2}}f_{s}+\frac{1}{z}f_{t}, $$

$$ \frac{\partial u}{\partial z}\;=\;\frac{\partial f}{\partial t}\cdot\frac{\partial t}{\partial z}\;=\;-\;\frac{y}{z^{2}}f_{t}. $$

（3）将中间变量 x, xy, xyz 依次编为 1, 2, 3 号，则

$$ \frac{\partial u}{\partial x}=f_{1}^{\prime}\cdot1+f_{2}^{\prime}\cdot y+f_{3}^{\prime}\cdot y z=f_{1}^{\prime}+y f_{2}^{\prime}+y z f_{3}^{\prime}, $$

$$ \frac{\partial u}{\partial y}=\int_{2}^{\prime}\cdot x+\int_{3}^{\prime}\cdot x z=x\int_{2}^{\prime}+x z f_{3}^{\prime}, $$

$$ \frac{\partial u}{\partial z}\;=\;f_{3}^{\prime}\cdot x y\;=\;x y f_{3}^{\prime}. $$

1. 设 $ z = xy + xF(u) $，而 $ u = \frac{y}{x} $， $ F(u) $ 为可导函数，证明

$$ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z+x y. $$

证

$$ \begin{align*}x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}&=x\Big[y+F\big(u\big)\;+x f^{\prime}\big(u\big)\frac{\partial u}{\partial x}\Big]+y\Big[x+x f^{\prime}\big(u\big)\frac{\partial u}{\partial y}\Big]\\&=x\Big[y+F\big(u\big)\;-\frac{y}{x}f^{\prime}\big(u\big)\Big]+y\big[x+f^{\prime}\big(u\big)\big]\\&=x y+x F\big(u\big)\;+x y\;=\;z+x y,\end{align*} $$

故等式成立.

1. 设 $ z = \frac{y}{f(x^{2} - y^{2})} $，其中 $ f(u) $ 为可导函数，验证

$$ \frac{1}{x}\frac{\partial z}{\partial x}+\frac{1}{y}\frac{\partial z}{\partial y}=\frac{z}{y^{2}}. $$

证

$$ \frac{\partial z}{\partial x}\;=\;\frac{-{\mathrm{~y~}}\cdot{\mathrm{~f~}}_{u}\cdot2x}{f^{2}({\mathrm{~}}u)}\;=\;-\frac{2x y f_{u}}{f^{2}({\mathrm{~}}u)}, $$

$$ \frac{\partial z}{\partial y}=\frac{f(u)-y f_{u}\cdot(-2y)}{f^{2}(u)}=\frac{1}{f(u)}+\frac{2y^{2}f_{u}}{f^{2}(u)}, $$

故

$$ \frac{1}{x}\frac{\partial z}{\partial x}+\frac{1}{y}\frac{\partial z}{\partial y}=-\frac{2y f_{u}}{f^{2}(u)}+\frac{1}{y f(u)}+\frac{2y f_{u}}{f^{2}(u)}=\frac{1}{y f(u)}=\frac{z}{y^{2}}. $$

1. 设 $ z = f(x^{2} + y^{2}) $，其中 f 具有二阶导数，求 $ \frac{\partial^{2}z}{\partial x^{2}} $， $ \frac{\partial^{2}z}{\partial x\partial y} $， $ \frac{\partial^{2}z}{\partial y^{2}} $。

解 令 $ u = x^{2} + y^{2} $，则 $ z = f(u) $。记 $ f' = f'(u) $， $ f'' = f''(u) $。

$$ \frac{\partial z}{\partial x}\;=\;f^{\prime}(u)\;\cdot\;\frac{\partial u}{\partial x}\;=\;2x f^{\prime}\;, $$

$$ \frac{\partial z}{\partial y}\;=\;f^{\prime}\left(u\right)\;\frac{\partial u}{\partial y}\;=\;2y f^{\prime}\;, $$

$$ \frac{\partial^{2}z}{\partial x^{2}}\;=\;2f^{\prime}\;+\;2x f^{\prime \prime}\cdot\frac{\partial u}{\partial x}\;=\;2f^{\prime}\;+\;4x^{2}f^{\prime \prime}, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=2x f^{\prime \prime}\cdot\frac{\partial u}{\partial y}=4x y f^{\prime \prime}, $$

$$ \frac{\partial^{2}z}{\partial y^{2}}\;=\;2f^{\prime}\;+\;2y f^{\prime \prime}\cdot\frac{\partial u}{\partial y}\;=\;2f^{\prime}\;+\;4y^{2}f^{\prime \prime}. $$

1. 求下列函数的 $ \frac{\partial^{2}z}{\partial x^{2}} $, $ \frac{\partial^{2}z}{\partial x\partial y} $, $ \frac{\partial^{2}z}{\partial y^{2}} $ (其中 f 具有二阶连续偏导数):

(1)

$$ z=f(x y,y); $$

(2)

$$ z=f\Big(x,\frac{x}{y}\Big); $$

$$ (3)z=f(x y^{2},x^{2}y);\qquad\quad(4)z=f(\sin x,\cos y,\mathrm{e}^{x+y}). $$

解 (1) 令 s = xy, t = y，则 $ z = f(s, t) $, s 和 t 是中间变量。将 s, t 依次编为 1, 2 号，则

$$ \frac{\partial z}{\partial x}\;=\;f_{1}^{\prime}\; \cdot\;\frac{\partial s}{\partial x}\;=\;y f_{1}^{\prime}\;,\quad\frac{\partial z}{\partial y}\;=\;f_{1}^{\prime}\; \cdot\;\frac{\partial s}{\partial y}\;+\;f_{2}^{\prime}\; \cdot\;\frac{\mathrm{d}t}{\mathrm{d}y}\;=\;x f_{1}^{\prime}\;+\;f_{2}^{\prime}. $$

因为 $ f(s,t) $是s和t的函数，所以 $ f'_{1} $和 $ f'_{2} $也是s和t的函数，从而 $ f'_{1} $和 $ f'_{2} $是以s和t为中间变量的x和y的函数。故

$$ \frac{\partial^{2}z}{\partial x^{2}}=\frac{\partial}{\partial x}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial x}(y f_{1}^{\prime})=y f_{11}^{\prime \prime}\cdot\frac{\partial s}{\partial x}=y^{2}f_{11}^{\prime \prime}, $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial y}(y f_{1}^{\prime})=f_{1}^{\prime}+y\bigg(f_{11}^{\prime \prime}\cdot\frac{\partial s}{\partial y}+f_{12}^{\prime \prime}\cdot\frac{\mathrm{d}t}{\mathrm{d}y}\bigg)\\&=f_{1}^{\prime}+x y f_{11}^{\prime \prime}+y f_{12}^{\prime \prime},\end{align*} $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial y^{2}}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial y}\bigg)=\frac{\partial}{\partial y}\big(xf^{\prime}_{~1}+f^{\prime}_{~2}\big)\\&=x\bigg(f^{\prime \prime}_{~11}\frac{\partial s}{\partial y}+f^{\prime \prime}_{~12}\frac{\mathrm{d}t}{\mathrm{d}y}\bigg)+f^{\prime \prime}_{~21}\frac{\partial s}{\partial y}+f^{\prime \prime}_{~22}\frac{\mathrm{d}t}{\mathrm{d}y}\\&=x^{2}f^{\prime \prime}_{~11}+2xf^{\prime \prime}_{~12}+f^{\prime \prime}_{~22}.\end{align*} $$

（2）令 $ s = x, t = \frac{x}{y} $，并将 s, t 依次编为 1, 2 号，则

$$ \frac{\partial z}{\partial x}\;=\;f_{1}^{\prime}\;\frac{\mathrm{d}s}{\mathrm{d}x}\;+\;f_{2}^{\prime}\;\frac{\partial t}{\partial x}\;=\;f_{1}^{\prime}\;+\;\frac{1}{y}f_{2}^{\prime}\;, $$

$$ \frac{\partial z}{\partial y}\;=\;f_{2}^{\prime}\frac{\partial t}{\partial y}\;=\;-\;\frac{x}{y^{2}}f_{2}^{\prime}. $$

因为 $ f(s,t) $是s和t的函数，所以 $ f'_{1} $和 $ f'_{2} $也是s和t的函数，从而 $ f'_{1} $和 $ f'_{2} $是以s和t为中间变量的x和y的函数。故

$$ \begin{align*}\frac{\partial^{2}z}{\partial x^{2}}&=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial x}\Big(f_{1}^{\prime}+\frac{1}{y}f_{2}^{\prime}\Big)=f_{11}^{\prime \prime}+f_{12}^{\prime \prime}\cdot\frac{\partial t}{\partial x}+\frac{1}{y}\Big(f_{21}^{\prime \prime}+f_{22}^{\prime \prime}\cdot\frac{\partial t}{\partial x}\Big)\\&=f_{11}^{\prime \prime}+\frac{2}{y}f_{12}^{\prime \prime}+\frac{1}{y^{2}}f_{22}^{\prime \prime},\end{align*} $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial y}\bigg(f_{1}^{\prime}+\frac{1}{y}f_{2}^{\prime}\bigg)=f_{12}^{\prime \prime}\cdot\frac{\partial t}{\partial y}-\frac{1}{y^{2}}f_{2}^{\prime}+\frac{1}{y}f_{22}^{\prime \prime}\frac{\partial t}{\partial y}\\&=-\frac{x}{y^{2}}f_{12}^{\prime \prime}-\frac{1}{y^{2}}f_{2}^{\prime}-\frac{x}{y^{3}}f_{22}^{\prime \prime},\end{align*} $$

$$ \frac{\partial^{2}z}{\partial y^{2}}=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial y}\bigg)=\frac{\partial}{\partial y}\bigg(-\frac{x}{y^{2}}f_{2}^{\prime}\bigg)=\frac{2x}{y^{3}}f_{2}^{\prime\prime}-\frac{x}{y^{2}}f_{22}^{\prime\prime}\frac{\partial t}{\partial y}=\frac{2x}{y^{3}}f_{2}^{\prime}+\frac{x^{2}}{y^{4}}f_{22}^{\prime\prime}. $$

（3）令 $ s = xy^{2} $， $ t = x^{2}y $，并将 s, t 依次编为 1, 2 号，则

$$ \frac{\partial z}{\partial x}\;=\;f_{1}^{\prime}\;\frac{\partial s}{\partial x}\;+\;f_{2}^{\prime}\;\frac{\partial t}{\partial x}\;=\;y^{2}f_{1}^{\prime}\;+\;2x y f_{2}^{\prime}\;, $$

$$ \frac{\partial z}{\partial y}\;=\;f_{1}^{\prime}\;\frac{\partial s}{\partial y}\;+\;f_{2}^{\prime}\;\frac{\partial t}{\partial y}\;=\;2xyf_{1}^{\prime}\;+\;x^{2}f_{2}^{\prime}. $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x^{2}}&=\frac{\partial}{\partial x}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial x}(y^{2}f_{1}^{\prime}+2xyf_{2}^{\prime})\\&=y^{2}\bigg(f_{11}^{\prime \prime}\cdot\frac{\partial s}{\partial x}+f_{12}^{\prime \prime}\cdot\frac{\partial t}{\partial x}\bigg)+2yf_{2}^{\prime}+2xy\bigg(f_{21}^{\prime \prime}\frac{\partial s}{\partial x}+f_{22}^{\prime \prime}\frac{\partial t}{\partial x}\bigg)\\&=y^{2}\left(y^{2}f_{11}^{\prime \prime}+2xyf_{12}^{\prime \prime}\right)+2yf_{2}^{\prime}+2xy\left(y^{2}f_{21}^{\prime \prime}+2xyf_{22}^{\prime \prime}\right)\\&=2yf_{2}^{\prime}+y^{4}f_{11}^{\prime \prime}+4xy^{3}f_{12}^{\prime \prime}+4x^{2}y^{2}f_{22}^{\prime \prime},\end{align*} $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial y}(y^{2}f_{1}^{\prime}+2xyf_{2}^{\prime})\\&=2yf_{1}^{\prime}+y^{2}\bigg(f_{11}^{\prime \prime}\cdot\frac{\partial s}{\partial y}+f_{12}^{\prime \prime}\cdot\frac{\partial t}{\partial y}\bigg)+\\&\quad2xf_{2}^{\prime}+2xy\bigg(f_{21}^{\prime \prime}\frac{\partial s}{\partial y}+f_{22}^{\prime \prime}\frac{\partial t}{\partial y}\bigg)\\&=2yf_{1}^{\prime}+y^{2}\big(2xyf_{11}^{\prime \prime}+x^{2}f_{12}^{\prime \prime}\big)+2xf_{2}^{\prime}\\&\quad+2xy\big(2xyf_{21}^{\prime \prime}+x^{2}f_{22}^{\prime \prime}\big)\\&=2yf_{1}^{\prime}+2xf_{2}^{\prime}+2xy^{3}f_{11}^{\prime \prime}+5x^{2}y^{2}f_{12}^{\prime \prime}+2x^{3}yf_{22}^{\prime \prime}\;,\end{align*} $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial y^{2}}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial y}\bigg)=\frac{\partial}{\partial y}(2xyf_{1}^{\prime}+x^{2}f_{2}^{\prime})\\&=2xf_{1}^{\prime}+2xy\bigg(f_{11}^{\prime \prime}\frac{\partial s}{\partial y}+f_{12}^{\prime \prime}\frac{\partial t}{\partial y}\bigg)+x^{2}\bigg(f_{21}^{\prime \prime}\frac{\partial s}{\partial y}+f_{22}^{\prime \prime}\frac{\partial t}{\partial y}\bigg)\\&=2xf_{1}^{\prime}+2xy(2xyf_{11}^{\prime \prime}+x^{2}f_{12}^{\prime \prime})+x^{2}(2xyf_{21}^{\prime \prime}+x^{2}f_{22}^{\prime \prime})\\&=2xf_{1}^{\prime}+4x^{2}y^{2}f_{11}^{\prime \prime}+4x^{3}yf_{12}^{\prime \prime}+x^{4}f_{22}^{\prime \prime}.\end{align*} $$

（4）令 $ u = \sin x, v = \cos y, w = e^{x+y} $，并将 u, v, w 依次编为 1,2,3 号，则

$$ \frac{\partial z}{\partial x}=\int_{1}^{\prime}\frac{\mathrm{d}u}{\mathrm{d}x}+\int_{3}^{\prime}\frac{\partial w}{\partial x}=\cos x f_{1}^{\prime}+\mathrm{e}^{x+y}f_{3}^{\prime}, $$

$$ \frac{\partial z}{\partial y}=f_{2}^{\prime}\frac{\mathrm{d}v}{\mathrm{d}y}+f_{3}^{\prime}\frac{\partial w}{\partial y}=-\sin yf_{2}^{\prime}+\mathrm{e}^{x+y}f_{3}^{\prime}, $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x^{2}}&=\frac{\partial}{\partial x}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial x}\big(\cos x f_{1}^{\prime}+e^{x+y}f_{3}^{\prime}\big)\\&=-\sin x f_{1}^{\prime}+\cos x\bigg(f_{11}^{\prime \prime}\frac{\mathrm{d}u}{\mathrm{d}x}+f_{13}^{\prime \prime}\frac{\mathrm{d}w}{\mathrm{d}x}\bigg)+e^{x+y}f_{3}^{\prime}+e^{x+y}\bigg(f_{31}^{\prime \prime}\frac{\mathrm{d}u}{\mathrm{d}x}+f_{33}^{\prime \prime}\frac{\mathrm{d}w}{\mathrm{d}x}\bigg)\\&=-\sin x f_{1}^{\prime}+\cos x\big(\cos x f_{11}^{\prime \prime}+e^{x+y}f_{13}^{\prime \prime}\big)+e^{x+y}f_{3}^{\prime}+e^{x+y}\big(\cos x f_{31}^{\prime \prime}+e^{x+y}f_{33}^{\prime \prime}\big)\end{align*} $$

$$ \begin{array}{r l}{=}&{{}\mathrm{e}^{x+y}f_{3}^{\prime}-\sin x f_{1}^{\prime}+\cos^{2}x f_{11}^{\prime\prime}+2\mathrm{e}^{x+y}\cos x f_{13}^{\prime\prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime\prime},}\end{array} $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial y}(\cos x f_{1}^{\prime}+\mathrm{e}^{x+y}f_{3}^{\prime})\\&=\cos x\bigg(f_{12}^{\prime \prime}\frac{\mathrm{d}v}{\mathrm{d}y}+f_{13}^{\prime \prime}\frac{\partial w}{\partial y}\bigg)+\mathrm{e}^{x+y}f_{3}^{\prime}+\mathrm{e}^{x+y}\bigg(f_{32}^{\prime \prime}\frac{\mathrm{d}v}{\mathrm{d}y}+f_{33}^{\prime \prime}\frac{\partial w}{\partial y}\bigg)\\&=\cos x(-\sin y f_{12}^{\prime \prime}+\mathrm{e}^{x+y}f_{13}^{\prime \prime})+\mathrm{e}^{x+y}f_{3}^{\prime}+\mathrm{e}^{x+y}(-\sin y f_{32}^{\prime \prime}+\mathrm{e}^{x+y}f_{33}^{\prime \prime})\\&=\mathrm{e}^{x+y}f_{3}^{\prime}-\cos x\sin y f_{12}^{\prime \prime}+\mathrm{e}^{x+y}\cos x f_{13}^{\prime \prime}-\mathrm{e}^{x+y}\sin y f_{32}^{\prime \prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime \prime},\end{align*} $$

$$ \begin{aligned}\frac{\partial^{2}z}{\partial y^{2}}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial y}\bigg)=\frac{\partial}{\partial y}(-\sin yf_{2}^{\prime}+\mathrm{e}^{x+y}f_{3}^{\prime})\\&=-\cos yf_{2}^{\prime}-\sin y\bigg(f_{22}^{\prime \prime}\frac{\mathrm{d}v}{\mathrm{d}y}+f_{23}^{\prime \prime}\frac{\partial w}{\partial y}\bigg)+\mathrm{e}^{x+y}f_{3}^{\prime}+\mathrm{e}^{x+y}\bigg(f_{32}^{\prime \prime}\frac{\mathrm{d}v}{\mathrm{d}y}+f_{33}^{\prime \prime}\frac{\partial w}{\partial y}\bigg)\\&=-\cos yf_{2}^{\prime}-\sin y(-\sin yf_{22}^{\prime \prime}+\mathrm{e}^{x+y}f_{23}^{\prime \prime})+\mathrm{e}^{x+y}f_{3}^{\prime}+\mathrm{e}^{x+y}(-\sin yf_{32}^{\prime \prime}+\mathrm{e}^{x+y}f_{33}^{\prime \prime})\\&=\mathrm{e}^{x+y}f_{3}^{\prime}-\cos yf_{2}^{\prime}+\sin^{2}yf_{22}^{\prime \prime}-2\mathrm{e}^{x+y}\sin yf_{23}^{\prime \prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime \prime}.\end{aligned} $$

