高等数学（第七版）同济大学 总习题九（前10题） 个人解答

高等数学（第七版）同济大学 总习题九（前10题）

 

$\begin{array}{rlr}& 1.在“充分”“必要”和“充分必要”三者中选择一个正确的填入下列空格内：& \end{array}$

$\begin{array}{rlr}& (1)f(x,y)在点(x,y)可微分是f(x,y)在该点连续的\_\_\_\_条件，f(x,y)在点(x,y)连续是f(x,y)在该点& \\ & & \\ & 可微分的\_\_\_\_条件；& \\ & & \\ & (2)z=f(x,y)在点(x,y)的偏导数\frac{\partial z}{\partial x}及\frac{\partial z}{\partial y}存在是f(x,y)在该点可微分的\_\_\_\_条件，z=f(x,y)在& \\ & & \\ & 点(x,y)可微分是函数在该点的偏导数\frac{\partial z}{\partial x}及\frac{\partial z}{\partial y}存在的\_\_\_\_条件；& \\ & & \\ & (3)z=f(x,y)的偏导数\frac{\partial z}{\partial x}及\frac{\partial z}{\partial y}在点(x,y)存在且连续是f(x,y)在该点可微分的\_\_\_\_条件；& \\ & & \\ & (4)函数z=f(x,y)的两个二阶混合偏导数\frac{{\partial }^{2}z}{\partial x\partial y}及\frac{{\partial }^{2}z}{\partial y\partial x}在区域D内连续是这两个二阶混合偏导数& \\ & & \\ & 在D内相等的\_\_\_\_条件.& \end{array}$

解：

$\begin{array}{rlr}& (1)充分，必要& \\ & & \\ & (2)必要，充分& \\ & & \\ & (3)充分& \\ & & \\ & (4)充分.& \end{array}$

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$\begin{array}{rlr}& 2.下题中给出了四个结论，从中选出一个正确的结论：& \end{array}$

$\begin{array}{rlr}& 设函数f(x,y)在点(0,0)的某领域内有定义，且f_x(0,0)=3，f_y(0,0)=-1，则有()：& \\ & & \\ & (A)dz{|}_{(0,0)}=3dx-dy& \\ & & \\ & (B)曲面z=f(x,y)在点(0,0,f(0,0))的一个法向量为(3,-1,1)& \\ & & \\ & (C)曲线\{\begin{array}{l}z=f(x,y)，\\ \\ y=0\end{array}在点(0,0,f(0,0))的一个切向量为(1,0,3)& \\ & & \\ & (D)曲线\{\begin{array}{l}z=f(x,y)，\\ \\ y=0\end{array}在点(0,0,f(0,0))的一个切向量为(3,0,1)& \end{array}$

解：

$\begin{array}{rlr}& 函数偏导数存在不一定可微分，从而不能保证曲面存在切平面，所以A、B不对，& \\ & & \\ & 取x为参数，曲线x=x，y=0，z=f(x,0)在点(0,0,f(0,0))处的一个切向量为(1,0,3)，选C.& \end{array}$

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$\begin{array}{rlr}& 3.求函数f(x,y)=\frac{\sqrt{4x-y^2}}{ln(1-x^2-y^2)}的定义域，并求\underset{(x,y)\to (\frac{1}{2},0)}{\lim }f(x,y).& \end{array}$

解：

$\begin{array}{rlr}& 函数的定义域为D=\{(x,y)|0<x^2+y^2<1,y^2\le 4x\}，因为点\left(\frac{1}{2},0\right)\in D，f(x,y)为初等函数，& \\ & & \\ & 所以\underset{(x,y)\to (\frac{1}{2},0)}{\lim }f(x,y)=f\left(\frac{1}{2},0\right)=\frac{\sqrt{2}}{ln\frac{3}{4}}=\frac{\sqrt{2}}{ln3-ln4}.& \end{array}$

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$\begin{array}{rlr}& 4.证明极限\underset{(x,y)\to (0,0)}{\lim }\frac{x{y}^2}{x^2+y^4}不存在.& \end{array}$

解：

$\begin{array}{rlr}& 取两条趋于(0,0)的路径，c_1:x=0，c_2:y^2=x，& \\ & & \\ & \underset{\genfrac{}{}{0ex}{}{(x,y)\to (0,0)}{(x,y)\to c_1}}{\lim }f(x,y)=\underset{\genfrac{}{}{0ex}{}{(x,y)\to (0,0)}{x=0}}{\lim }\frac{x{y}^2}{x^2+y^4}=0，\underset{\genfrac{}{}{0ex}{}{(x,y)\to (0,0)}{(x,y)\to c_2}}{\lim }f(x,y)=\underset{\genfrac{}{}{0ex}{}{(x,y)\to (0,0)}{y^2=x}}{\lim }\frac{x{y}^2}{x^2+y^4}=\underset{x\to 0}{\lim }\frac{x^2}{x^2+x^2}=\frac{1}{2}，& \\ & & \\ & 因为(x,y)分别沿c_1，c_2趋于(0,0)时f(x,y)的极限不相等，所以\underset{(x,y)\to (0,0)}{\lim }\frac{x{y}^2}{x^2+y^4}不存在.& \end{array}$