1. 设 $ u = f(x, y) $ 的所有二阶偏导数连续，而

$$ x~=~\frac{s~-\sqrt{3}t}{2},\quad y~=~\frac{\sqrt{3}s~+t}{2}, $$

证明

$$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2} 及 \frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}=\frac{\partial^{2}u}{\partial s^{2}}+\frac{\partial^{2}u}{\partial t^{2}}. $$

证 因为

$$ \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}, $$

$$ \frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}, $$

所以

$$ \begin{align*}\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2}&=\left(\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}\right)^{2}+\left(-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}\right)^{2}\\&=\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}.\end{align*} $$

又因为

$$ \begin{align*}\frac{\partial^{2}u}{\partial s^{2}}&=\frac{\partial}{\partial s}\bigg(\frac{\partial u}{\partial s}\bigg)=\frac{\partial}{\partial s}\bigg(\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}\bigg)\\&=\frac{1}{2}\bigg(\frac{\partial^{2}u}{\partial x^{2}}\frac{\partial x}{\partial s}+\frac{\partial^{2}u}{\partial x\partial y}\frac{\partial y}{\partial s}\bigg)+\frac{\sqrt{3}}{2}\bigg(\frac{\partial^{2}u}{\partial y\partial x}\frac{\partial x}{\partial s}+\frac{\partial^{2}u}{\partial y^{2}}\frac{\partial y}{\partial s}\bigg)\end{align*} $$

$$ \begin{aligned}&=\frac{1}{2}\bigg(\frac{1}{2}\frac{\partial^{2}u}{\partial x^{2}}+\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial x\partial y}\bigg)+\frac{\sqrt{3}}{2}\bigg(\frac{1}{2}\frac{\partial^{2}u}{\partial y\partial x}+\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial y^{2}}\bigg)\\&=\frac{1}{4}\frac{\partial^{2}u}{\partial x^{2}}+\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial x\partial y}+\frac{3}{4}\frac{\partial^{2}u}{\partial y^{2}},\end{aligned} $$

$$ \begin{align*}\frac{\partial^{2}u}{\partial t^{2}}&=\frac{\partial}{\partial t}\bigg(\frac{\partial u}{\partial t}\bigg)=\frac{\partial}{\partial t}\bigg(-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}\bigg)\\&=-\frac{\sqrt{3}}{2}\bigg(\frac{\partial^{2}u}{\partial x^{2}}\frac{\partial x}{\partial t}+\frac{\partial^{2}u}{\partial x\partial y}\frac{\partial y}{\partial t}\bigg)+\frac{1}{2}\bigg(\frac{\partial^{2}u}{\partial y\partial x}\frac{\partial x}{\partial t}+\frac{\partial^{2}u}{\partial y^{2}}\frac{\partial y}{\partial t}\bigg)\\&=-\frac{\sqrt{3}}{2}\bigg(-\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial x^{2}}+\frac{1}{2}\frac{\partial^{2}u}{\partial x\partial y}\bigg)+\frac{1}{2}\bigg(-\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial y\partial x}+\frac{1}{2}\frac{\partial^{2}u}{\partial y^{2}}\bigg)\\&=\frac{3}{4}\frac{\partial^{2}u}{\partial x^{2}}-\frac{\sqrt{3}}{2}\frac{\partial^{2}u}{\partial x\partial y}+\frac{1}{4}\frac{\partial^{2}u}{\partial y^{2}},\end{align*} $$

所以 $ \frac{\partial^{2}u}{\partial s^{2}}+\frac{\partial^{2}u}{\partial t^{2}}=\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}. $

习题9-5

隐函数的求导公式

1. 设 $ \sin y + e^{x} - xy^{2} = 0 $，求 $ \frac{dy}{dx} $.

解 设 $ F(x,y) = \sin y + e^{x} - xy^{2} $，则

$$ F_{_{x}}=e^{^{x}}-y^{2},\quad F_{_{y}}=\cos y-2xy. $$

当 $ F_{y}\neq0 $时，有

$$ \begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{F_{x}}{F_{y}}=-\frac{\mathrm{e}^{x}-y^{2}}{\cos y-2xy}\\&=\frac{y^{2}-\mathrm{e}^{x}}{\cos y-2xy}.\end{aligned} $$

1. 设 $ \ln\sqrt{x^{2}+y^{2}}=\arctan\frac{y}{x} $，求 $ \frac{dy}{dx} $

解 设 $ F(x,y) = \ln\sqrt{x^{2} + y^{2}} - \arctan\frac{y}{x} $，则一阶偏导数分别为

$$ F_{x}=\frac{1}{\sqrt{x^{2}+y^{2}}}\cdot\frac{2x}{2\sqrt{x^{2}+y^{2}}}-\frac{1}{1+\left(\frac{y}{x}\right)^{2}}\cdot\left(-\frac{y}{x^{2}}\right)=\frac{x+y}{x^{2}+y^{2}}, $$

$$ F_{y}=\frac{1}{\sqrt{x^{2}+y^{2}}}\cdot\frac{2y}{2\sqrt{x^{2}+y^{2}}}-\frac{1}{1+\left(\frac{y}{x}\right)^{2}}\cdot\frac{1}{x}=\frac{y-x}{x^{2}+y^{2}}. $$

当 $ F_{y}\neq0 $时，有

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=-\left.\frac{F_{x}}{F_{y}}\right.=-\left.\frac{x+y}{x^{2}+y^{2}}\right/\frac{y-x}{x^{2}+y^{2}}\;=\;\frac{x+y}{x-y}. $$

1. 设 $ x + 2y + z - 2\sqrt{xyz} = 0 $，求 $ \frac{\partial z}{\partial x} $ 及 $ \frac{\partial z}{\partial y} $.

解法一 设 $ F(x,y,z)=x+2y+z-2\sqrt{xyz} $，则

$$ F_{x}=1-\frac{y z}{\sqrt{x y z}},\quad F_{y}=2-\frac{x z}{\sqrt{x y z}},\quad F_{z}=1-\frac{x y}{\sqrt{x y z}}. $$

于是当 $ F_{z}\neq0 $时，有

$$ \frac{\partial z}{\partial x}\;=\;-\;\frac{F_{_{x}}}{F_{_{z}}}\;=\;\frac{y z\;-\;\sqrt{x y z}}{\sqrt{x y z}\;-\;x y}, $$

$$ \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}=\frac{xz-2\sqrt{xyz}}{\sqrt{xyz}-xy}. $$

解法二 在所给方程两端分别对 x 求偏导数，并注意 $ z = z(x, y) $，得

$$ 1+\frac{\partial z}{\partial x}-\frac{1}{\sqrt{xyz}}\Big(yz+xy\frac{\partial z}{\partial x}\Big)=0, $$

当 $ 1 - \frac{xy}{\sqrt{xyz}} \neq 0 $ 时，解得 $ \frac{\partial z}{\partial x} = \frac{\frac{yz}{\sqrt{xyz}} - 1}{1 - \frac{xy}{\sqrt{xyz}}} = \frac{yz - \sqrt{xyz}}{\sqrt{xyz} - xy} $.

同理，方程两端分别对 y 求偏导数，得

$$ 2+\frac{\partial z}{\partial y}-\frac{1}{\sqrt{xyz}}\Big(xz+xy\frac{\partial z}{\partial y}\Big)=0, $$

当 $ 1 - \frac{xy}{\sqrt{xyz}} \neq 0 $ 时，解得 $ \frac{\partial z}{\partial y} = \frac{\frac{xz}{\sqrt{xyz}} - 2}{1 - \frac{xy}{\sqrt{xyz}}} = \frac{xz - 2\sqrt{xyz}}{\sqrt{xyz} - xy} $.

解法三 对所给方程两端分别求全微分，得

$$ \mathrm{d}x+2\mathrm{d}y+\mathrm{d}z-\frac{1}{\sqrt{xyz}}(yz\mathrm{d}x+xz\mathrm{d}y+xy\mathrm{d}z)=0, $$

即

$$ \left(1-\frac{x y}{\sqrt{x y z}}\right)\mathrm{d}z=\left(\frac{y z}{\sqrt{x y z}}-1\right)\mathrm{d}x+\left(\frac{x z}{\sqrt{x y z}}-2\right)\mathrm{d}y. $$

当 $ \sqrt{xyz} - xy \neq 0 $ 时，解得 $ \mathrm{d}z = \frac{yz - \sqrt{xyz}}{\sqrt{xyz} - xy}dx + \frac{xz - 2\sqrt{xyz}}{\sqrt{xyz} - xy}dy $。所以

$$ \frac{\partial z}{\partial x}=\frac{yz-\sqrt{xyz}}{\sqrt{xyz}-xy},\quad\frac{\partial z}{\partial y}=\frac{xz-2\sqrt{xyz}}{\sqrt{xyz}-xy}. $$

1. 设 $ \frac{x}{z} = \ln \frac{z}{y} $，求 $ \frac{\partial z}{\partial x} $ 及 $ \frac{\partial z}{\partial y} $.

解 令 $ F(x, y, z) = \frac{x}{z} - \ln \frac{z}{y} $，则

$$ F_{x}=\frac{1}{z},\quad F_{y}=-\frac{1}{\frac{z}{y}}\cdot\left(-\frac{z}{y^{2}}\right)=\frac{1}{y}, $$

$$ F_{z}\;=\;-\;\frac{x}{z^{2}}\;-\;\frac{1}{\frac{z}{y}}\;{\cdot}\;\frac{1}{y}\;=\;-\;\frac{x+z}{z^{2}}. $$

于是当 $ F_{2}\neq0 $时，有

$$ \frac{\partial z}{\partial x}\;=\;-\;\frac{F_{x}}{F_{z}}\;=\;-\;\left.-\frac{1}{z}\right/\left(\;-\;\frac{x\;+\;z}{z^{2}}\right)=\;\frac{z}{x\;+\;z}\,, $$

$$ \frac{\partial z}{\partial y}=-\left.\frac{F_{y}}{F_{z}}\right.=-\left.\frac{1}{y}\middle/\left(\begin{array}{c}{x+z}\\ {z^{2}}\\ \end{array}\right)\right.=\frac{z^{2}}{y\left(x+z\right)}. $$

1. 设 $ 2\sin(x+2y-3z)=x+2y-3z $，证明 $ \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1 $.

证 设 $ F(x,y,z)=2\sin(x+2y-3z)-x-2y+3z $，则

$$ F_{x}=2\cos(x+2y-3z)-1, $$

$$ F_{y}=2\cos(x+2y-3z)\cdot2-2=2F_{x}, $$

$$ F_{z}=2\cos(x+2y-3z)\cdot(-3)+3=-3F_{x}, $$

故当 $ F_{z}\neq0 $时，有

$$ \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=-\frac{F_{x}}{F_{z}}-\frac{F_{y}}{F_{z}}=\frac{1}{3}+\frac{2}{3}=1. $$

1. 设 $ x = x(y, z) $, $ y = y(x, z) $, $ z = z(x, y) $ 都是由方程 $ F(x, y, z) = 0 $ 所确定的具有连续偏导数的函数，证明： $ \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} \cdot \frac{\partial z}{\partial x} = -1 $.

证 因为

$$ \frac{\partial x}{\partial y}=-\frac{F_{y}}{F_{x}},\quad\frac{\partial y}{\partial z}=-\frac{F_{z}}{F_{x}},\quad\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}, $$

所以

$$ \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=\left(\begin{array}{c}F_{x}\\ -\frac{F_{y}}{F_{x}}\end{array}\right)\cdot\left(\begin{array}{c}F_{z}\\ -\frac{F_{y}}{F_{z}}\end{array}\right)\cdot\left(\begin{array}{c}F_{x}\\ -\frac{F_{y}}{F_{z}}\end{array}\right)=-1. $$

1. 设 $ \Phi(u,v) $ 具有连续偏导数，证明由方程 $ \Phi(cx-az,cy-bz)=0 $ 所确定的函数 $ z=f(x,y) $ 满足 $ a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=c $.

证令 $ u = cx - az, v = cy - bz $，则

$$ \phi_{_{x}}=\phi_{_{u}}\cdot\frac{\partial u}{\partial x}=c\phi_{_{u}}, $$

$$ \Phi_{y}\;=\;\Phi_{v}\;.\;{\frac{\partial v}{\partial y}}\;=\;c\;\Phi_{v}\;, $$

$$ \Phi_{z}=\Phi_{u}\cdot\frac{\partial u}{\partial z}+\Phi_{v}\cdot\frac{\partial v}{\partial z}=-a\Phi_{u}-b\Phi_{v}. $$

故当 $ \Phi_{z}\neq0 $时，有

$$ \frac{\partial z}{\partial x}=-\frac{\phi_{_{x}}}{\phi_{_{z}}}=\frac{c\phi_{_{u}}}{a\phi_{_{u}}+b\phi_{_{v}}}, $$

$$ \frac{\partial z}{\partial y}=-\frac{\varPhi_{y}}{\varPhi_{z}}=\frac{c\varPhi_{v}}{a\varPhi_{u}+b\varPhi_{v}}. $$

干是

$$ a\ \frac{\partial z}{\partial x}\ +\ b\ \frac{\partial z}{\partial y}\ =\ a\ \cdot\ \frac{c\ \Phi_{u}}{a\ \Phi_{u}\ +\ b\ \Phi_{v}}\ +\ b\ \cdot\ \frac{c\ \Phi_{v}}{a\ \Phi_{u}\ +\ b\ \Phi_{v}}\ =\ c. $$

1. 设 $ e^z - xyz = 0 $, 求 $ \frac{\partial^2 z}{\partial x^2} $.

解 设 $ F(x,y,z)=\mathrm{e}^{z}-xyz $，则 $ F_{x}=-yz $， $ F_{z}=\mathrm{e}^{z}-xy $。于是当 $ F_{z}\neq0 $ 时，有

$$ \frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}=\frac{yz}{\mathrm{e}^{z}-xy}, $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x^{2}}&=\frac{\partial}{\partial x}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{y\frac{\partial z}{\partial x}(\mathrm{e}^{z}-xy)-yz\bigg(\mathrm{e}^{z}\frac{\partial z}{\partial x}-y\bigg)}{(\mathrm{e}^{z}-xy)^{2}}\\&=\frac{y^{2}z-yz\bigg(\mathrm{e}^{z}\cdot\frac{yz}{\mathrm{e}^{z}-xy}-y\bigg)}{(\mathrm{e}^{z}-xy)^{2}}\\&=\frac{2y^{2}ze^{z}-2xy^{3}z-y^{2}z^{2}\mathrm{e}^{z}}{(\mathrm{e}^{z}-xy)^{3}}.\end{align*} $$

☒ $ ^{*}9 $. 设 $ z^{3}-3xyz=a^{3} $，求 $ \frac{\partial^{2}z}{\partial x\partial y} $.

解 设 $ F(x,y,z)=z^{3}-3xyz-a^{3} $，则

$$ F_{x}=-3y z,\quad F_{y}=-3x z,\quad F_{z}=3z^{2}-3x y. $$

于是当 $ F_{z} \neq 0 $ 时，有 $ \frac{\partial z}{\partial x} = -\frac{F_{x}}{F_{z}} = \frac{yz}{z^{2} - xy}, \frac{\partial z}{\partial y} = -\frac{F_{y}}{F_{z}} = \frac{xz}{z^{2} - xy} $

$$ \begin{aligned}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}\bigg(\frac{\partial z}{\partial x}\bigg)=\frac{\partial}{\partial y}\bigg(\frac{yz}{z^{2}-xy}\bigg)\\&=\frac{\bigg(z+y\frac{\partial z}{\partial y}\bigg)\big(z^{2}-xy\big)-yz\bigg(2z\frac{\partial z}{\partial y}-x\bigg)}{\big(z^{2}-xy\big)^{2}}\end{aligned} $$

$$ \begin{aligned}&=\frac{\left(z+\frac{xyz}{z^{2}-xy}\right)\cdot\left(z^{2}-xy\right)-yz\left(\frac{2xz^{2}}{z^{2}-xy}-x\right)}{\left(z^{2}-xy\right)^{2}}\\&=\frac{z\left(z^{4}-2xyz^{2}-x^{2}y^{2}\right)}{\left(z^{2}-xy\right)^{3}}.\end{aligned} $$

1. 求由下列方程组所确定的函数的导数或偏导数：

（1）设 $ \{\begin{aligned}&z=x^{2}+y^{2},\\ &x^{2}+2y^{2}+3z^{2}=20,\end{aligned}. $ 求 $ \frac{dy}{dx} $， $ \frac{dz}{dx} $;

（2）设 $ \{\begin{aligned}x+y+z&=0,\\ x^{2}+y^{2}+z^{2}&=1,\end{aligned}. $求 $ \frac{dx}{dz} $， $ \frac{dy}{dz} $;

（3）设 $ \{\begin{aligned}u&=f(ux,v+y),\\ v&=g(u-x,v^{2}y),\end{aligned}. $ 其中f,g具有一阶连续偏导数，求 $ \frac{\partial u}{\partial x} $， $ \frac{\partial v}{\partial x} $;

（4）设 $ \{\begin{aligned}x&=\mathrm{e}^{u}+u\sin v,\\ y&=\mathrm{e}^{u}-u\cos v,\end{aligned}. $求 $ \frac{\partial u}{\partial x} $， $ \frac{\partial u}{\partial y} $， $ \frac{\partial v}{\partial x} $， $ \frac{\partial v}{\partial y} $.

解（1）分别在两个方程两端对 x 求导，得

$$ \{\begin{aligned}\frac{\mathrm{d}z}{\mathrm{d}x}&=2x+2y\frac{\mathrm{d}y}{\mathrm{d}x},\\ 2x&+4y\frac{\mathrm{d}y}{\mathrm{d}x}+6z\frac{\mathrm{d}z}{\mathrm{d}x}=0.\end{aligned}. $$

移项，得

$$ \{\begin{aligned}&2y\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{\mathrm{d}z}{\mathrm{d}x}=-2x,\\ &2y\frac{\mathrm{d}y}{\mathrm{d}x}+3z\frac{\mathrm{d}z}{\mathrm{d}x}=-x.\end{aligned}. $$

当 $ D=\begin{vmatrix}2y & -1 \\ 2y & 3z\end{vmatrix}=6yz+2y\neq0 $ 时，解方程组得

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\left|\begin{array}{cc}-2x&-1\\-x&3z\end{array}\right|}{D}=\frac{-6xz-x}{6yz+2y}=\frac{-x(6z+1)}{2y(3z+1)}, $$

$$ \frac{\mathrm{d}z}{\mathrm{d}x}=\frac{\left|\begin{array}{cc}2y&-2x\\2y&-x\end{array}\right|}{D}=\frac{2xy}{6yz+2y}=\frac{x}{3z+1}. $$

（2）所给方程组确定两个一元隐函数： $ x=x(z) $ 和 $ y=y(z) $，将所给方程的两端分别对 z 求导并移项，得

$$ \{\begin{aligned}&\frac{\mathrm{d}x}{\mathrm{d}z}+\frac{\mathrm{d}y}{\mathrm{d}z}=-1,\\ &2x\frac{\mathrm{d}x}{\mathrm{d}z}+2y\frac{\mathrm{d}y}{\mathrm{d}z}=-2z.\end{aligned}. $$

当 $ D=\left|\begin{matrix}1&1\\ 2x&2y\end{matrix}\right|=2(y-x)\neq0 $ 时，解方程组得

$$ \frac{\mathrm{d}x}{\mathrm{d}z}=\frac{\left|\begin{array}{cc}-1&1\\-2z&2y\end{array}\right|}{D}=\frac{-2y+2z}{2\left(y-x\right)}=\frac{y-z}{x-y}, $$

$$ \frac{\mathrm{d}y}{\mathrm{d}z}=\frac{\left|\begin{array}{cc}1&-1\\ 2x&-2z\end{array}\right|}{D}=\frac{-2z+2x}{2\left(y-x\right)}=\frac{z-x}{x-y}. $$

（3）此方程组可以确定两个二元隐函数： $ u=u(x,y) $, $ v=v(x,y) $. 分别在方程两端对x求偏导数，得

$$ \{\begin{aligned}\frac{\partial u}{\partial x}&=f_{1}^{\prime}\cdot(u+x\frac{\partial u}{\partial x})+f_{2}^{\prime}\cdot\frac{\partial v}{\partial x},\\\frac{\partial v}{\partial x}&=g_{1}^{\prime}\cdot(\frac{\partial u}{\partial x}-1)+2g_{2}^{\prime}yv\cdot\frac{\partial v}{\partial x}.\end{aligned}. $$

移项整理后得

$$ \{\begin{aligned}&(xf_{1}^{\prime}-1)\frac{\partial u}{\partial x}+f_{2}^{\prime}\frac{\partial v}{\partial x}=-uf_{1}^{\prime},\\ &\\ &g_{1}^{\prime}\frac{\partial u}{\partial x}+(2yvg_{2}^{\prime}-1)\frac{\partial v}{\partial x}=g_{1}^{\prime}.\\ \end{aligned}. $$

当 $ D = \left| \begin{array}{ccc} xf_{1}^{\prime} - 1 & f_{2}^{\prime} \\ g_{1}^{\prime} & 2yvg_{2}^{\prime} - 1 \end{array} \right| = (xf_{1}^{\prime} - 1)(2yvg_{2}^{\prime} - 1) - f_{2}^{\prime}g_{1}^{\prime} \neq 0 $ 时，解方程组

得

$$ \begin{align*}\frac{\partial u}{\partial x}&=\frac{1}{D}\begin{vmatrix}-\ u f_{1}^{\prime}&\ f_{2}^{\prime}\\g_{1}^{\prime}&2y v g_{2}^{\prime}-1\end{vmatrix}\\&=\frac{-\ u f_{1}^{\prime}\left(2y v g_{2}^{\prime}-1\right)-f_{2}^{\prime}g_{1}^{\prime}}{\left(x f_{1}^{\prime}-1\right)\left(2y v g_{2}^{\prime}-1\right)-f_{2}^{\prime}g_{1}^{\prime}},\end{align*} $$

$$ \frac{\partial v}{\partial x}\;=\;\frac{1}{D}\left|\begin{array}{c c}{x f_{1}^{\prime}-1}&{-u f_{1}^{\prime}}\\ {g_{1}^{\prime}}&{g_{1}^{\prime}}\end{array}\right|\;=\;\frac{g_{1}^{\prime}\left(x f_{1}^{\prime}+u f_{1}^{\prime}-1\right)}{\left(x f_{1}^{\prime}-1\right)\left(2y v g_{2}^{\prime}-1\right)\;-f_{2}^{\prime}g_{1}^{\prime}}. $$

（4）此方程组确定的两个二元隐函数 $ u = u(x, y) $, $ v = v(x, y) $ 是已知函数的反函数，令

$$ F(x,y,u,v)=x-\mathrm{e}^{u}-u\sin v, $$

$$ G(x,y,u,v)=y-\mathrm{e}^{u}+u\cos v. $$

则

$$ F_{x}=1,\quad F_{y}=0,\quad F_{u}=-\mathrm{e}^{u}-\sin v,\quad F_{v}=-u\cos v, $$

$$ G_{x}=0,\quad G_{y}=1,\quad G_{u}=-\mathrm{e}^{u}+\cos v,\quad G_{v}=-u\sin v. $$

当 $ J = \frac{\partial(F, G)}{\partial(u, v)} = \begin{vmatrix} -e^{u} - \sin v & -u \cos v \\ -e^{u} + \cos v & -u \sin v \end{vmatrix} = u e^{u} (\sin v - \cos v) + u \neq 0 $ 时，由

隐函数求导公式得

$$ \begin{aligned}\frac{\partial u}{\partial x}&=-\left.\frac{1}{J}\frac{\partial\left(F,G\right)}{\partial\left(x,v\right)}=-\left.\frac{1}{J}\right|1\quad-u\cos v\right.\\&=\left.\frac{\sin v}{e^{u}\left(\sin v-\cos v\right)+1}\right.,\end{aligned} $$

$$ \begin{align*}\frac{\partial u}{\partial y}&=-\left.\frac{1}{J}\frac{\partial\left(F,G\right)}{\partial\left(y,v\right)}=-\left.\frac{1}{J}\right|0\quad-u\cos v\right|\\&=\frac{-\cos v}{e^{u}\left(\sin v-\cos v\right)+1},\end{align*} $$

$$ \begin{align*}\frac{\partial v}{\partial x}&=-\left.\frac{1}{J}\frac{\partial\left(F,G\right)}{\partial\left(u,x\right)}=-\left.\frac{1}{J}\right|-\mathrm{e}^{u}-\sin v\quad1\right|\\&=\frac{\cos v-\mathrm{e}^{u}}{u\left[\mathrm{e}^{u}\left(\sin v-\cos v\right)+1\right]},\end{align*} $$

$$ \begin{align*}\frac{\partial v}{\partial y}&=-\left.\frac{1}{J}\frac{\partial\left(F,G\right)}{\partial\left(u,y\right)}=-\left.\frac{1}{J}\right|-\mathrm{e}^{u}-\sin v\quad0\right|\\&=\frac{\sin v+\mathrm{e}^{u}}{u\left[\mathrm{e}^{u}\left(\sin v-\cos v\right)+1\right]}.\end{align*} $$

1. 设 $ y = f(x, t) $，而 $ t = t(x, y) $ 是由方程 $ F(x, y, t) = 0 $ 所确定的函数，其中 f, F 都具有一阶连续偏导数。试证明

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\partial f}{\partial x}\frac{\partial F}{\partial t}-\frac{\partial f}{\partial t}\frac{\partial F}{\partial x}}{\frac{\partial f}{\partial t}\frac{\partial F}{\partial y}+\frac{\partial F}{\partial t}}. $$

证法一　由方程组 $ \{\begin{aligned}y&=f(x,t),\\ F(x,y,t)&=0\end{aligned}. $ 可确定两个一元隐函数 $ y=y(x),t=t(x) $. 分别在两个方程两端对 x 求导可得

$$ \{\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x},\\ \frac{\partial F}{\partial x}&+\frac{\partial F}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}=0.\end{aligned}. $$

移项得

$$ \{\begin{aligned}&\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{\partial f}{\partial t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{\partial f}{\partial x},\\ &\\ &\frac{\partial F}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}=-\frac{\partial F}{\partial x}.\end{aligned}. $$

当 $ D = \begin{vmatrix} 1 & -\frac{\partial f}{\partial t} \\ \frac{\partial F}{\partial y} & \frac{\partial F}{\partial t} \end{vmatrix} = \frac{\partial F}{\partial t} + \frac{\partial f}{\partial t} \cdot \frac{\partial F}{\partial y} \neq 0 $ 时，解方程组得

$$ \frac{\mathrm{d}y}{\mathrm{d}x}\;=\;\frac{1}{D}\;.\;\left|\;\begin{array}{c c}{\displaystyle\frac{\partial f}{\partial x}}&{-\;\displaystyle\frac{\partial f}{\partial t}}\\ {}&{}\\ {-\;\displaystyle\frac{\partial F}{\partial x}}&{\displaystyle\frac{\partial F}{\partial t}}\\ \end{array}\right|\;=\;\frac{\displaystyle\frac{\partial f}{\partial x}\cdot\frac{\partial F}{\partial t}-\frac{\partial f}{\partial t}\cdot\frac{\partial F}{\partial x}}{\displaystyle\frac{\partial F}{\partial t}+\frac{\partial f}{\partial t}\cdot\frac{\partial F}{\partial y}}. $$

证法二 分别在 $ y=f(x,t) $ 及 $ F(x,y,t)=0 $ 两端求全微分，得

$$ \begin{array}{r}{\int\mathrm{d}\boldsymbol{y}\;=\;f_{x}\mathrm{d}x\;+\;f_{t}\mathrm{d}t,}\end{array} $$

$$ \begin{array}{r}{\left\lfloor F_{x}\mathrm{d}x\;+\;F_{y}\mathrm{d}y\;+\;F_{t}\mathrm{d}t\right.\;=\;0.}\end{array} $$

由(2)，得

$$ F_{t}\mathrm{d}t=-\left(F_{x}\mathrm{d}x+F_{y}\mathrm{d}y\right). $$

将 $ F_{t} $乘(1)式两端，并以(3)式代入，得

即

$$ \begin{aligned}&F_{t}\mathrm{d}y=f_{x}F_{t}\mathrm{d}x-f_{t}\left(F_{x}\mathrm{d}x+F_{y}\mathrm{d}y\right),\\&\left(F_{t}+f_{t}F_{y}\right)\mathrm{d}y=\left(f_{x}F_{t}-f_{t}F_{x}\right)\mathrm{d}x.\\ \end{aligned} $$

故当 $ F_{t}+f_{t}F_{y}\neq0 $时，有

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{f_{x}F_{t}-f_{t}F_{x}}{F_{t}+f_{t}F_{y}}. $$

习题9-6 多元函数微分学的几何应用

1. 设 $ f(t) = f_1(t)i + f_2(t)j + f_3(t)k $, $ g(t) = g_1(t)i + g_2(t)j + g_3(t)k $, $ \lim_{t \to t_0} f(t) = u $, $ \lim_{t \to t_0} g(t) = v $, 证明 $ \lim_{t \to t_0} [f(t) \times g(t)] = u \times v $.

证

$$ \begin{aligned}\lim_{t\to t_{0}}[f(t)\times\boldsymbol{g}(t)]&=\lim_{t\to t_{0}}\left|\begin{matrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\f_{1}(t)&f_{2}(t)&f_{3}(t)\\g_{1}(t)&g_{2}(t)&g_{3}(t)\end{matrix}\right|\\&=\lim_{t\to t_{0}}\left(f_{2}(t)g_{3}(t)-f_{3}(t)g_{2}(t),f_{3}(t)g_{1}(t)-f_{1}(t)g_{3}(t),\right.\\&\quad\left.f_{1}(t)g_{2}(t)-f_{2}(t)g_{1}(t)\right)\\&=\left(\lim_{t\to t_{0}}[f_{2}(t)g_{3}(t)-f_{3}(t)g_{2}(t)],\lim_{t\to t_{0}}[f_{3}(t)g_{1}(t)-f_{1}(t)g_{3}(t)],\lim_{t\to t_{0}}[f_{1}(t)g_{2}(t)-f_{2}(t)g_{1}(t)]\right)\\&=\left|\begin{matrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\lim_{t\to t_{0}}f_{1}(t)&\lim_{t\to t_{0}}f_{2}(t)&\lim_{t\to t_{0}}f_{3}(t)\\\lim_{t\to t_{0}}g_{1}(t)&\lim_{t\to t_{0}}g_{2}(t)&\lim_{t\to t_{0}}g_{3}(t)\end{matrix}\right|=\boldsymbol{u}\times\boldsymbol{v}.\end{aligned} $$

这个结果表示：两个向量值函数的向量积的极限等于它们各自的极限（向量）的向量积，即

$$ \begin{array}{c c l}{\displaystyle\lim_{t\to t_{0}}\bigl[\boldsymbol{f}(\mathbf{\theta}(t)\mathbf{\theta})\times\boldsymbol{g}(\mathbf{\theta}(t)\mathbf{\theta})\bigr]}&{=}&{\displaystyle\bigl[\lim_{t\to t_{0}}\boldsymbol{f}(\mathbf{\theta}(t)\mathbf{\theta})\bigr]\times\bigl[\lim_{t\to t_{0}}\boldsymbol{g}(\mathbf{\theta}(t)\mathbf{\theta})\bigr].}\end{array} $$

1. 下列各题中， $ r=f(t) $ 是空间中的质点 M 在时刻 t 的位置，求质点 M 在时刻 $ t_{0} $ 的速度向量和加速度向量，以及在任意时刻 t 的速率.

(1) $ \boldsymbol{r} = \boldsymbol{f}(t) = (t+1)\boldsymbol{i} + (t^2-1)\boldsymbol{j} + 2t\boldsymbol{k}, t_0 = 1 $;

(2) $ \boldsymbol{r} = \boldsymbol{f}(t) = (2\cos t)\boldsymbol{i} + (3\sin t)\boldsymbol{j} + 4t\boldsymbol{k}, t_0 = \frac{\pi}{2}; $

(3) $ \boldsymbol{r} = \boldsymbol{f}(t) = \left[ 2 \ln(t + 1) \right] \boldsymbol{i} + t^2 \boldsymbol{j} + \frac{1}{2} t^2 \boldsymbol{k}, t_0 = 1. $

解 （1）速度向量 $ \boldsymbol{v}_{0} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\bigg|_{t=1} = (i + 2tj + 2k)\bigg|_{t=1} = i + 2j + 2k $;

加速度向量 $ a_{0}=\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}\bigg|_{t=1}=2j $;

速率 $ \left|v(t)\right|=\left|i+2tj+2k\right|=\sqrt{5+4t^{2}} $

（2）速度向量 $ \boldsymbol{v}_{0} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\bigg|_{t=\frac{\pi}{2}} = \left[\begin{array}{c} (-2\sin t)i + (3\cos t)j + 4k \end{array}\right]_{t=\frac{\pi}{2}} = -2i + 4k $;

加速度向量 $ \boldsymbol{a}_{0} = \frac{\mathrm{d}^{2}\boldsymbol{r}}{\mathrm{d}t^{2}}\bigg|_{t=\frac{\pi}{2}} = \left[\begin{array}{c}(-2\cos t)i-(3\sin t)j\end{array}\right]_{t=\frac{\pi}{2}} = -3\boldsymbol{j} $;

速率 $ \left|v(t)\right|=\left|(-2\sin t)i+(3\cos t)j+4k\right|=\sqrt{9\cos^{2}t+4\sin^{2}t+16} $

$ =20+5\cos^{2}t $

（3）速度向量 $ \boldsymbol{v}_{0} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} \bigg|_{t=1} = \left( \frac{2}{t+1} \boldsymbol{i} + 2t \boldsymbol{j} + t \boldsymbol{k} \right) \bigg|_{t=1} = \boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k} $;

加速度向量 $ a_{0}=\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}\bigg|_{t=1}=\left[-\frac{2}{(t+1)^{2}}i+2j+k\right]_{t=1}=-\frac{1}{2}i+2j+k $;

速率 $ \left|v(t)\right|=\left|\frac{2}{t+1}i+2tj+tk\right|=\sqrt{5t^{2}+\frac{4}{(t+1)^{2}}} $

1. 求曲线 $ r = f(t) = (t - \sin t)i + (1 - \cos t)j + \left(4\sin\frac{t}{2}\right)k $ 在与 $ t_{0} = \frac{\pi}{2} $ 相应的点处的切线及法平面方程.

解 与 $ t_{0}=\frac{\pi}{2} $ 相应的点为 $ \left(\frac{\pi}{2}-1,1,2\sqrt{2}\right) $，曲线在该点处的切向量为 $ T=f'(t_{0})=(1,1,\sqrt{2}) $，于是所求切线方程为

$$ \frac{x-\left(\frac{\pi}{2}-1\right)}{1}=\frac{y-1}{1}=\frac{z-2\sqrt{2}}{\sqrt{2}}, $$

法平面方程为

$$ 1\cdot\left(x-\frac{\pi}{2}+1\right)+1\cdot(y-1)+\sqrt{2}(z-2\sqrt{2})=0, $$

即

$$ x+y+\sqrt{2}z=\frac{\pi}{2}+4. $$

1. 求曲线 $ x = \frac{t}{1 + t} $, $ y = \frac{1 + t}{t} $, $ z = t^{2} $ 在对应于 t = 1 的点处的切线及法平面方程.

解 曲线在对应于 t=1 的点为 $ \left(\frac{1}{2},2,1\right) $，该点处的切向量

$$ \boldsymbol{T}=\left(x^{\prime}(1),y^{\prime}(1),z^{\prime}(1)\right)=\left(\frac{1}{\left(1+t\right)^{2}},-\frac{1}{t^{2}},2t\right)\bigg|_{t=1}=\left(\frac{1}{4},-1,2\right), $$

于是曲线在该点处的切线方程为

$$ \frac{x-\frac{1}{2}}{\frac{1}{4}}=\frac{y-2}{-1}=\frac{z-1}{2}, $$

即

$$ \frac{x-\frac{1}{2}}{1}=\frac{y-2}{-4}=\frac{z-1}{8}. $$

所求法平面方程为

$$ \frac{1}{4}\left(x-\frac{1}{2}\right)-\left(y-2\right)+2\left(z-1\right)=0, $$

即

$$ 2x-8y+16z-1=0. $$

1. 求曲线 $ y^{2}=2mx, z^{2}=m-x $ 在点 $ (x_{0}, y_{0}, z_{0}) $ 处的切线及法平面方程.

解 设曲线的参数方程中的参数为 x，将方程 $ y^{2}=2mx $ 和 $ z^{2}=m-x $ 两端分别对 x 求导，得

$$ 2y\frac{\mathrm{d}y}{\mathrm{d}x}=2m,\quad2z\frac{\mathrm{d}z}{\mathrm{d}x}=-1,\quad 即 \quad\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{m}{y},\quad\frac{\mathrm{d}z}{\mathrm{d}x}=-\frac{1}{2z}. $$

所以曲线在点 $ (x_{0},y_{0},z_{0}) $的切向量为

$$ T=\Big(1,\frac{m}{y_{0}},-\frac{1}{2z_{0}}\Big). $$

于是在点 $ (x_{0},y_{0},z_{0}) $处的切线方程为

$$ \frac{x-x_{0}}{1}=\frac{y-y_{0}}{\frac{m}{y_{0}}}=\frac{z-z_{0}}{-\frac{1}{2z_{0}}}. $$

法平面方程为 $ (x - x_{0}) + \frac{m}{y_{0}}(y - y_{0}) - \frac{1}{2z_{0}}(z - z_{0}) = 0 $.

1. 求曲线 $ \{\begin{aligned}x^{2}+y^{2}+z^{2}-3x=0,\\ 2x-3y+5z-4=0\end{aligned}. $，在点 $ (1,1,1) $处的切线及法平面方程.