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$\begin{array}{rlr}& 5.设f(x,y)=\{\begin{array}{l}\frac{x^{2}y}{x^2+y^2}，x^2+y^2\ne 0，\\ \\ 0，x^2+y^2=0.\end{array}，求f_x(x,y)及f_y(x,y).& \end{array}$

解：

$\begin{array}{rlr}& 当x^2+y^2\ne 0时，& \\ & & \\ & f_x(x,y)=\frac{\partial }{\partial x}\left(\frac{x^{2}y}{x^2+y^2}\right)=\frac{2xy(x^2+y^2)-x^{2}y\cdot 2x}{(x^2+y^2{)}^2}=\frac{2x{y}^3}{(x^2+y^2{)}^2}，& \\ & & \\ & f_y(x,y)=\frac{\partial }{\partial y}\left(\frac{x^{2}y}{x^2+y^2}\right)=\frac{x^2(x^2+y^2)-x^{2}y\cdot 2y}{(x^2+y^2{)}^2}=\frac{x^2(x^2-y^2)}{(x^2+y^2{)}^2}，& \\ & & \\ & 当x^2+y^2=0时，& \\ & & \\ & f_x(0,0)=\underset{\Delta x\to 0}{\lim }\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{0}{\Delta x}=0，& \\ & & \\ & f_y(0,0)=\underset{\Delta y\to 0}{\lim }\frac{f(0+\Delta y,0)-f(0,0)}{\Delta y}=\underset{\Delta y\to 0}{\lim }\frac{0}{\Delta y}=0，& \\ & & \\ & 所以，f_x(x,y)=\{\begin{array}{l}\frac{2x{y}^3}{(x^2+y^2{)}^2}，x^2+y^2\ne 0，\\ \\ 0，x^2+y^2=0.\end{array}，f_y(x,y)=\{\begin{array}{l}\frac{x^2(x^2-y^2)}{(x^2+y^2{)}^2}，x^2+y^2\ne 0，\\ \\ 0，x^2+y^2=0.\end{array}& \end{array}$

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$\begin{array}{rlr}& 6.求下列函数的一阶和二阶偏导数:& \end{array}$

$\begin{array}{rlr}& (1)z=ln(x+y^2)；(2)z=x^y.& \end{array}$

解：

$\begin{array}{rlr}& (1)\frac{\partial z}{\partial x}=\frac{1}{x+y^2}，\frac{{\partial }^{2}z}{\partial x^2}=-\frac{1}{(x+y^2{)}^2}，\frac{\partial z}{\partial y}=\frac{2y}{x+y^2}，\frac{{\partial }^{2}z}{\partial y^2}=\frac{2(x+y^2)-4y^2}{(x+y^2{)}^2}=\frac{2(x-y^2)}{(x+y^2{)}^2}，& \\ & & \\ & \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial y}\left(\frac{1}{x+y^2}\right)=-\frac{2y}{(x+y^2{)}^2}.& \\ & & \\ & (2)\frac{\partial z}{\partial x}=y{x}^{y-1}，\frac{{\partial }^{2}z}{\partial x^2}=y(y-1)x^{y-2}，\frac{\partial z}{\partial y}=x^{y}lnx，\frac{{\partial }^{2}z}{\partial y^2}=x^{y}l{n}^{2}x，& \\ & & \\ & \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial y}(y{x}^{y-1})=x^{y-1}+y\cdot x^{y-1}lnx.& \end{array}$

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$\begin{array}{rlr}& 7.求函数z=\frac{xy}{x^2-y^2}当x=2，y=1，\Delta x=0.01，\Delta y=0.03时的全增量和全微分.& \end{array}$