解法一 为了求 $ \frac{dy}{dx} $， $ \frac{dz}{dx} $，在所给方程两端分别对 x 求导，得

$$ \{\begin{aligned}&2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}+2z\frac{\mathrm{d}z}{\mathrm{d}x}-3=0,\\ &2-3\frac{\mathrm{d}y}{\mathrm{d}x}+5\frac{\mathrm{d}z}{\mathrm{d}x}=0.\end{aligned}. $$

即

$$ \{\begin{aligned}&2y\frac{\mathrm{d}y}{\mathrm{d}x}+2z\frac{\mathrm{d}z}{\mathrm{d}x}=-2x+3,\\ &3\frac{\mathrm{d}y}{\mathrm{d}x}-5\frac{\mathrm{d}z}{\mathrm{d}x}=2.\end{aligned}. $$

当 $ D=\left|\begin{array}{cc}2y & 2z \\ 3 & -5\end{array}\right|=-10y-6z\neq0 $ 时，解方程组得

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{D}\left|\begin{array}{cc}-2x+3&2z\\2&-5\end{array}\right|=\frac{10x-4z-15}{-10y-6z}, $$

$$ \frac{\mathrm{d}z}{\mathrm{d}x}=\frac{1}{D}\left|\begin{array}{cc}2y&-2x+3\\3&2\end{array}\right|=\frac{6x+4y-9}{-10y-6z}. $$

$$ \frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,1,1)}=\frac{9}{16},\quad\frac{\mathrm{d}z}{\mathrm{d}x}\bigg|_{(1,1,1)}=-\frac{1}{16}. $$

于是在点 $ (1,1,1) $处的切线方程为

$$ \frac{x-1}{1}=\frac{y-1}{\frac{9}{16}}=\frac{z-1}{-\frac{1}{16}}, $$

即

$$ \frac{x-1}{16}=\frac{y-1}{9}=\frac{z-1}{-1}. $$

法平面方程为

$$ (x-1)+\frac{9}{16}(y-1)-\frac{1}{16}(z-1)=0, $$

即

$$ 16x+9y-z-24=0. $$

解法二 所求曲线的切线，也就是曲面 $ x^{2} + y^{2} + z^{2} - 3x = 0 $ 在点 (1,1,1) 处的切平面与平面 $ 2x - 3y + 5z = 4 $ 的交线，利用曲面的切平面方程得所求切线为

$$ \{\begin{aligned}&-(x-1)+2(y-1)+2(z-1)=0,\\ &2x-3y+5z=4.\end{aligned}. $$

即

$$ \{\begin{aligned}&-x+2y+2z=3,\\ &2x-3y+5z=4.\end{aligned}. $$

这切线的方向向量为 $ (16,9,-1) $，于是所求法平面方程为

$$ 16\left(x-1\right)+9\left(y-1\right)-\left(z-1\right)=0, $$

即

$$ 16x+9y-z-24=0. $$

1. 求出曲线 x = t, $ y = t^{2} $, z = t^{3} 上的点，使在该点的切线平行于平面 x + 2y + z = 4.

解 因为 $ x_{t}=1, y_{t}=2t, z_{t}=3t^{2} $，设所求点对应的参数为 $ t_{0} $，于是曲线在该点处的切向量可取为 $ \boldsymbol{T}=(1,2t_{0},3t_{0}^{2}) $。已知平面的法向量为 $ \boldsymbol{n}=(1,2,1) $，由切线与平面平行，得 $ \boldsymbol{T}\cdot\boldsymbol{n}=0 $，即 $ 1+4t_{0}+3t_{0}^{2}=0 $，解得 $ t_{0}=-1 $ 和 $ -\frac{1}{3} $。于是所求点为 $ (-1,1,-1) $ 或 $ \left(-\frac{1}{3},\frac{1}{9},-\frac{1}{27}\right) $。

1. 求曲面 $ e^{z} - z + xy = 3 $ 在点 $ (2,1,0) $ 处的切平面及法线方程.

解 令 $ F(x, y, z) = e^{z} - z + xy - 3 $，则

$$ \boldsymbol{n}=\left(F_{x},F_{y},F_{z}\right)=\left(y,x,\mathrm{e}^{z}-1\right),\quad\boldsymbol{n}\mid_{(2,1,0)}=(1,2,0). $$

曲面在点 $ (2,1,0) $处的切平面方程为

$$ 1\cdot(x-2)+2(y-1)+0\cdot(z-0)=0, $$

即

$$ x+2y-4=0. $$

曲面在点 $ (2,1,0) $处的法线方程为

$$ \{\begin{aligned}\frac{x-2}{1}&=\frac{y-1}{2},\\ z&=0.\end{aligned}. $$

1. 求曲面 $ ax^{2} + by^{2} + cz^{2} = 1 $ 在点 $ (x_{0}, y_{0}, z_{0}) $ 处的切平面及法线方程.

解 令 $ F(x,y,z)=ax^{2}+by^{2}+cz^{2}-1 $，则曲面在点 $ (x,y,z) $ 处的一个法向量

$$ \begin{aligned}\boldsymbol{n}&=\left(F_{x},F_{y},F_{z}\right)=\left(2a x,2b y,2c z\right)\\&=2\left(a x,b y,c z\right),\end{aligned} $$

在点 $ (x_{0},y_{0},z_{0}) $处的一个法向量为 $ (ax_{0},by_{0},cz_{0}) $，故曲面在该点处的切平面方程为

$$ a x_{0}\left(x-x_{0}\right)+b y_{0}\left(y-y_{0}\right)+c z_{0}\left(z-z_{0}\right)=0, $$

即

$$ ax_{0}x+by_{0}y+cz_{0}z=ax_{0}^{2}+by_{0}^{2}+cz_{0}^{2}=1. $$

法线方程为

$$ \frac{x-x_{0}}{ax_{0}}=\frac{y-y_{0}}{by_{0}}=\frac{z-z_{0}}{cz_{0}}. $$

1. 求椭球面 $ x^{2}+2y^{2}+z^{2}=1 $ 上平行于平面 x-y+2z=0 的切平面方程.

解 设 $ F(x,y,z)=x^{2}+2y^{2}+z^{2}-1 $，则曲面在点 $ (x,y,z) $ 处的一个法向量 $ \boldsymbol{n}=(F_{x},F_{y},F_{z})=(2x,4y,2z) $。已知平面的法向量为 $ (1,-1,2) $，由已知平面与所求切平面平行，得

$$ \frac{2x}{1}=\frac{4y}{-1}=\frac{2z}{2},\quad 即 \quad x=\frac{1}{2}z,\quad y=-\frac{1}{4}z. $$

代入椭球而方程得

$$ \left(\frac{z}{2}\right)^{2}+2\left(\left.-\frac{z}{4}\right)^{2}+z^{2}=1\right.. $$

解得 $ z = \pm 2\sqrt{\frac{2}{11}} $，则 $ x = \pm\sqrt{\frac{2}{11}} $， $ y = \mp\frac{1}{2}\sqrt{\frac{2}{11}} $。所以切点为

$$ \left(\pm\sqrt{\frac{2}{11}},\mp\frac{1}{2}\sqrt{\frac{2}{11}},\pm2\sqrt{\frac{2}{11}}\right). $$

所求切平面方程为

$$ \left(x\pm\sqrt{\frac{2}{11}}\right)-\left(y\mp\frac{1}{2}\sqrt{\frac{2}{11}}\right)+2\left(z\pm2\sqrt{\frac{2}{11}}\right)=0, $$

即

$$ x-y+2z=\pm\sqrt{\frac{11}{2}}. $$

1. 求旋转椭球面 $ 3x^{2} + y^{2} + z^{2} = 16 $ 上点 $ (-1, -2, 3) $ 处的切平面与 xOy 面的夹角的余弦.

解 令 $ F(x,y,z)=3x^{2}+y^{2}+z^{2}-16 $，曲面的法向量为

$$ \boldsymbol{n}=(\boldsymbol{F}_{x},\boldsymbol{F}_{y},\boldsymbol{F}_{z})=(6x,2y,2z), $$

曲面在点 $ (-1,-2,3) $处的法向量为 $ n_{1}=n\left|(-1,-2,3)\right.=\left(-6,-4,6\right) $，xOy面的法向量为 $ n_{2}=(0,0,1) $，记 $ n_{1} $与 $ n_{2} $的夹角为 $ \gamma $，则所求的余弦值为

$$ \cos\gamma=\frac{\boldsymbol{n}_{1}\cdot\boldsymbol{n}_{2}}{\left|\boldsymbol{n}_{1}\right|\left|\boldsymbol{n}_{2}\right|}=\frac{6}{\sqrt{6^{2}+4^{2}+6^{2}}\cdot1}=\frac{3}{\sqrt{22}}. $$

1. 试证曲面 $ \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a} (a > 0) $上任何点处的切平面在各坐标轴上的截距之和等于a.

证 设 $ F(x,y,z)=\sqrt{x}+\sqrt{y}+\sqrt{z}-\sqrt{a} $，则曲面在点 $ (x,y,z) $ 处的一个法向量

$$ n=\left(\frac{1}{2\sqrt{x}},\frac{1}{2\sqrt{y}},\frac{1}{2\sqrt{z}}\right). $$

在曲面上任取一点 $ M(x_{0},y_{0},z_{0}) $，则曲面在点 M 处的切平面方程为

$$ \frac{1}{2\sqrt{x_{0}}}(x-x_{0})+\frac{1}{2\sqrt{y_{0}}}(y-y_{0})+\frac{1}{2\sqrt{z_{0}}}(z-z_{0})=0, $$

即

$$ \frac{x}{\sqrt{x_{0}}}+\frac{y}{\sqrt{y_{0}}}+\frac{z}{\sqrt{z_{0}}}=\sqrt{x_{0}}+\sqrt{y_{0}}+\sqrt{z_{0}}=\sqrt{a}, $$

化为截距式，得

$$ \frac{x}{\sqrt{ax_{0}}}+\frac{y}{\sqrt{ay_{0}}}+\frac{z}{\sqrt{az_{0}}}=1, $$

所以截距之和为

$$ \sqrt{a x_{0}}+\sqrt{a y_{0}}+\sqrt{a z_{0}}=\sqrt{a}(\sqrt{x_{0}}+\sqrt{y_{0}}+\sqrt{z_{0}})=a. $$

1. 设 $ u(t) $, $ v(t) $ 是可导的向量值函数, 证明：

(1)

$$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\boldsymbol{u}\left(t\right)\pm\boldsymbol{v}\left(t\right)\right]=\boldsymbol{u}^{\prime}\left(t\right)\pm\boldsymbol{v}^{\prime}\left(t\right); $$

(2)

$$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t\right)\right]=\boldsymbol{u}^{\prime}\left(t\right)\cdot\boldsymbol{v}\left(t\right)+\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}^{\prime}\left(t\right); $$

(3)

$$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t\right)\right]=\boldsymbol{u}^{\prime}\left(t\right)\times\boldsymbol{v}\left(t\right)+\boldsymbol{u}\left(t\right)\times\boldsymbol{v}^{\prime}\left(t\right). $$

证（1）

$$ \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\big[\boldsymbol{u}\left(t\right)\pm\boldsymbol{v}\left(t\right)\big]\\ &=\lim_{\Delta t\rightarrow0}\frac{\big[\boldsymbol{u}\left(t+\Delta t\right)\pm\boldsymbol{v}\left(t+\Delta t\right)\big]-\big[\boldsymbol{u}\left(t\right)\pm\boldsymbol{v}\left(t\right)\big]}{\Delta t}\\ &=\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)}{\Delta t}\pm\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{v}\left(t\right)}{\Delta t}\\ &=\boldsymbol{u}^{\prime}\big(t\big)\pm\boldsymbol{v}^{\prime}\big(t\big),\\ \end{aligned} $$

其中用到了向量值函数的极限的四则运算法则.

(2)

$$ \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\big[\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t\right)\big]\\ &=\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)\cdot\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t\right)}{\Delta t}\\ &=\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)\cdot\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t+\Delta t\right)}{\Delta t}\quad+\\ &\quad\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}\left(t\right)}{\Delta t}\\ &=\left[\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)}{\Delta t}\right]\cdot\left[\lim_{\Delta t\rightarrow0}\boldsymbol{v}\left(t+\Delta t\right)\right]\quad+\\ &\quad\left[\lim_{\Delta t\rightarrow0}\boldsymbol{u}\left(t\right)\right]\cdot\left[\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{v}\left(t\right)}{\Delta t}\right]\\ &=\boldsymbol{u}^{\prime}\left(t\right)\cdot\boldsymbol{v}\left(t\right)+\boldsymbol{u}\left(t\right)\cdot\boldsymbol{v}^{\prime}\left(t\right),\\ \end{aligned} $$

其中用到了向量值函数极限的四则运算法则以及数量积与极限运算次序的交换.

$$ \begin{aligned}&(3)\frac{\mathrm{d}}{\mathrm{d}t}[\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t\right)]=\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)\times\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t\right)}{\Delta t}\\ &=\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)\times\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t+\Delta t\right)+\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)\times\boldsymbol{v}\left(t\right)}{\Delta t}\\ &=\lim_{\Delta t\rightarrow0}\left[\frac{\boldsymbol{u}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)}{\Delta t}\times\boldsymbol{v}\left(t+\Delta t\right)\right]+\lim_{\Delta t\rightarrow0}\left[\boldsymbol{u}\left(t\right)\times\frac{\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{v}\left(t\right)}{\Delta t}\right]\\ &=\left[\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{u}\left(t+\Delta t\right)-\boldsymbol{u}\left(t\right)}{\Delta t}\right]\times\left[\lim_{\Delta t\rightarrow0}\boldsymbol{v}\left(t+\Delta t\right)\right]+\left[\lim_{\Delta t\rightarrow0}\boldsymbol{u}\left(t\right)\right]\times\left[\lim_{\Delta t\rightarrow0}\frac{\boldsymbol{v}\left(t+\Delta t\right)-\boldsymbol{v}\left(t\right)}{\Delta t}\right]\\ &=\boldsymbol{u}^{\prime}\left(t\right)\times\boldsymbol{v}\left(t\right)+\boldsymbol{u}\left(t\right)\times\boldsymbol{v}^{\prime}\left(t\right),\\ \end{aligned} $$

其中用到了向量值函数极限的四则运算法则以及向量积与极限运算次序的交换.

习题9-7

1. 求函数 $ z = x^{2} + y^{2} $ 在点 (1,2) 处沿从点 (1,2) 到点 (2,2 + $ \sqrt{3} $) 的方向的方向导数.

解 按题意，方向 $ l = (1, \sqrt{3}) $， $ e_{l} = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $.

又

$$ \frac{\partial z}{\partial x}=2x,\quad\frac{\partial z}{\partial y}=2y,\quad\frac{\partial z}{\partial x}\bigg|_{(1,2)}=2,\quad\frac{\partial z}{\partial y}\bigg|_{(1,2)}=4, $$

故

$$ \frac{\partial z}{\partial l}\bigg|_{(1,2)}=2\cdot\frac{1}{2}+4\cdot\frac{\sqrt{3}}{2}=1+2\sqrt{3}. $$

1. 求函数 $ z = \ln(x + y) $ 在抛物线 $ y^{2} = 4x $ 上点 (1,2) 处，沿着这抛物线在该点处偏向 x 轴正向的切线方向的方向导数.

解 先求切线斜率：在 $ y^{2}=4x $ 两端分别对 x 求导，得

$$ 2y\frac{\mathrm{d}y}{\mathrm{d}x}=4. $$

于是

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{y},\quad k=\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{(1,2)}=1, $$

切线方向 $ l = (1,1) $, $ e_l = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

又

$$ \frac{\partial z}{\partial x}\bigg|_{(1,2)}=\frac{1}{x+y}\bigg|_{(1,2)}=\frac{1}{3}, $$

$$ \frac{\partial z}{\partial y}\left|\begin{array}{c}\\\left(1,2\right)\\\end{array}\right.=\left.\frac{1}{x+y}\right|_{\left(1,2\right)}=\frac{1}{3}. $$

故

$$ \frac{\partial z}{\partial l}\bigg|_{(1,2)}=\frac{1}{3}\cdot\frac{\sqrt{2}}{2}+\frac{1}{3}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{3}. $$

1. 求函数 z = 1 - $ \left(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}\right) $ 在点 $ \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right) $ 处沿曲线 $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 $ 在这点的内法线方向的方向导数.

解 先求切线斜率：在 $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $两端分别对x求导，得

$$ \frac{2x}{a^{2}}+\frac{2y}{b^{2}}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}=0. $$

于是

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{b^{2}x}{a^{2}y},\quad k=\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)}=-\frac{b}{a}, $$

法线斜率为

$$ k^{\prime}=-\frac{1}{k}=\frac{a}{b}, $$

内法线方向 $ l = (-b, -a), e_l = \left( -\frac{b}{\sqrt{a^2 + b^2}}, -\frac{a}{\sqrt{a^2 + b^2}} \right) $.

又

$$ \frac{\partial z}{\partial x}\bigg|_{\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)}=-\frac{\sqrt{2}}{a},\quad\frac{\partial z}{\partial y}\bigg|_{\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)}=-\frac{\sqrt{2}}{b}. $$

故

$$ \begin{align*}\frac{\partial z}{\partial l}\bigg|_{\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)}&=-\frac{\sqrt{2}}{a}\cdot\left(-\frac{b}{\sqrt{a^{2}+b^{2}}}\right)-\frac{\sqrt{2}}{b}\cdot\left(-\frac{a}{\sqrt{a^{2}+b^{2}}}\right)\\&=\frac{1}{ab}\sqrt{2\left(a^{2}+b^{2}\right)}.\end{align*} $$

1. 求函数 $ u = xy^{2} + z^{3} - xyz $ 在点 (1,1,2) 处沿方向角为 $ \alpha = \frac{\pi}{3}, \beta = \frac{\pi}{4}, \gamma = \frac{\pi}{3} $ 的方向的方向导数.

解 因为 $ \frac{\partial u}{\partial x} = y^{2} - yz $, $ \frac{\partial u}{\partial y} = 2xy - xz $, $ \frac{\partial u}{\partial z} = 3z^{2} - xy $,

$$ \left.\frac{\partial u}{\partial x}\right|_{(1,1,2)}=-\left.1\right.,\quad\left.\frac{\partial u}{\partial y}\right|_{(1,1,2)}=0,\quad\left.\frac{\partial u}{\partial z}\right|_{(1,1,2)}=11. $$

$$ e_{l}=\left(\cos\frac{\pi}{3},\cos\frac{\pi}{4},\cos\frac{\pi}{3}\right)=\left(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2}\right), $$

所以

$$ \frac{\partial u}{\partial l}\bigg|_{(1,1,2)}=-\ 1\cdot\frac{1}{2}+0+1 1\cdot\frac{1}{2}=5. $$

1. 求函数 u = xyz 在点 (5,1,2) 处沿从点 (5,1,2) 到点 (9,4,14) 的方向的方向导数.

解 按题意，方向 $ l = (4, 3, 12) $， $ e_{l} = \left( \frac{4}{13}, \frac{3}{13}, \frac{12}{13} \right) $.

又

$$ \frac{\partial u}{\partial x}\;=\;y z\,,\quad\frac{\partial u}{\partial y}\;=\;x z\,,\quad\frac{\partial u}{\partial z}\;=\;x y\,, $$

$$ \left.\frac{\partial u}{\partial x}\right|_{(5,1,2)}~=~2,\quad\left.\frac{\partial u}{\partial y}\right|_{(5,1,2)}~=~10,\quad\left.\frac{\partial u}{\partial z}\right|_{(5,1,2)}~=~5, $$

故

$$ \frac{\partial u}{\partial l}\bigg|_{(5,1,2)}=2\cdot\frac{4}{13}+10\cdot\frac{3}{13}+5\cdot\frac{12}{13}=\frac{98}{13}. $$

1. 求函数 $ u = x^{2} + y^{2} + z^{2} $ 在曲线 x = t, y = t^{2}, z = t^{3} $ 上点 (1, 1, 1) 处，沿曲线在该点的切线正方向（对应于 t 增大的方向）的方向导数.$

解 先求曲线在给定点的切线方向.

因为 $ x_{t}=1, y_{t}=2t, z_{t}=3t^{2} $，所以曲线在点 $ (1,1,1) $ 处的切线的方向向量可取为 $ T=(1,2,3) $， $ e_{T}=\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right) $。又

$$ \left.\frac{\partial u}{\partial x}\right|_{(1,1,1)}=2,\quad\left.\frac{\partial u}{\partial y}\right|_{(1,1,1)}=2,\quad\left.\frac{\partial u}{\partial z}\right|_{(1,1,1)}=2, $$

故

$$ \left.\frac{\partial u}{\partial\boldsymbol{T}}\right|_{(1,1,1)}=2\cdot\frac{1}{\sqrt{14}}+2\cdot\frac{2}{\sqrt{14}}+2\cdot\frac{3}{\sqrt{14}}=\frac{6}{7}\sqrt{14}. $$

1. 求函数 u = x + y + z 在球面 $ x^{2} + y^{2} + z^{2} = 1 $ 上点 $ (x_{0}, y_{0}, z_{0}) $ 处，沿球面在该点的外法线方向的方向导数.

解 设 $ F(x,y,z)=x^{2}+y^{2}+z^{2}-1 $，则 $ F_{x}=2x, F_{y}=2y, F_{z}=2z $，于是球面在 $ (x_{0}, y_{0}, z_{0}) $ 处的外法线方向向量可取为

$$ l=\left(F_{x},F_{y},F_{z}\right)\left|\begin{array}{c}\\\left(x_{0},y_{0},z_{0}\right)\end{array}\right.=\left(2x_{0},2y_{0},2z_{0}\right), $$

l 的方向余弦为

$$ \cos\alpha=\frac{x_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}},\quad\cos\beta=\frac{y_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}},\quad\cos\gamma=\frac{z_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}}, $$

又

$$ \frac{\partial u}{\partial x}\;=\;1\;,\quad\frac{\partial u}{\partial y}\;=\;1\;,\quad\frac{\partial u}{\partial z}\;=\;1. $$

故

$$ \begin{aligned}\left.\frac{\partial u}{\partial l}\right|_{\left(x_{0},y_{0},z_{0}\right)}&=\left(\frac{\partial u}{\partial x}\cos\alpha+\frac{\partial u}{\partial y}\cos\beta+\frac{\partial u}{\partial z}\cos\gamma\right)\bigg|_{\left(x_{0},y_{0},z_{0}\right)}\\&=1\cdot\frac{x_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}}+1\cdot\frac{y_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}}+1\cdot\frac{z_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}}\\&=\frac{x_{0}+y_{0}+z_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}+z_{0}^{2}}}\\&=x_{0}+y_{0}+z_{0}.\end{aligned} $$

1. 设 $ f(x,y,z) = x^2 + 2y^2 + 3z^2 + xy + 3x - 2y - 6z $，求 $ \text{grad } f(0,0,0) $ 及 $ \text{grad } f(1,1,1) $.

$$ \begin{aligned}\operatorname{grad}f(x,y,z)&=f_{x}\boldsymbol{i}+f_{y}\boldsymbol{j}+f_{z}\boldsymbol{k}\\&=\left(2x+y+3\right)\boldsymbol{i}+\left(4y+x-2\right)\boldsymbol{j}+\left(6z-6\right)\boldsymbol{k},\\\quad&\operatorname{grad}f(0,0,0)=3\boldsymbol{i}-2\boldsymbol{j}-6\boldsymbol{k},\\\quad&\operatorname{grad}f(1,1,1)=6\boldsymbol{i}+3\boldsymbol{j}.\end{aligned} $$

1. 设函数 $ u(x, y, z) $, $ v(x, y, z) $ 的各个偏导数都存在且连续，证明：

(1) $ \nabla(cu) = c\nabla u $ (其中 c 为常数);

(2) $ \nabla(u \pm v) = \nabla u \pm \nabla v $;

(3) $ \nabla(uv) = v\nabla u + u\nabla v $;

(4) $ \nabla\left(\frac{u}{v}\right) = \frac{v\nabla u - u\nabla v}{v^{2}}. $

证（1）

$$ \begin{align*}\nabla(cu)&=\left(c\frac{\partial u}{\partial x},c\frac{\partial u}{\partial y},c\frac{\partial u}{\partial z}\right)=c\bigg(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\bigg)\\&=c\nabla u.\end{align*} $$

(2)

$$ \begin{aligned}\nabla(u\pm v)&=\left(\frac{\partial u}{\partial x}\pm\frac{\partial v}{\partial x},\frac{\partial u}{\partial y}\pm\frac{\partial v}{\partial y},\frac{\partial u}{\partial z}\pm\frac{\partial v}{\partial z}\right)\\&=\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)\pm\left(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y},\frac{\partial v}{\partial z}\right)\end{aligned} $$

$$ =~\nabla u~\pm~\nabla v. $$

(3)

$$ \begin{aligned}\nabla(uv)&=\left(\frac{\partial}{\partial x}(uv),\frac{\partial}{\partial y}(uv),\frac{\partial}{\partial z}(uv)\right)\\&=\left(\frac{\partial u}{\partial x}v+u\frac{\partial v}{\partial x},\frac{\partial u}{\partial y}v+u\frac{\partial v}{\partial y},\frac{\partial u}{\partial z}v+u\frac{\partial v}{\partial z}\right)\\&=v\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)+u\left(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y},\frac{\partial v}{\partial z}\right)\\&=v\nabla u+u\nabla v.\\ \end{aligned} $$

(4)

$$ \begin{align*}\nabla\bigg(\frac{u}{v}\bigg)&=\bigg(\frac{\partial}{\partial x}\bigg(\frac{u}{v}\bigg),\frac{\partial}{\partial y}\bigg(\frac{u}{v}\bigg),\frac{\partial}{\partial z}\bigg(\frac{u}{v}\bigg)\bigg)\\&=\left(\frac{v\frac{\partial u}{\partial x}-u\frac{\partial v}{\partial x}}{v^{2}},\frac{v\frac{\partial u}{\partial y}-u\frac{\partial v}{\partial y}}{v^{2}},\frac{v\frac{\partial u}{\partial z}-u\frac{\partial v}{\partial z}}{v^{2}}\right)\\&=\frac{1}{v}\bigg(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\bigg)-\frac{u}{v^{2}}\bigg(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y},\frac{\partial v}{\partial z}\bigg)\\&=\frac{v\nabla u-u\nabla v}{v^{2}}.\end{align*} $$

1. 求函数 $ u = xy^{2}z $ 在点 $ P_{0}(1, -1, 2) $ 处变化最快的方向，并求沿这个方向的方向导数.

解

$$ \nabla u=\frac{\partial u}{\partial x}\boldsymbol{i}+\frac{\partial u}{\partial y}\boldsymbol{j}+\frac{\partial u}{\partial z}\boldsymbol{k}=y^{2}z\boldsymbol{i}+2x y z\boldsymbol{j}+x y^{2}\boldsymbol{k}, $$

$$ \nabla u\mid_{P_{0}}=2\boldsymbol{i}-4\boldsymbol{j}+\boldsymbol{k}. $$

由方向导数与梯度的关系可知， $ u = xy^{2}z $ 在 $ P_{0} $ 处沿 $ n = \nabla u \big|_{P_{0}} = 2i - 4j + k $ 的方向增加最快，其方向导数为

$$ \frac{\partial u}{\partial n}\big|_{P_{0}}.=|\begin{array}{c}\nabla u\\ \end{array}|_{P_{0}}|\begin{array}{l}\end{array}|=|\begin{array}{l}2i-4j+k\\ \end{array}|=\sqrt{21}; $$

沿 $ n_{1}=-\nabla u\big|_{P_{0}}=-2i+4j-k $方向减少最快，其方向导数为

$$ \frac{\partial u}{\partial n_{1}}\left|\begin{array}{l}\end{array}\right|_{P_{0}}=-\sqrt{21}. $$

习题9-8

多元函数的极值及其求法

1. 已知函数 $ f(x,y) $ 在点 $ (0,0) $ 的某个邻域内连续，且

$$ \lim_{(x,y)\to(0,0)}\frac{f(x,y)-xy}{\left(x^{2}+y^{2}\right)^{2}}=1, $$

则下述四个选项中正确的是( ).

(A) 点(0,0)不是 $ f(x,y) $的极值点

(B) 点(0,0)是 $ f(x,y) $的极大值点

(C) 点(0,0)是 $ f(x,y) $的极小值点

（D）根据所给条件无法判断（0,0）是否为 $ f(x,y) $的极值点

解 令 $ \rho=\sqrt{x^{2}+y^{2}} $，则由题设可知

$$ f(x,y)=xy+\rho^{4}+o(\rho^{4})， $$

当 $ (x,y)\to(0,0) $时, $ \rho\to0 $

由于 $ f(x,y) $在 $ (0,0) $附近的值主要由xy决定，而xy在 $ (0,0) $附近符号不定，故点 $ (0,0) $不是 $ f(x,y) $的极值点，即应选(A).

本题也可以取两条路径 y = x 和 y = -x 来考虑. 当 |x| 充分小时,

$$ f(x,x)=x^{2}+4x^{4}+o(x^{4})>0,\quad f(x,-x)=-x^{2}+4x^{4}+o(x^{4})<0, $$

故点 $ (0,0) $不是 $ f(x,y) $的极值点，即应选(A).

1. 求函数 $ f(x, y) = 4(x - y) - x^{2} - y^{2} $ 的极值.

解方程组

$$ \{\begin{aligned}f_{x}&=4-2x=0,\\ f_{y}&=-4-2y=0,\end{aligned}. $$

求得驻点 $ (2, -2) $.

メ

$$ A=f_{x x}(2,-2)=-2<0,\quad B=f_{x y}(2,-2)=0, $$

$$ C~=~f_{yy}(2,~-2)~=-~2~,\quad AC~-B^{2}~>~0~, $$

由判定极值的充分条件知：在点 $ (2,-2) $处，函数取得极大值 $ f(2,-2)=8 $。

1. 求函数 $ f(x,y)=(6x-x^{2})(4y-y^{2}) $ 的极值.

解方程组

$$ \{\begin{aligned}f_{x}&=(6-2x)(4y-y^{2})&=0,\\ f_{y}&=(6x-x^{2})(4-2y)&=0.\end{aligned}. $$

求得以下五组解：

$$ \{\begin{aligned}x_{1}&=0,\\ y_{1}&=0,\end{aligned}.\quad\{\begin{aligned}x_{2}&=0,\\ y_{2}&=4,\end{aligned}.\quad\{\begin{aligned}x_{3}&=3,\\ y_{3}&=2,\end{aligned}.\quad\{\begin{aligned}x_{4}&=6,\\ y_{4}&=0,\end{aligned}.\quad\{\begin{aligned}x_{5}&=6,\\ y_{5}&=4.\end{aligned}. $$

于是，得驻点 $ (0,0) $， $ (0,4) $， $ (3,2) $， $ (6,0) $， $ (6,4) $.

メ

$$ f_{x x}(x,y)=-2(4y-y^{2}), $$

$$ f_{x y}(x,y)\;=\;4\left(3-x\right)\left(2-y\right), $$

$$ f_{v v}(x,y)=-2(6x-x^{2}). $$

由判定极值的充分条件知：

在点 $ (0,0) $处， $ A=f_{xx}(0,0)=0 $， $ B=f_{x}(0,0)=24 $， $ C=f_{yy}(0,0)=0 $，AC- $ B^{2}=-24^{2}<0 $，故 $ f(0,0) $不是极值；

在点 $ (0,4) $处， $ A=f_{xx}(0,4)=0 $， $ B=f_{xy}(0,4)=-24 $， $ C=f_{yy}(0,4)=0 $，AC- $ B^{2}=-(-24)^{2}<0 $，故 $ f(0,4) $不是极值；

在点(3,2)处, $ A=f_{xx}(3,2)=-8<0 $, $ B=f_{xy}(3,2)=0 $, $ C=f_{yy}(3,2)=-18 $, $ AC-B^{2}=144>0 $,故函数在点(3,2)处取得极大值,极大值为 $ f(3,2)=36 $;

在点(6,0)处, $ A=f_{xx}(6,0)=0 $, $ B=f_{xy}(6,0)=-24 $, $ C=f_{yy}(6,0)=0 $, $ AC-B^{2}=-(-24)^{2}<0 $,故 $ f(6,0) $不是极值;

在点(6,4)处, $ A=f_{xx}(6,4)=0 $, $ B=f_{xy}(6,4)=24 $, $ C=f_{yy}(6,4)=0 $, $ AC-B^{2}=-24^{2}<0 $,故 $ f(6,4) $不是极值.

1. 求函数 $ f(x,y)=e^{ex}(x+y^{2}+2y) $的极值.

解 解方程组

$ \{\begin{aligned}f_{x}&=e^{2x}(2x+2y^{2}+4y+1)=0,\\ f_{y}&=e^{2x}(2y+2)=0,\end{aligned}. $

求得驻点 $ \left(\frac{1}{2},-1\right) $.

又 $ A=f_{xx}\left(\frac{1}{2},-1\right)=2e>0 $, $ B=f_{xy}\left(\frac{1}{2},-1\right)=0 $,

$ C=f_{yy}\left(\frac{1}{2},-1\right)=2e $, $ AC-B^{2}=4e^{2}>0 $,

由判定极值的充分条件知,在点 $ \left(\frac{1}{2},-1\right) $处,函数取得极小值

$ f\left(\frac{1}{2},-1\right)=-\frac{e}{2} $.

1. 求函数 $ z=xy $在适合附加条件 $ x+y=1 $下的极大值.

解 本题属条件极值问题,易将它化为无条件极值问题.

条件 $ x+y=1 $可表示成y=1-x,代入z=xy,则问题化为求 $ z=x(1-x) $的极大值.

由 $ \frac{\mathrm{d}z}{\mathrm{d}x}=1-2x=0 $,得 $ x=\frac{1}{2} $.又

$ \left.\frac{\mathrm{d}^{2}z}{\mathrm{d}x^{2}}\right|_{x=\frac{1}{2}}=-2<0 $.

由一元函数取得极值的充分条件知, $ x=\frac{1}{2} $为极大值点,极大值为

$ z=\frac{1}{2}\left(1-\frac{1}{2}\right)=\frac{1}{4} $.

1. 从斜边之长为 l 的一切直角三角形中，求有最大周长的直角三角形.

解 设直角三角形的两直角边之长分别为 x, y，则周长

$ S = x + y + l $ $ (0 < x < l, 0 < y < l) $

本题是求周长 S 在 $ x^{2}+y^{2}=l^{2} $ 条件下的条件极值问题.

作拉格朗日函数

$$ L\left(x,y\right)=x+y+l+\lambda\left(x^{2}+y^{2}-l^{2}\right). $$

令

$$ \{\begin{aligned}L_{x}&=1+2\lambda x=0,\\ L_{y}&=1+2\lambda y=0.\end{aligned}. $$

解得 x = y = - $ \frac{1}{2\lambda} $. 代入 $ x^{2} + y^{2} = l^{2} $，得 $ \lambda = -\frac{\sqrt{2}}{2l} $，于是 x = y = $ \frac{l}{\sqrt{2}} $， $ \left(\frac{l}{\sqrt{2}}, \frac{l}{\sqrt{2}}\right) $ 是唯一的极值点，根据问题性质可知这种最大周长的直角三角形一定存在，所以在斜边之长为 l 的一切直角三角形中，周长最大的是等腰直角三角形.

注 条件极值的解法，一般是采用拉格朗日乘数法求解。但要注意利用乘数法所得到的点只是可能极值点，究竟这些点是否为极值点以及是极大点还是极小点尚需进一步判断。在实际问题中往往可根据问题本身的性质来判定。在特殊情形下，条件极值问题可化为无条件极值问题求解。

1. 要造一个容积等于定数 k 的长方体无盖水池，应如何选择水池的尺寸，方可使它的表面积最小.

解 设水池的长为 a，宽为 b，高为 c，则水池的表面积为

$$ A=ab+2ac+2bc(a>0,b>0,c>0). $$

约束条件 abc = k.

作拉格朗日函数 $ L(a,b,c)=ab+2ac+2bc+\lambda(abc-k) $. 由

$$ \{\begin{aligned}L_{a}&=b+2c+\lambda b c=0,\\ L_{b}&=a+2c+\lambda a c=0,\\ L_{c}&=2a+2b+\lambda a b=0,\\ a b c&=k,\end{aligned}. $$

解得 $ a = b = \sqrt[3]{2k} $, $ c = \frac{1}{2}\sqrt[3]{2k} $, $ \lambda = -\sqrt[3]{\frac{32}{k}} $.

$ \left(\sqrt[3]{2k},\sqrt[3]{2k},\frac{1}{2}\sqrt[3]{2k}\right) $是唯一可能的极值点，由问题本身可知A一定有最小值，所以表面积最小的水池的长和宽都应为 $ \sqrt[3]{2k} $，高为 $ \frac{1}{2}\sqrt[3]{2k} $.

1. 在平面 xOy 上求一点，使它到 x=0, y=0 及 $ x+2y-16=0 $ 三直线的距离平方之和为最小.

解 设所求点为 $ (x,y) $，则此点到三直线的距离依次为： $ \left|x\right|,\left|y\right|,\frac{\left|x+2y-16\right|}{\sqrt{5}} $

三距离平方之和为

$$ z=x^{2}+y^{2}+\frac{1}{5}(x+2y-16)^{2}. $$

山

$$ \begin{cases}\frac{\partial z}{\partial x}=2x+\frac{2}{5}(x+2y-16)=0,\\\\\frac{\partial z}{\partial y}=2y+\frac{4}{5}(x+2y-16)=0\end{cases} $$

求得唯一可能的极值点 $ \left(\frac{8}{5},\frac{16}{5}\right) $. 根据问题本身可知，距离平方和最小的点必定存在，故所求点即为 $ \left(\frac{8}{5},\frac{16}{5}\right) $.