解：

$\begin{array}{rlr}& \Delta z=\frac{2.01\cdot 1.03}{2.01^2-1.03^2}-\frac{2}{3}=0.03，因\frac{\partial z}{\partial x}=\frac{-(y^3+x^{2}y)}{(x^2-y^2{)}^2}，\frac{\partial z}{\partial y}=\frac{x^3+x{y}^2}{(x^2-y^2{)}^2}，\frac{\partial z}{\partial x}{|}_{(2,1)}=-\frac{5}{9}，\frac{\partial z}{\partial y}{|}_{(2,1)}=\frac{10}{9}，& \\ & & \\ & 所以dz{|}_{\genfrac{}{}{0ex}{}{x=2,\Delta x=0.01}{y=1,\Delta y=0.03}}=\frac{\partial z}{\partial x}{|}_{(2,1)}\cdot \Delta x+\frac{\partial z}{\partial y}{|}_{(2,1)}\cdot \Delta y=\mathrm{0.03.}& \\ & & \end{array}$

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$\begin{array}{rlr}& 8.设f(x,y)=\{\begin{array}{l}\frac{x^2{y}^2}{(x^2+y^2{)}^{\frac{3}{2}}}，x^2+y^2\ne 0，\\ \\ 0，x^2+y^2=0.\end{array}证明:f(x,y)在点(0,0)处连续且偏导数存在，但不可微分.& \end{array}$

解：

$\begin{array}{rlr}& 因为0\le \frac{x^2{y}^2}{(x^2+y^2{)}^{\frac{3}{2}}}\le \frac{(x^2+y^2{)}^2}{(x^2+y^2{)}^{\frac{3}{2}}}=\sqrt{x^2+y^2}，\underset{(x,y)\to (0,0)}{\lim }\sqrt{x^2+y^2}=0，所以\underset{(x,y)\to (0,0)}{\lim }f(x,y)=0，& \\ & & \\ & 又因f(0,0)=0，所以\underset{(x,y)\to (0,0)}{\lim }f(x,y)=f(0,0)，即f(x,y)在点(0,0)处连续，& \\ & & \\ & f_x(0,0)=\underset{\Delta x\to 0}{\lim }\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{0}{\Delta x}=0，& \\ & & \\ & f_y(0,0)=\underset{\Delta y\to 0}{\lim }\frac{f(0+\Delta y,0)-f(0,0)}{\Delta y}=\underset{\Delta y\to 0}{\lim }\frac{0}{\Delta y}=0，& \\ & & \\ & \Delta z-[f_x(0,0)\Delta x+f_y(0,0)\Delta y]=\frac{(\Delta x{)}^2\cdot (\Delta y{)}^2}{[(\Delta x{)}^2+(\Delta y{)}^2{]}^{\frac{3}{2}}}，& \\ & & \\ & \underset{\genfrac{}{}{0ex}{}{\Delta x\to 0}{\Delta y=\Delta x}}{\lim }\frac{\frac{(\Delta x{)}^2\cdot (\Delta y{)}^2}{[(\Delta x{)}^2+(\Delta y{)}^2{]}^{\frac{3}{2}}}}{\rho }=\underset{\Delta x\to 0}{\lim }\frac{(\Delta x{)}^4}{[2(\Delta x{)}^2{]}^2}=\frac{1}{4}\ne 0，其中\rho =\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}，& \\ & & \\ & 所以f(x,y)在点(0,0)处偏导数存在，但不可微分.& \end{array}$

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$\begin{array}{rlr}& 9.设u=x^y，而x=\phi (t)，y=\psi (t)都是可微函数，求\frac{du}{dt}.& \end{array}$

解：

$\begin{array}{rlr}& \frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y}\frac{dy}{dt}=y{x}^{y-1}\cdot {\phi }^{\prime }(t)+x^{y}lnx\cdot {\psi }^{\prime }(t).& \end{array}$

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$\begin{array}{rlr}& 10.设z=f(u,v,w)具有连续偏导数，而u=\eta -\zeta ，v=\zeta -\xi ，w=\xi -\eta ，求\frac{\partial z}{\partial \xi }，\frac{\partial z}{\partial \eta }，\frac{\partial z}{\partial \zeta }.& \end{array}$

解：

$\begin{array}{rlr}& \frac{\partial z}{\partial \xi }=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial \xi }+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial \xi }+\frac{\partial z}{\partial w}\cdot \frac{\partial w}{\partial \xi }=-\frac{\partial z}{\partial v}+\frac{\partial z}{\partial w}，& \\ & & \\ & \frac{\partial z}{\partial \eta }=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial \eta }+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial \eta }+\frac{\partial z}{\partial w}\cdot \frac{\partial w}{\partial \eta }=\frac{\partial z}{\partial u}-\frac{\partial z}{\partial w}，& \\ & & \\ & \frac{\partial z}{\partial \zeta }=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial \zeta }+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial \zeta }+\frac{\partial z}{\partial w}\cdot \frac{\partial w}{\partial \zeta }=-\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}.& \end{array}$