1. 将周长为2p的矩形绕它的一边旋转而构成一个圆柱体。问矩形的边长各为多少时，才可使圆柱体的体积为最大？

解 设矩形的一边长为 x，则另一边长为 p - x，假设矩形绕长为 p - x 的一边旋转，则旋转所成圆柱体的体积为 $ V = \pi x^{2}(p - x) $. 由

$$ \frac{\mathrm{d}V}{\mathrm{d}x}=2\pi x(p-x)-\pi x^{2}=\pi x(2p-3x)=0, $$

求得驻点为 $ x=\frac{2}{3}p $.

由于驻点唯一，由题意又可知这种圆柱体一定有最大值，所以当矩形的边长为 $ \frac{2p}{3} $和 $ \frac{p}{3} $时，绕短边旋转所得圆柱体体积最大.

1. 求内接于半径为 a 的球且有最大体积的长方体.

解 设球面方程为 $ x^{2}+y^{2}+z^{2}=a^{2} $， $ (x,y,z) $ 是它的内接长方体在第一卦限内的一个顶点，则此长方体的长、宽、高分别为 2x,2y,2z，体积为

$$ V=2x\cdot2y\cdot2z=8xyz. $$

令

$$ L\left(x,y,z\right)=8xyz+\lambda\left(x^{2}+y^{2}+z^{2}-a^{2}\right), $$

由

$$ \{\begin{aligned}L_{x}&=8yz+2\lambda x=0,\\ L_{y}&=8xz+2\lambda y=0,\\ L_{z}&=8xy+2\lambda z=0,\end{aligned}. $$

$$ \{\begin{aligned}4yz+\lambda x&=0,\\ 4xz+\lambda y&=0,\\ 4xy+\lambda z&=0,\end{aligned}. $$

解得 x = y = z = - $ \frac{\lambda}{4} $，代入 $ x^{2} + y^{2} + z^{2} = a^{2} $，得 $ \lambda = -\frac{4}{\sqrt{3}}a $，故 $ \left(\frac{a}{\sqrt{3}}, \frac{a}{\sqrt{3}}, \frac{a}{\sqrt{3}}\right) $ 为唯一可能的极值点。由于内接于球且有最大体积的长方体必定存在，所以当长方体的长、宽、高都为 $ \frac{2a}{\sqrt{3}} $ 时其体积最大。

1. 抛物面 $ z = x^{2} + y^{2} $ 被平面 x + y + z = 1 截成一椭圆，求这椭圆上的点到原点的距离的最大值与最小值.

解 设椭圆上的点为 $ (x,y,z) $，则椭圆上的点到原点的距离平方为

$$ d^{2}=x^{2}+y^{2}+z^{2}. $$

x,y,z 满足条件： $ z=x^{2}+y^{2} $， $ x+y+z=1 $。

作拉格朗日函数

$$ {\cal L}\;=\;x^{2}\;+\;y^{2}\;+\;z^{2}\;+\;\lambda\left(z-x^{2}\;-y^{2}\right)\;+\;\mu\left(x+y+z-1\right). $$

令

$$ ,L_{x}=2x-2\lambda x+\mu=0, $$

$$ L_{y}=2y-2\lambda y+\mu=0, $$

$$ l_{L_{z}}=2z+\lambda+\mu=0. $$

(1)-(2)，得

$$ (1-\lambda)(x-y)=0. $$

故有 $ \lambda = 1 $ 或 x = y.

由 $ \lambda = 1 \Rightarrow \mu = 0, z = -\frac{1}{2} $，不合题意，故舍去.

将 x = y 代入 $ z = x^{2} + y^{2} $ 和 $ x + y + z = 1 $，得

$$ z=2x^{2},2x+z=1\Longrightarrow2x^{2}+2x-1=0. $$

解得

$$ x=y=\frac{-1\pm\sqrt{3}}{2},\quad z=2\mp\sqrt{3}. $$

于是得到两个可能的极值点：

$$ M_{1}\left(\frac{-1+\sqrt{3}}{2},\frac{-1+\sqrt{3}}{2},2-\sqrt{3}\right),\quad M_{2}\left(\frac{-1-\sqrt{3}}{2},\frac{-1-\sqrt{3}}{2},2+\sqrt{3}\right). $$

由题意可知这种距离的最大值和最小值一定存在，所以距离的最大值和最小值分别在这两点处取得.而

$$ 2\left(\frac{-1\pm\sqrt{3}}{2}\right)^{2}+\left(2\mp\sqrt{3}\right)^{2}=9\mp5\sqrt{3}, $$

故最大值与最小值分别为

$$ d_{max}=d_{M_{2}}=\sqrt{9+5\sqrt{3}},\quad d_{min}=d_{M_{1}}=\sqrt{9-5\sqrt{3}}. $$

1. 设有一圆板占有平面闭区域 $ \{(x,y)\mid x^{2}+y^{2}\leqslant1\} $. 该圆板被加热，以致在点 $ (x,y) $ 的温度是 $ T=x^{2}+2y^{2}-x $，求该圆板的最热点和最冷点.

解 解方程组

$$ \{\begin{aligned}\frac{\partial T}{\partial x}&=2x-1=0,\\ \frac{\partial T}{\partial y}&=4y=0,\end{aligned}. $$

求得驻点 $ \left(\frac{1}{2},0\right) $. $ T_{1}=T $ $ \left(\frac{1}{2},0\right)=-\frac{1}{4} $

在边界 $ x^{2} + y^{2} = 1 $ 上，

$$ T=2-\left(x^{2}+x\right)=\frac{9}{4}-\left(x+\frac{1}{2}\right)^{2}, $$

当 $ x = -\frac{1}{2} $ 时，有边界上的最大值 $ T_{2} = \frac{9}{4} $，x = 1 时，有边界上的最小值 $ T_{3} = 0 $.

比较 $ T_{1}, T_{2} $ 及 $ T_{3} $ 的值知，最热点在 $ \left(-\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right), T_{\max} = \frac{9}{4} $，最冷点在 $ \left(\frac{1}{2}, 0\right) $， $ T_{\min} = -\frac{1}{4} $。

1. 形状为椭球 $ 4x^{2} + y^{2} + 4z^{2} \leq 16 $ 的空间探测器进入地球大气层，其表面开始受热，1 小时后在探测器的点 $ (x, y, z) $ 处的温度 $ T = 8x^{2} + 4yz - 16z + 600 $，求探测器表面最热的点。

解 作拉格朗日函数

$$ L=8x^{2}+4y z-16z+600+\lambda\left(4x^{2}+y^{2}+4z^{2}-16\right). $$

令

$$ \begin{aligned},\boldsymbol{L}_{x}=16x+8\lambda x=0,\end{aligned} $$

$$ L_{y}=4z+2\lambda y=0, $$

$$ L_{z}=4y-16+8\lambda z=0. $$

由(1)得 x=0 或 $ \lambda = -2 $.

若 $ \lambda = -2 $，代入(2)(3)，得 y = z = - $ \frac{4}{3} $。再将 y = z = - $ \frac{4}{3} $ 代入约束条件

$$ 4x^{2}+y^{2}+4z^{2}=16, $$

得 $ x = \pm \frac{4}{3} $. 于是得到两个可能的极值点： $ M_{1}\left(\frac{4}{3}, -\frac{4}{3}, -\frac{4}{3}\right) $， $ M_{2}\left(-\frac{4}{3}, -\frac{4}{3}, -\frac{4}{3}\right) $.

若 x=0 ，由 (2) (3) (4) 解得 $ \lambda=0, y=4, z=0 $; $ \lambda=\sqrt{3}, y=-2, z=\sqrt{3} $; $ \lambda=-\sqrt{3}, y=-2, z=-\sqrt{3} $. 于是得到另外三个可能极值点： $ M_{3}(0,4,0) $, $ M_{4}(0,-2,\sqrt{3}) $, $ M_{5}(0,-2,-\sqrt{3}) $.

比较 T 在上述五个可能极值点处的数值知： $ T \mid M_{1} = T \mid M_{2} = \frac{1928}{3} $ 为最大，故探测器表面最热的点为 $ M\left(\pm \frac{4}{3}, -\frac{4}{3}, -\frac{4}{3}\right) $.

习题9-9

二 元函数的泰勒公式

1. 求函数 $ f(x,y)=2x^{2}-xy-y^{2}-6x-3y+5 $ 在点 $ (1,-2) $ 的泰勒公式.

$$ 解 \quad f(1,-2)=5,f_{x}(1,-2)=(4x-y-6)\mid_{(1,-2)}=0, $$

$$ f_{y}(1,-2)=(-x-2y-3)\left.\right|_{(1,-2)}=0, $$

$$ f_{x x}(1,-2)=4,f_{x y}(1,-2)=-1,f_{y y}(1,-2)=-2. $$

函数为2次多项式，三阶及三阶以上的各偏导数均为零.又

$$ h=x-1,\quad k=y+2. $$

将以上各项代入泰勒公式，便得

$$ \begin{aligned}f(x,y)&=f(1,-2)+(x-1)f_{x}(1,-2)+(y+2)f_{y}(1,-2)+\frac{1}{2!}[(x-1)^{2}\cdot\\&\quad f_{xx}(1,-2)+2(x-1)(y+2)f_{xy}(1,-2)+(y+2)^{2}f_{yy}(1,-2)]\\&=5+\frac{1}{2}[4(x-1)^{2}-2(x-1)(y+2)-2(y+2)^{2}]\\&=5+2(x-1)^{2}-(x-1)(y+2)-(y+2)^{2}.\end{aligned} $$

1. 求函数 $ f(x,y)=\mathrm{e}^{x}\ln(1+y) $ 在点 $ (0,0) $ 的三阶泰勒公式.

解

$$ f_{x}\left(x,y\right)=\mathrm{e}^{x}\ln\left(1+y\right),\quad f_{y}\left(x,y\right)=\frac{\mathrm{e}^{x}}{1+y}, $$

$$ f_{x x}\left(x,y\right)=\mathrm{e}^{x}\ln\left(1+y\right),\quad f_{x y}\left(x,y\right)=\frac{\mathrm{e}^{x}}{1+y}, $$

$$ f_{y y}\left(x,y\right)=-\frac{\mathrm{e}^{x}}{\left(1+y\right)^{2}},\quad f_{x x x}\left(x,y\right)=\mathrm{e}^{x}\ln\left(1+y\right), $$

$$ f_{yyy}(x,y)=\frac{2\mathrm{e}^{x}}{\left(1+y\right)^{3}}. $$

于是

$$ \bigg(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\bigg)f(0,0)=h f_{x}(0,0)+k f_{y}(0,0)=k, $$

$$ \begin{align*}\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^{2}f(0,0)&=h^{2}f_{xx}\left(0,0\right)+2hk f_{xy}\left(0,0\right)+k^{2}f_{yy}\left(0,0\right)\\&=2hk-k^{2},\end{align*} $$

$$ \begin{align*}\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^{3}f(0,0)&=h^{3}f_{xxx}\left(0,0\right)+3h^{2}k f_{xxy}\left(0,0\right)+3hk^{2}f_{xyy}\left(0,0\right)\\&\quad+k^{3}f_{yyy}\left(0,0\right)\\&=3h^{2}k-3hk^{2}+2k^{3}.\end{align*} $$

又

$$ f(0,0)=0,\quad h=x,\quad k=y. $$

将以上各项代入三阶泰勒公式，便得

$$ \mathrm{e}^{x}\ln(1+y)=y+\frac{1}{2!}(2xy-y^{2})+\frac{1}{3!}(3x^{2}y-3xy^{2}+2y^{3})+R_{3}, $$

其中

$$ \begin{aligned}R_{3}&=\frac{1}{4!}\Big[\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^{4}f(\theta h,\theta k)\Big]_{h=x,k=y}\\&=\frac{\mathrm{e}^{\theta x}}{24}\Big[x^{4}\ln\left(1+\theta y\right)+\frac{4x^{3}y}{1+\theta y}-\frac{6x^{2}y^{2}}{\left(1+\theta y\right)^{2}}+\frac{8xy^{3}}{\left(1+\theta y\right)^{3}}-\frac{6y^{4}}{\left(1+\theta y\right)^{4}}\Big]\quad(0<\theta<1).\end{aligned} $$

1. 求函数 $ f(x, y) = \sin x \sin y $ 在点 $ \left( \frac{\pi}{4}, \frac{\pi}{4} \right) $ 的二阶泰勒公式.

解

$$ f_{x}(x,y)=\cos x\sin y,\quad f_{y}(x,y)=\sin x\cos y, $$

$$ f_{x x}\left(x,y\right)=-\sin x\sin y,\quad f_{x y}\left(x,y\right)=\cos x\cos y, $$

$$ f_{y y}\left(x,y\right)=-\sin x\sin y,\quad f_{x x x}\left(x,y\right)=-\cos x\sin y, $$

$$ f_{x x y}\left(x,y\right)=-\sin x\cos y,\quad f_{x y y}\left(x,y\right)=-\cos x\sin y, $$

$$ f_{yy}(x,y)=-\sin x\cos y. $$

于是

$$ \left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=h f_{x}\left(\frac{\pi}{4},\frac{\pi}{4}\right)+k f_{y}\left(\frac{\pi}{4},\frac{\pi}{4}\right)=\frac{1}{2}h+\frac{1}{2}k, $$

$$ \begin{align*}\bigg(h\ \frac{\partial}{\partial x}+k\ \frac{\partial}{\partial y}\bigg)^{2}f\bigg(\frac{\pi}{4}\ ,\frac{\pi}{4}\bigg)&=h^{2}f_{xx}\bigg(\frac{\pi}{4}\ ,\frac{\pi}{4}\bigg)\ +2hk f_{xy}\bigg(\frac{\pi}{4}\ ,\frac{\pi}{4}\bigg)\ +k^{2}f_{yy}\bigg(\frac{\pi}{4}\ ,\frac{\pi}{4}\bigg)\\&=-\frac{1}{2}h^{2}\ +hk-\frac{1}{2}k^{2}.\end{align*} $$

又

$$ f\Big(\frac{\pi}{4},\frac{\pi}{4}\Big)=\frac{1}{2},\quad h=x-\frac{\pi}{4},\quad k=y-\frac{\pi}{4}. $$

将以上各项代入二阶泰勒公式，便得

$$ \begin{align*}\sin x\sin y&=\frac{1}{2}+\frac{1}{2}\bigg(x-\frac{\pi}{4}\bigg)+\frac{1}{2}\bigg(y-\frac{\pi}{4}\bigg)+\frac{1}{2!}\bigg[-\frac{1}{2}\bigg(x-\frac{\pi}{4}\bigg)^{2}+\\&\quad\bigg(x-\frac{\pi}{4}\bigg)\bigg(y-\frac{\pi}{4}\bigg)-\frac{1}{2}\bigg(y-\frac{\pi}{4}\bigg)^{2}\bigg]+R_{2}\\&=\frac{1}{2}+\frac{1}{2}\bigg(x-\frac{\pi}{4}\bigg)+\frac{1}{2}\bigg(y-\frac{\pi}{4}\bigg)-\frac{1}{4}\bigg[\bigg(x-\frac{\pi}{4}\bigg)^{2}-\end{align*} $$

$$ 2\left(x-\frac{\pi}{4}\right)\left(y-\frac{\pi}{4}\right)+\left(y-\frac{\pi}{4}\right)^{2} $$

其中

$$ \begin{aligned}R_{2}&=\frac{1}{3!}\Big[\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^{3}f(\xi,\eta)\Big]_{h=x-\frac{\pi}{4},k=y-\frac{\pi}{4}}\\&=-\frac{1}{6}\Big[\cos\xi\sin\eta\cdot\left(x-\frac{\pi}{4}\right)^{3}+3\sin\xi\cos\eta\cdot\left(x-\frac{\pi}{4}\right)^{2}\left(y-\frac{\pi}{4}\right)+\end{aligned} $$

$$ 3\cos\xi\sin\eta\cdot\left(x-\frac{\pi}{4}\right)\left(y-\frac{\pi}{4}\right)^{2}+\sin\xi\cos\eta\cdot\left(y-\frac{\pi}{4}\right)^{3}, $$

$$ \xi=\frac{\pi}{4}+\theta\bigg(x-\frac{\pi}{4}\bigg),\quad\eta=\frac{\pi}{4}+\theta\bigg(y-\frac{\pi}{4}\bigg),\quad0<\theta<1. $$

1. 利用函数 $ f(x, y) = x^{y} $ 的三阶泰勒公式，计算 $ 1.1^{1.02} $ 的近似值.

解 先求函数 $ f(x,y)=x^{y} $ 在点 (1,1) 的三阶泰勒公式.

$$ f_{x}\left(1,1\right)=yx^{y-1}\left|\begin{array}{l}\left(1,1\right)=1,\\ \end{array}\right.,\quad f_{y}\left(1,1\right)=x^{y}\ln x\left|\begin{array}{l}\left(1,1\right)=0,\\ \end{array}\right., $$

$$ f_{x x}(1,1)=y(y-1)x^{y-2}\big|_{(1,1)}=0, $$

$$ f_{x y}\left(1,1\right)=\left(x^{y-1}+y x^{y-1}\ln x\right)\left.\middle|_{(1,1)}=1\right., $$

$$ f_{yy}(1,1)=x^{y}\ln^{2}x\mid_{(1,1)}=0, $$

$$ f_{xxx}(1,1)=y(y-1)(y-2)x^{y-3}\mid_{(1,1)}=0, $$

$$ f_{x x y}\left(1,1\right)=\left[\left(2y-1\right)x^{y-2}+y\left(y-1\right)x^{y-2}\ln x\right]\big|_{(1,1)}=1, $$

$$ f_{x y y}(1,1)=(2x^{y-1}\ln x+y x^{y-1}\ln^{2}x)\left.\right|_{(1,1)}=0, $$

$$ f_{yyy}(1,1)=x^{y}\ln^{3}x\left.\right|_{(1,1)}=0. $$

又

$$ f(1,1)=1,\quad h=x-1,\quad k=y-1. $$

将以上各项代入三阶泰勒公式，便得

$$ \begin{aligned}x^{y}&=1+\left(x-1\right)+\frac{1}{2\dot{1}}\big[2\left(x-1\right)\left(y-1\right)\big]+\frac{1}{3\dot{1}}\big[3\left(x-1\right)^{2}\left(y-1\right)\big]+R_{3}\\&=1+\left(x-1\right)+\left(x-1\right)\left(y-1\right)+\frac{1}{2}\big(x-1\big)^{2}\left(y-1\right)+R_{3}.\end{aligned} $$

因此

$$ \begin{aligned}1.1^{1.02}&\approx1+0.1+0.1\times0.02+\frac{1}{2}\times0.1^{2}\times0.02\\&=1+0.1+0.002+0.0001=1.1021.\end{aligned} $$

1. 求函数 $ f(x, y) = e^{x+y} $ 在点 $ (0,0) $ 的 n 阶泰勒公式.

$$ f(0,0)=1,\quad f_{x}(0,0)=\mathrm{e}^{x+y}\mid_{(0,0)}=1,\quad f_{y}(0,0)=\mathrm{e}^{x+y}\mid_{(0,0)}=1,\cdots, $$

$$ f_{x^{x+1}}^{(n)}(0,0)=\mathrm{e}^{x+y}\mid_{(0,0)}=1(m=0,1,\cdots,n). $$

又

$$ h=x,\quad k=y. $$

将以上各项代入 n 阶泰勒公式，便得

$$ \begin{aligned}\mathrm{e}^{x+y}&=1+\left(x+y\right)+\frac{1}{2\mathrm{!}}\left(x^{2}+2xy+y^{2}\right)+\frac{1}{3\mathrm{!}}\left(x^{3}+3x^{2}y+3xy^{2}+y^{3}\right)\\&\quad+\cdots+\frac{1}{n\mathrm{!}}\left(x+y\right)^{n}+R_{n}=\sum_{k=0}^{n}\frac{\left(x+y\right)^{k}}{k\mathrm{!}}+R_{n},\end{aligned} $$

其中

$$ R_{n}=\frac{\left(x+y\right)^{n+1}}{\left(n+1\right)!}\mathrm{e}^{\theta\left(x+y\right)}\quad\left(0<\theta<1\right). $$

$ ^{*} $习题9-10

1. 某种合金的含铅量百分比（%）为 p，其熔解温度（℃）为 $ \theta $，由实验测得 p 与 $ \theta $ 的数据如下表：

试用最小二乘法建立 $ \theta $与p之间的经验公式 $ \theta=ap+b $

解 设 M 是各个数据的偏差平方和，即

$$ \begin{array}{l l l}{M}&{=}&{\displaystyle\sum_{i=1}^{6}\left[\theta_{i}\;-\;\left(a p_{i}\;+\;b\right)\right]^{2}.}\\ \end{array} $$

令

$$ \{\begin{aligned}\frac{\partial M}{\partial a}&=-\sum_{i=1}^{6}2p_{i}[\theta_{i}-(ap_{i}+b)]=0,\\ \frac{\partial M}{\partial b}&=-\sum_{i=1}^{6}2[\theta_{i}-(ap_{i}+b)]=0.\end{aligned}. $$

整理，得

$$ \{\begin{aligned}&a\sum_{i=1}^{6}p_{i}^{2}+b\sum_{i=1}^{6}p_{i}=\sum_{i=1}^{6}\theta_{i}p_{i},\\ &a\sum_{i=1}^{6}p_{i}+6b=\sum_{i=1}^{6}\theta_{i}.\end{aligned}. $$

计算，得

$$ \sum_{i=1}^{6}p_{i}^{2}\;=\;28365.28,\qquad\sum_{i=1}^{6}p_{i}\;=\;396.6, $$

$$ \sum_{i=1}^{6}\theta_{i}p_{i}=101176.3,\quad\sum_{i=1}^{6}\theta_{i}=1458. $$

代入方程组，得

$$ \{\begin{aligned}&28~365.~28a~+~396.~6b~=~101~176.~3,\\ &396.~6a~+~6b~=~1~458.\\ \end{aligned}. $$

解得

$$ a=\frac{4802.5}{2150.02}=2.234, $$

$$ b=\frac{572.0}{6}=95.33. $$

所以经验公式为 $ \theta=2.234p+95.33 $.

1. 已知一组实验数据为 $ (x_{1}, y_{1}), (x_{2}, y_{2}), \cdots, (x_{n}, y_{n}) $. 现若假定经验公式是

$$ y=a x^{2}+b x+c. $$

试按最小二乘法建立 a, b, c 应满足的三元一次方程组.

解 设 M 是各个数据的偏差平方和，即

$$ \begin{array}{r l r}{M}&{=}&{\displaystyle\sum_{i=1}^{n}\left[\;y_{i}\;-\;(\;a x_{i}^{2}\;+\;b x_{i}\;+\;c\;)\;\right]^{2}.}\end{array} $$

令

$$ \{\frac{\partial M}{\partial a}\;=\;-2\sum_{i=1}^{n}[y_{i}\;-(a x_{i}^{2}+b x_{i}+c)]\cdot x_{i}^{2}\;=\;0., $$

$$ \{\frac{\partial M}{\partial b}\;=\;-2\sum_{i=1}^{n}[y_{i}\;-(a x_{i}^{2}+b x_{i}+c)]\;\cdot x_{i}\;=\;0., $$

$$ \left\lfloor\frac{\partial M}{\partial c}\right.={}-\left.2\sum_{i=1}^{n}\left[y_{i}\mathrm{~-~}(a x_{i}^{2}+b x_{i}+c)\right]\right\rfloor=0. $$

整理，得a,b,c应满足的三元一次方程组如下：

$$ \left\lceil a\sum_{i=1}^{n}x_{i}^{4}\ +\ b\sum_{i=1}^{n}x_{i}^{3}\ +\ c\sum_{i=1}^{n}x_{i}^{2}\ =\ \sum_{i=1}^{n}x_{i}^{2}y_{i}\right., $$

$$ \{a\sum_{i=1}^{n}x_{i}^{3}+b\sum_{i=1}^{n}x_{i}^{2}+c\sum_{i=1}^{n}x_{i}=\sum_{i=1}^{n}x_{i}y_{i},. $$

$$ \left\lfloor a\sum_{i=1}^{n}x_{i}^{2}\right.+\left.b\sum_{i=1}^{n}x_{i}+nc\right.=\sum_{i=1}^{n}y_{i}. $$

总习题九

1. 在“充分”“必要”和“充分必要”三者中选择一个正确的填入下列空格内：

(1) $ f(x,y) $ 在点 $ (x,y) $ 可微分是 $ f(x,y) $ 在该点连续的 ___ 条件， $ f(x,y) $ 在点 $ (x,y) $ 连续是 $ f(x,y) $ 在该点可微分的 ___ 条件；

(2) $ z = f(x, y) $ 在点 $ (x, y) $ 的偏导数 $ \frac{\partial z}{\partial x} $ 及 $ \frac{\partial z}{\partial y} $ 存在是 $ f(x, y) $ 在该点可微分的 ___ 条件， $ z = f(x, y) $ 在点 $ (x, y) $ 可微分是函数在该点的偏导数 $ \frac{\partial z}{\partial x} $ 及 $ \frac{\partial z}{\partial y} $ 存在的 ___ 条件；

(3) $ z = f(x, y) $ 的偏导数 $ \frac{\partial z}{\partial x} $ 及 $ \frac{\partial z}{\partial y} $ 在点 $ (x, y) $ 存在且连续是 $ f(x, y) $ 在该点可微分的 ___ 条件；

（4）函数 $ z = f(x, y) $ 的两个二阶混合偏导数 $ \frac{\partial^{2}z}{\partial x\partial y} $ 及 $ \frac{\partial^{2}z}{\partial y\partial x} $ 在区域 D 内连续是这两个二阶混合偏导数在 D 内相等的 ___ 条件.

解（1）充分，必要。（2）必要，充分.

（3）充分. （4）充分.

注 本题结果给出了二元函数连续、可偏导（两个偏导数均存在）、可微分及具有连续偏导数之间的联系，用图表可表示为

1. 下题中给出了四个结论，从中选出一个正确的结论：

设函数 $ f(x,y) $ 在点 $ (0,0) $ 的某邻域内有定义，且 $ f_{x}(0,0)=3,f_{y}(0,0)=-1 $ ，则有（）.

$$ \mathrm{d}z\mid_{(0,0)}=3\mathrm{d}x-\mathrm{d}y $$

（B）曲面 $ z = f(x, y) $ 在点 $ (0, 0, f(0, 0)) $ 的一个法向量为 $ (3, -1, 1) $

（C）曲线 $ \{\begin{aligned}z&=f(x,y)\\ y&=0\end{aligned}. $，在点 $ (0,0,f(0,0)) $的一个切向量为 $ (1,0,3) $

(D) 曲线 $ \{\begin{aligned}z&=f(x,y),\\ y&=0\end{aligned}. $，在点 $ (0,0,f(0,0)) $的一个切向量为 $ (3,0,1) $

解 函数 $ f(x,y) $ 在点 $ (0,0) $ 处的两个偏导数存在，不一定可微分，故 (A) 不对.

由于函数存在偏导数不能保证可微分，从而不能保证曲面 $ z = f(x, y) $ 在点 $ (0, 0, f(0, 0)) $ 处存在切平面，因而 (B) 不对；若 $ z = f(x, y) $ 在点 $ (0, 0, f(0, 0)) $ 处存在连续偏导数，曲面在该点处有切平面，其法向量是 $ (3, -1, -1) $，而不是 $ (3, -1, 1) $，故 (B) 也不对.

取 x 为参数，则曲线 x = x, y = 0, z = f(x, 0) 在点 (0, 0, f(0, 0)) 处的一个切向量为 (1, 0, 3)，故 (C) 正确.

1. 求函数 $ f(x,y)=\frac{\sqrt{4x-y^{2}}}{\ln(1-x^{2}-y^{2})} $ 的定义域，并求 $ \lim_{(x,y)\to(\frac{1}{2},0)}f(x,y) $.

解 函数的定义域为 $ D=\{(x,y)\mid 0

因为点 $ \left(\frac{1}{2},0\right)\in D $, $ f(x,y) $ 为初等函数, 所以

$$ \lim_{(x,y)\to\left(\frac{1}{2},0\right)}f(x,y)=f\left(\frac{1}{2},0\right)=\frac{\sqrt{2}}{\ln\frac{3}{4}}=\frac{\sqrt{2}}{\ln3-\ln4}. $$

$ ^{*} $4. 证明极限 $ \lim_{(x,y)\to(0,0)}\frac{xy^{2}}{x^{2}+y^{4}} $ 不存在.

证 取两条趋于 $ (0,0) $的路径， $ c_{1}:x=0,c_{2}:y^{2}=x $.

$$ \begin{array}{c}\lim\limits_{(x,y)\to(0,0)}f(x,y)\\\text{↓}(x,y)\text{↓}(x,y)\\\text{↑}(x,y)\text{↑}(x,y)\end{array}=\lim\limits_{(x,y)\to(0,0)}\frac{xy^{2}}{x^{2}+y^{4}}=0, $$

$$ \lim_{(x,y)\to(x,y)}f(x,y)=\lim_{(x,y)\to(x,0)}\frac{xy^{2}}{x^{2}+y^{4}}=\lim_{x\to0}\frac{x^{2}}{x^{2}+x^{2}}=\frac{1}{2}. $$

由于 $ (x,y) $分别沿 $ c_{1},c_{2} $趋于 $ (0,0) $时 $ f(x,y) $的极限不相等，故 $ \lim_{(x,y)\to(0,0)}\frac{xy^{2}}{x^{2}+y^{4}} $不存在.

1. 设

$$ f(x,y)=\{\begin{aligned}&\frac{x^{2}y}{x^{2}+y^{2}},&x^{2}+y^{2}\neq0,\\ &0,&x^{2}+y^{2}=0.\end{aligned}. $$

求 $ f_{x}(x,y) $ 及 $ f_{y}(x,y) $.

解 当 $ x^{2}+y^{2}\neq0 $ 时，

$$ f_{x}(x,y)=\frac{\partial}{\partial x}\bigg(\frac{x^{2}y}{x^{2}+y^{2}}\bigg)=\frac{2xy(x^{2}+y^{2})-x^{2}y\cdot2x}{\left(x^{2}+y^{2}\right)^{2}}=\frac{2xy^{3}}{\left(x^{2}+y^{2}\right)^{2}}, $$

$$ f_{y}(x,y)=\frac{\partial}{\partial y}\biggl(\frac{x^{2}y}{x^{2}+y^{2}}\biggr)=\frac{x^{2}\left(x^{2}+y^{2}\right)-x^{2}y\cdot2y}{\left(x^{2}+y^{2}\right)^{2}}=\frac{x^{2}\left(x^{2}-y^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}. $$

当 $ x^{2}+y^{2}=0 $时，

$$ f_{x}(0,0)=\lim_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\lim_{\Delta x\to0}\frac{0}{\Delta x}=0, $$

$$ f_{y}(0,0)=\lim_{\Delta y\to0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=\lim_{\Delta y\to0}\frac{0}{\Delta y}=0, $$

故

$$ f_{x}(x,y)=\{\begin{array}{l}\displaystyle\frac{2xy^{3}}{(x^{2}+y^{2})^{2}},\quad x^{2}+y^{2}\neq0,\\ \\ 0,\quad x^{2}+y^{2}=0.\end{array}. $$

$$ f_{y}(x,y)=\{\begin{array}{l}\dfrac{x^{2}(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}},\quad x^{2}+y^{2}\neq0,\\0,\quad x^{2}+y^{2}=0.\end{array}. $$

1. 求下列函数的一阶和二阶偏导数：

(1) $ z = \ln(x + y^2) $; (2) $ z = x^y $.

解（1）

$$ \frac{\partial z}{\partial x}=\frac{1}{x+y^{2}},\quad\frac{\partial^{2}z}{\partial x^{2}}=-\frac{1}{\left(x+y^{2}\right)^{2}}, $$

$$ \frac{\partial z}{\partial y}=\frac{2y}{x+y^{2}},\quad\frac{\partial^{2}z}{\partial y^{2}}=\frac{2\left(x+y^{2}\right)-4y^{2}}{\left(x+y^{2}\right)^{2}}=\frac{2\left(x-y^{2}\right)}{\left(x+y^{2}\right)^{2}}, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial}{\partial y}\left(\frac{1}{x+y^{2}}\right)=-\frac{2y}{\left(x+y^{2}\right)^{2}}. $$

(2)

$$ \frac{\partial z}{\partial x}=yx^{y-1},\quad\frac{\partial^{2}z}{\partial x^{2}}=y(y-1)x^{y-2}, $$

$$ \frac{\partial z}{\partial y}=x^{y}\ln x,\quad\frac{\partial^{2}z}{\partial y^{2}}=x^{y}\ln^{2}x, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\frac{\partial}{\partial y}(y x^{y-1})=x^{y-1}+y\cdot x^{y-1}\ln x. $$

1. 求函数 $ z = \frac{xy}{x^{2} - y^{2}} $ 当 x = 2, y = 1, $ \Delta x = 0.01 $, $ \Delta y = 0.03 $ 时的全增量和全微分.

解

$$ \Delta z\;=\;\frac{2.01\cdot1.03}{2.01^{2}-1.03^{2}}\;-\;\frac{2}{3}\;=\;0.03. $$

又

$$ \frac{\partial z}{\partial x}=\frac{-\left(y^{3}+x^{2}y\right)}{\left(x^{2}-y^{2}\right)^{2}},\quad\frac{\partial z}{\partial y}=\frac{x^{3}+x y^{2}}{\left(x^{2}-y^{2}\right)^{2}}, $$

$$ \frac{\partial z}{\partial x}\bigg|_{(2,1)}=-\frac{5}{9},\quad\frac{\partial z}{\partial y}\bigg|_{(2,1)}=\frac{10}{9}. $$

故

$$ \mathrm{d}z\left|\begin{array}{c}\frac{\partial z}{\partial x}=2,\Delta x=0.01\\ \frac{\partial x}{y}=1,\Delta y=0.03\end{array}\right.=\left.\frac{\partial z}{\partial x}\right|_{(2,1)}\cdot\Delta x+\left.\frac{\partial z}{\partial y}\right|_{(2,1)}\cdot\Delta y=0.03. $$

$ ^{8} $. 设

$$ f(x,y)=\{\begin{aligned}&\frac{x^{2}y^{2}}{(x^{2}+y^{2})^{3/2}},&x^{2}+y^{2}&\neq0,\\ &0,&x^{2}+y^{2}&=0.\end{aligned}. $$

证明： $ f(x,y) $ 在点 $ (0,0) $ 处连续且偏导数存在，但不可微分.

证 因为

$$ 0\leqslant\frac{x^{2}y^{2}}{\left(x^{2}+y^{2}\right)^{3/2}}\leqslant\frac{\left(x^{2}+y^{2}\right)^{2}}{\left(x^{2}+y^{2}\right)^{3/2}}=\sqrt{x^{2}+y^{2}}, $$

$$ \lim_{(x,y)\to(0,0)}\sqrt{x^{2}+y^{2}}=0, $$

所以

$$ \lim_{(x,y)\to(0,0)}f(x,y)=0. $$

又 $ f(0,0)=0 $，故 $ \lim_{(x,y)\to(0,0)}f(x,y)=f(0,0) $，即 $ f(x,y) $在点 $ (0,0) $处连续.

$$ f_{x}(0,0)=\lim_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\lim_{\Delta x\to0}\frac{0}{\Delta x}=0, $$

$$ f_{y}(0,0)=\lim_{\Delta y\to0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=\lim_{\Delta x\to0}\frac{0}{\Delta y}=0. $$

$$ \Delta z-\left[f_{x}\left(0,0\right)\Delta x+f_{y}\left(0,0\right)\Delta y\right]=\frac{\left(\Delta x\right)^{2}\cdot\left(\Delta y\right)^{2}}{\left[\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}\right]^{3/2}}, $$

$$ \lim_{\substack{\Delta x\to0\\\Delta y\to\Delta x}}\frac{\overline{\left[\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}\right]^{3/2}}}{\rho}=\lim_{\Delta x\to0}\frac{\left(\Delta x\right)^{4}}{\left[2\left(\Delta x\right)^{2}\right]^{2}}=\frac{1}{4}\neq0, $$

其中 $ \rho=\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}} $，故 $ f(x,y) $ 在点 $ (0,0) $ 处偏导数存在，但不可微分。

1. 设 $ u=x^{y} $，而 $ x=\varphi(t) $， $ y=\psi(t) $ 都是可微函数，求 $ \frac{du}{dt} $

解

$$ \frac{\mathrm{d}u}{\mathrm{d}t}=\frac{\partial u}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial u}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}=y x^{y-1}\cdot\varphi^{\prime}(t)+x^{y}\ln x\cdot\psi^{\prime}(t). $$

1. 设 $ z = f(u, v, w) $ 具有连续偏导数，而

$$ u=\eta-\zeta,\quad v=\zeta-\xi,\quad w=\xi-\eta, $$

求 $ \frac{\partial z}{\partial\xi},\frac{\partial z}{\partial\eta},\frac{\partial z}{\partial\zeta} $

解

$$ \frac{\partial z}{\partial\xi}\;=\;\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial\xi}\;+\;\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial\xi}\;+\;\frac{\partial z}{\partial w}\cdot\frac{\partial w}{\partial\xi}\;=\;-\;\frac{\partial z}{\partial v}\;+\;\frac{\partial z}{\partial w}\;, $$

$$ \frac{\partial z}{\partial\eta}\;=\;\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial\eta}\;+\;\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial\eta}\;+\;\frac{\partial z}{\partial w}\cdot\frac{\partial w}{\partial\eta}\;=\;\frac{\partial z}{\partial u}\;-\;\frac{\partial z}{\partial w}, $$

$$ \frac{\partial z}{\partial\zeta}\;=\;\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial\zeta}\;+\;\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial\zeta}\;+\;\frac{\partial z}{\partial w}\cdot\frac{\partial w}{\partial\zeta}\;=\;-\;\frac{\partial z}{\partial u}\;+\;\frac{\partial z}{\partial v}. $$

1. 设 $ z = f(u, x, y) $, $ u = xe^{y} $，其中 f 具有连续的二阶偏导数，求 $ \frac{\partial^{2}z}{\partial x\partial y} $.

解

$$ \frac{\partial z}{\partial x}\;=\;f_{u}\cdot\frac{\partial u}{\partial x}\;+\;f_{x}\;=\;f_{u}\cdot\mathrm{e}^{y}\;+\;f_{x}\;, $$

$$ \begin{align*}\frac{\partial^{2}z}{\partial x\partial y}&=\frac{\partial}{\partial y}(f_{u}\cdot\mathbf{e}^{y}+f_{x})=\left(\frac{\partial}{\partial y}f_{u}\right)\cdot\mathbf{e}^{y}+f_{u}\cdot\mathbf{e}^{y}+\frac{\partial}{\partial y}f_{x}\\&=\left(f_{uu}\cdot\frac{\partial u}{\partial y}+f_{uy}\right)\mathbf{e}^{y}+f_{u}\cdot\mathbf{e}^{y}+\left(f_{xu}\cdot\frac{\partial u}{\partial y}+f_{xy}\right)\\&=\left(f_{uu}\cdot x\mathbf{e}^{y}+f_{uy}\right)\mathbf{e}^{y}+f_{u}\cdot\mathbf{e}^{y}+f_{xu}\cdot x\mathbf{e}^{y}+f_{xy}\\&=x\mathbf{e}^{2y}f_{uu}+\mathbf{e}^{y}f_{uy}+x\mathbf{e}^{y}f_{xu}+f_{xy}+\mathbf{e}^{y}f_{u}.\end{align*} $$

1. 设 $ x = e^{u} \cos v $, $ y = e^{u} \sin v $, z = uv, 试求 $ \frac{\partial z}{\partial x} $ 和 $ \frac{\partial z}{\partial y} $.

解 $ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} = v \frac{\partial u}{\partial x} + u \frac{\partial v}{\partial x} $.

分别在 $ x = e^{u} \cos v $, $ y = e^{u} \sin v $ 的两端对 x 求偏导数，得

$$ \{\begin{aligned}\mathrm{e}^{u}\cos v\frac{\partial u}{\partial x}-\mathrm{e}^{u}\sin v\frac{\partial v}{\partial x}&=1,\\ \mathrm{e}^{u}\sin v\frac{\partial u}{\partial x}+\mathrm{e}^{u}\cos v\frac{\partial v}{\partial x}&=0.\end{aligned}. $$

由以上方程组解得

$$ \frac{\partial u}{\partial x}=\mathrm{e}^{-u}\cos v,\quad\frac{\partial v}{\partial x}=-\mathrm{e}^{-u}\sin v. $$

从而

$$ \frac{\partial z}{\partial x}\;=\;\mathrm{e}^{-u}\left(v\cos\;v\;-\;u\sin\;v\right). $$

同理

$$ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y}=v\frac{\partial u}{\partial y}+u\frac{\partial v}{\partial y}. $$

分别在 $ x = e^{u} \cos v $, $ y = e^{u} \sin v $ 的两端对 y 求偏导数，得

$$ \{\begin{aligned}\mathrm{e}^{u}\cos v\frac{\partial u}{\partial y}-\mathrm{e}^{u}\sin v\frac{\partial v}{\partial y}&=0,\\ \mathrm{e}^{u}\sin v\frac{\partial u}{\partial y}+\mathrm{e}^{u}\cos v\frac{\partial v}{\partial y}&=1.\end{aligned}. $$

由以上方程组解得

$$ \frac{\partial u}{\partial y}=\mathrm{e}^{-u}\sin v,\quad\frac{\partial v}{\partial y}=\mathrm{e}^{-u}\cos v. $$

从而

$$ \frac{\partial z}{\partial y}\;=\;\mathrm{e}^{-u}\left(u\cos v\;+\;v\sin v\right). $$

1. 求螺旋线 $ x = a \cos \theta, y = a \sin \theta, z = b \theta $ 在点 $ (a, 0, 0) $ 处的切线及法平面方程.

解

$$ \frac{\mathrm{d}x}{\mathrm{d}\theta}=-a\sin\theta,\quad\frac{\mathrm{d}y}{\mathrm{d}\theta}=a\cos\theta,\quad\frac{\mathrm{d}z}{\mathrm{d}\theta}=b. $$

点 $ (a,0,0) $所对应的参数 $ \theta=0 $，故曲线在给定点的切向量

$$ T=\left(0,a,b\right). $$

于是切线方程为

$$ \frac{x-a}{0}=\frac{y}{a}=\frac{z}{b}, $$

即

$$ \{\begin{aligned}x&=a,\\ by&=az=0.\end{aligned}. $$

法平面方程为

$$ a(y-0)+b(z-0)=0, $$

即

$$ ay+b z=0. $$

1. 在曲面 z = xy 上求一点，使这点处的法线垂直于平面 $ x + 3y + z + 9 = 0 $，并写出这法线的方程.

解 设所求点为 $ M(x_{0},y_{0},z_{0}) $，曲面在该点处的一个法向量为 $ \boldsymbol{n}=(y_{0},x_{0},-1) $，平面的法向量为 $ (1,3,1) $.

按题意，n 垂直于平面，故有

$$ \frac{y_{0}}{1}=\frac{x_{0}}{3}=\frac{-1}{1}. $$

求得 $ x_{0}=-3, y_{0}=-1, z_{0}=x_{0}y_{0}=3 $. 于是所求点为 $ M(-3,-1,3) $，法线方程为

$$ \frac{x+3}{1}=\frac{y+1}{3}=\frac{z-3}{1}. $$

1. 设 $ e_{l} = (\cos \theta, \sin \theta) $，求函数

$$ f(x,y)=x^{2}-xy+y^{2} $$

在点(1,1)沿方向l的方向导数，并分别确定角 $ \theta $，使这导数有(1)最大值；(2)最小值；(3)等于0.

解

$$ \frac{\partial f}{\partial x}\;=\;2x\;-\;y\;,\;\frac{\partial f}{\partial y}\;=\;-x\;+\;2y\;, $$

$$ \frac{\partial f}{\partial x}\bigg|_{(1,1)}=1,\frac{\partial f}{\partial y}\bigg|_{(1,1)}=1. $$

$$ \frac{\partial f}{\partial l}\bigg|_{(1,1)}=\frac{\partial f}{\partial x}\bigg|_{(1,1)}\cos\theta+\frac{\partial f}{\partial y}\bigg|_{(1,1)}\sin\theta=\cos\theta+\sin\theta. $$

因为 $ \cos\theta+\sin\theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right) $，所以

（1）当 $ \theta = \frac{\pi}{4} $ 时，方向导数最大，其最大值为 $ \sqrt{2} $.

（2）当 $ \theta = \frac{5}{4}\pi $ 时，方向导数最小，其最小值为 $ -\sqrt{2} $

（3）当 $ \theta = \frac{3}{4}\pi $ 或 $ \frac{7}{4}\pi $ 时，方向导数为 0.

1. 求函数 $ u = x^{2} + y^{2} + z^{2} $ 在椭球面 $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 $ 上点 $ M_{0}(x_{0}, y_{0}, z_{0}) $ 处沿外法线方向的方向导数.

解 椭球面在点 $ M_{0} $ 处的沿外法线方向的一个向量为 $ \boldsymbol{n}=\left(\frac{x_{0}}{a^{2}},\frac{y_{0}}{b^{2}},\frac{z_{0}}{c^{2}}\right) $

$$ e_{_{n}}=\frac{1}{\sqrt{\frac{x_{0}^{2}}{a^{4}}+\frac{y_{0}^{2}}{b^{4}}+\frac{z_{0}^{2}}{c^{4}}}}\bigg(\frac{x_{0}}{a^{2}},\frac{y_{0}}{b^{2}},\frac{z_{0}}{c^{2}}\bigg). $$

$$ \begin{array}{l}\frac{\partial z}{\partial n}\bigg|_{(x_{0},y_{0},z_{0})}=\frac{1}{\sqrt{\frac{x_{0}^{2}}{a^{4}}+\frac{y_{0}^{2}}{b^{4}}+\frac{z_{0}^{2}}{c^{4}}}}\bigg(2x_{0}\cdot\frac{x_{0}}{a^{2}}+2y_{0}\cdot\frac{y_{0}}{b^{2}}+2z_{0}\cdot\frac{z_{0}}{c^{2}}\bigg)\\=\frac{2}{\sqrt{\frac{x_{0}^{2}}{a^{4}}+\frac{y_{0}^{2}}{b^{4}}+\frac{z_{0}^{2}}{c^{4}}}}\end{array} $$

1. 求平面 $ \frac{x}{3} + \frac{y}{4} + \frac{z}{5} = 1 $ 和柱面 $ x^{2} + y^{2} = 1 $ 的交线上与 xOy 平面距离最短的点.

解 设交线上的点为 $ M(x,y,z) $，它到 xOy 面上距离的平方为 $ z^{2} $. 问题就成为求函数 $ z^{2} $ 在约束条件 $ \frac{x}{3} + \frac{y}{4} + \frac{z}{5} = 1 $ 和 $ x^{2} + y^{2} = 1 $ 下的最小值问题. 作拉格朗日函数

$$ {\cal L}\;=\;z^{2}\;+\;\lambda\left(\frac{x}{3}\;+\;\frac{y}{4}\;+\;\frac{z}{5}\;-\;1\right)+\mu(x^{2}\;+\;y^{2}\;-\;1)\;. $$

令

$$ \{\begin{aligned}L_{x}&=\frac{\lambda}{3}+2\mu x=0,\\ \\ L_{y}&=\frac{\lambda}{4}+2\mu y=0,\\ \\ L_{z}&=2z+\frac{\lambda}{5}=0.\end{aligned}. $$

又由约束条件，有

$$ \frac{x}{3}+\frac{y}{4}+\frac{z}{5}=1, $$

$$ x^{2}+y^{2}=1. $$

解此方程组，得 $ x=\frac{4}{5} $， $ y=\frac{3}{5} $， $ z=\frac{35}{12} $。于是，得可能的极值点 $ M_{0}\left(\frac{4}{5},\frac{3}{5},\frac{35}{12}\right) $。由问题本身可知，距离最短的点必定存在，因此 $ M_{0} $ 就是所求的点。

1. 在第一卦限内作椭球面 $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $ 的切平面，使该切平面与三坐标面所围成的四面体的体积最小。求这切平面的切点，并求此最小体积.

解 设切点为 $ M(x_{0},y_{0},z_{0}) $， $ F(x,y,z)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1 $，

$$ \boldsymbol{n}=(\boldsymbol{F}_{x},\boldsymbol{F}_{y},\boldsymbol{F}_{z})=\left(\frac{2x}{a^{2}},\frac{2y}{b^{2}},\frac{2z}{c^{2}}\right). $$

曲面在点 M 处的切平面方程为

$$ \frac{x_{0}}{a^{2}}(x-x_{0})+\frac{y_{0}}{b^{2}}(y-y_{0})+\frac{z_{0}}{c^{2}}(z-z_{0})=0, $$

即

$$ \frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^{2}}+\frac{z_{0}z}{c^{2}}=1. $$

于是，切平面在三个坐标轴上的截距依次为 $ \frac{a^{2}}{x_{0}},\frac{b^{2}}{y_{0}},\frac{c^{2}}{z_{0}} $，切平面与三个坐标面所围成的四面体的体积为

$$ V=\frac{1}{6}\cdot\frac{a^{2}b^{2}c^{2}}{x_{0}y_{0}z_{0}}. $$

在 $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 $ 的条件下，求 V 的最小值，即求分母 xyz 的最大值。作拉格朗日函数

$$ L(x,y,z)~=~xyz~+~\lambda\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1\right). $$

令

$$ \left(L_{x}=y z+\frac{2\lambda x}{a^{2}}=0,\right. $$

$$ L_{_{y}}=xz+\frac{2\lambda y}{b^{2}}=0, $$

$$ L_{z}=x y+\frac{2\lambda z}{c^{2}}=0. $$

(1) $ \cdot x + (2) \cdot y + (3) \cdot z $，并由约束条件 $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 $，得

$$ \frac{x^{2}}{a^{2}}=\frac{y^{2}}{b^{2}}=\frac{z^{2}}{c^{2}}=\frac{1}{3}, $$

从而

$$ x~=~\frac{a}{\sqrt{3}},\quad y~=~\frac{b}{\sqrt{3}},\quad z~=~\frac{c}{\sqrt{3}}. $$

于是，得可能极值点 $ M\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right) $. 由此问题的性质知，所求的切点为 $ M\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right) $，四面体的最小体积为

$$ V_{min}\;=\;\frac{\sqrt{3}}{2}abc. $$

1. 某厂家生产的一种产品同时在两个市场销售，售价分别为 $ p_{1} $ 和 $ p_{2} $，销售量分别为 $ q_{1} $ 和 $ q_{2} $，需求函数分别为

$$ q_{1}=24-0.2p_{1},\quad q_{2}=10-0.05p_{2}, $$

总成本函数为

$$ C~=~35~+40(q_{1}+q_{2}). $$

试问：厂家如何确定两个市场的售价，能使其获得的总利润最大？最大总利润为多少？

解法一 总收入函数为

$$ R~=~p_{1}q_{1}~+~p_{2}q_{2}~=~24p_{1}~-~0.2p_{1}^{2}~+~10p_{2}~-~0.05p_{2}^{2}, $$

总利润函数为

$$ {\cal L}\;=\;R\;-\;C\;=\;32p_{1}\;-\;0.2p_{1}^{2}\;-0.05p_{2}^{2}\;+\;12p_{2}\;-1395. $$

由极值的必要条件，得方程组

$$ \{\begin{aligned}\frac{\partial L}{\partial p_{1}}&=32-0.4p_{1}=0,\\ \frac{\partial L}{\partial p_{2}}&=12-0.1p_{2}=0.\end{aligned}. $$

解此方程组，得 $ p_{1}=80, p_{2}=120 $.

由问题的实际意义可知，厂家获得总利润最大的市场售价必定存在，故当 $ p_{1}=80,p_{2}=120 $时，厂家所获得的总利润最大，其最大总利润为

$$ L\begin{aligned}\left|\begin{array}{l}&=605.\\p_{1}=80,p_{2}=120\\\end{array}\right.\end{aligned} $$

解法二 两个市场的价格函数分别为

$$ p_{1}~=~120~-5q_{1}\,,\quad p_{2}~=~200~-20q_{2}\,, $$

总收入函数为

$$ \begin{array}{r}{R~=~p_{1}q_{1}~+~p_{2}q_{2}~=~(120~-5q_{1})q_{1}~+~(200~-20q_{2})q_{2},}\end{array} $$

总利润函数为

$$ \begin{aligned}&L=R-C=\left(120-5q_{1}\right)q_{1}+\left(200-20q_{2}\right)q_{2}-\left\lfloor35+40\left(q_{1}+q_{2}\right)\right\rfloor\\&\quad=80q_{1}-5q_{1}^{2}+160q_{2}-20q_{2}^{2}-35.\\ \end{aligned} $$

由极值的必要条件，得方程组

$$ \begin{aligned}\{\begin{aligned}\frac{\partial L}{\partial q_{1}}&=80-10q_{1}=0,\\ \frac{\partial L}{\partial q_{2}}&=160-40q_{2}=0.\end{aligned}.\end{aligned} $$

解此方程组得 $ q_{1}=8, q_{2}=4 $.

由问题的实际意义可知，当 $ q_{1}=8, q_{2}=4 $ ，即 $ p_{1}=80, p_{2}=120 $ 时，厂家所获得的总利润最大，其最大总利润为

$$ L\mid_{q_{1}=8,q_{2}=4}=605. $$

1. 设有一小山，取它的底面所在的平面为 xOy 坐标面，其底部所占的闭区域为 $ D=\{(x,y)\mid x^{2}+y^{2}-xy\leqslant75\} $，小山的高度函数为 $ h=f(x,y)=75-x^{2}-y^{2}+xy $.

（1）设 $ M(x_{0},y_{0})\in D $，问 $ f(x,y) $在该点沿平面上什么方向的方向导数最大？若记此方向导数的最大值为 $ g(x_{0},y_{0}) $，试写出 $ g(x_{0},y_{0}) $的表达式.

（2）现欲利用此小山开展攀岩活动，为此需要在山脚找一上山坡度最大的点作为攀岩的起点，也就是说，要在D的边界线 $ x^{2}+y^{2}-xy=75 $上找出(1)中的 $ g(x,y) $达到最大值的点.试确定攀岩起点的位置.

解 （1）由梯度与方向导数的关系知， $ h=f(x,y) $ 在点 $ M(x_{0},y_{0}) $ 处沿梯度

$$ \mathbf{grad}f(x_{0},y_{0})=(y_{0}-2x_{0})i+(x_{0}-2y_{0})j $$

方向的方向导数最大，方向导数的最大值为该梯度的模，所以

$$ g(x_{0},y_{0})=\sqrt{(y_{0}-2x_{0})^{2}+(x_{0}-2y_{0})^{2}}=\sqrt{5x_{0}^{2}+5y_{0}^{2}-8x_{0}y_{0}}. $$

（2）欲在D的边界上求 $ g(x,y) $达到最大值的点，只需求 $ F(x,y)=g^{2}(x,y)=5x^{2}+5y^{2}-8xy $达到最大值的点。因此，作拉格朗日函数

$$ L=5x^{2}+5y^{2}-8x y+\lambda\left(75-x^{2}-y^{2}+x y\right). $$

令

$$ \left\lceil L_{x}=10x-8y+\lambda(y-2x)=0\right\rceil, $$

$$ \begin{array}{r l}{\mathcal{L}_{L_{y}}}&{=10y-8x+\lambda\left(x-2y\right)=0.}\end{array} $$

又由约束条件，有

$$ 75\ -\ x^{2}\ -\ y^{2}\ +\ x y\ =0. $$

(1) $ (+2) $，得

$$ \left(x+y\right)\left(2-\lambda\right)=0, $$

解得 y = -x 或 $ \lambda = 2 $.

若 $ \lambda=2 $，则由（1）得 y=x，再由（3）得 $ x=y=\pm5\sqrt{3} $

若 y = -x，则由（3）得 $ x = \pm 5, y = \mp 5 $.

于是得到四个可能的极值点：

$$ M_{1}\left(5,-5\right),\quad M_{2}\left(-5,5\right),\quad M_{3}\left(5\sqrt{3},5\sqrt{3}\right),\quad M_{4}\left(-5\sqrt{3},-5\sqrt{3}\right). $$

由于 $ F(M_{1}) = F(M_{2}) = 450 $, $ F(M_{3}) = F(M_{4}) = 150 $, 故 $ M_{1}(5, -5) $ 或 $ M_{2}(-5, 5) $ 可作为攀岩的起点.