《高等数学》习题6.8选做

在本节习题中所涉及的函数 $\displaystyle f,F$ 都是有一阶连续偏导数的函数。 1. 求由下列方程确定的隐函数 $\displaystyle z=z( x,y)$ 的所有一阶偏导数： (1) $\displaystyle x^{3} z+z^{3} x-2yz=0$ 解：记为 $\displaystyle f( x,y,z) =0$，那么 $\displaystyle \frac{\partial z}{\partial x} =-\frac{f_{x}}{f_{z}} =-\frac{3x^{2} z+z^{3}}{x^{3} +3z^{2} x-2y}$。而 $\displaystyle \frac{\partial z}{\partial y} =-\frac{f_{y}}{f_{z}} =\frac{2z}{x^{3} +3z^{2} x-2y}$ (3) $\displaystyle x+z-\epsilon \sin z=y\quad ( 0< \epsilon < 1)$ 解：设 $\displaystyle f( x,y,z) =x-y+z-\epsilon \sin z$，那么 $\displaystyle \frac{\partial z}{\partial x} =-\frac{f_{x}}{f_{z}} =\frac{1}{\epsilon \cos z-1}$，而 $\displaystyle \frac{\partial z}{\partial y} =-\frac{f_{y}}{f_{z}} =\frac{-1}{\epsilon \cos z-1}$。 (4) $\displaystyle z^{x} =y^{z}$ 解：设 $\displaystyle f( x,y,z) =z^{x} -y^{z}$，那么 $\displaystyle \frac{\partial z}{\partial x} =-\frac{f_{x}}{f_{z}} =\frac{-z^{x}\ln z}{xz^{x-1} -y^{z}\ln y} =\frac{-z\ln z}{x-z\ln y}$，而 $\displaystyle \frac{\partial z}{\partial y} =-\frac{f_{y}}{f_{z}} =\frac{zy^{z-1}}{xz^{x-1} -y^{z}\ln y} =\frac{z^{2}}{y( x-z\ln y)}$。 (5) $\displaystyle x\cos y+y\cos z+z\cos x=1$ 解：$\displaystyle \frac{\partial z}{\partial x} =-\frac{f_{x}}{f_{z}} =\frac{\cos y-z\sin x}{y\sin z-\cos x}$，$\displaystyle \frac{\partial z}{\partial y} =\frac{\cos z-x\sin y}{y\sin z-\cos x}$。 2. 设由方程 $\displaystyle f\left( xy^{2} ,x+y\right) =0$ 确定隐函数为 $\displaystyle y=y( x)$，求 $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$。 解：$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =-\frac{\frac{\partial }{\partial x}\left[ f\left( xy^{2} ,x+y\right)\right]}{\frac{\partial }{\partial y}\left[ f\left( xy^{2} ,x+y\right)\right]} =-\frac{y^{2} f_{x}\left( xy^{2} ,x+y\right) +f_{y}\left( xy^{2} ,x+y\right)}{2xyf_{x}\left( xy^{2} ,x+y\right) +f_{y}\left( xy^{2} ,x+y\right)}$。 6. 计算球坐标变换的 Jacobi 行列式 $\displaystyle \begin{cases} x=r\cos \theta \sin \varphi \\ y=r\sin \theta \sin \varphi \\ z=r\cos \varphi \end{cases}$。 解：Jacobi 行列式即为 $\displaystyle \begin{vmatrix} \cos \theta \sin \varphi & -r\sin \theta \sin \varphi & r\cos \theta \cos \varphi \\ \sin \theta \sin \varphi & r\cos \theta \sin \varphi & r\sin \theta \cos \varphi \\ \cos \varphi & & -r\sin \varphi \end{vmatrix}$，做消元，变为 $\displaystyle r^{2}\begin{vmatrix} \cos \theta \sin \varphi & -\sin \theta \sin \varphi & \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi & \\ \cos \varphi & & -\sin \varphi -\frac{\cos^{2} \varphi }{\sin \varphi } \end{vmatrix}$，展开得到 $\displaystyle =\frac{-r^{2}}{\sin \varphi }\begin{vmatrix} \cos \theta \sin \varphi & -\sin \theta \sin \varphi \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi \end{vmatrix} =-r^{2}\sin \varphi \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} =-r^{2}\sin \varphi $。（和标答差个符号是正常现象，这是符号排列顺序原因） 7. 设有 $\displaystyle x=u+v,y=u^{2} +v^{2} ,z=u^{3} +v^{3}$ 确定函数 $\displaystyle z=z( x,y)$，求当 $\displaystyle x=0,y=u=\frac{1}{2} ,v=-\frac{1}{2}$ 时，$\displaystyle \frac{\partial z}{\partial x}$ 与 $\displaystyle \frac{\partial z}{\partial y}$ 的值。 解：由全微分公式，$\displaystyle \mathrm{d} z=3\left( u^{2}\mathrm{d} u+v^{2}\mathrm{d} v\right)$，而 $\displaystyle \begin{cases} \mathrm{d} u+\mathrm{d} v & =\mathrm{d} x\\ 2u\mathrm{d} u+2v\mathrm{d} v & =\mathrm{d} y \end{cases}$，有 $\displaystyle \begin{cases} \mathrm{d} u & =\frac{\mathrm{d} x+\mathrm{d} y}{2}\\ \mathrm{d} v & =\frac{\mathrm{d} x-\mathrm{d} y}{2} \end{cases}$，可得 $\displaystyle \mathrm{d} z=\frac{3}{4}\mathrm{d} x$，因此 $\displaystyle \frac{\partial z}{\partial x} =\frac{3}{4} ,\frac{\partial z}{\partial y} =0$。 11. 设 $\displaystyle u=u( x,y)$ 及 $\displaystyle v=v( x,y)$ 有一阶连续偏导数，又设 $\displaystyle x=x( \xi ,\eta )$ 及 $\displaystyle y=y( \xi ,\eta )$ 也有一阶连续偏导数，且使复合函数 $\displaystyle u=u( x( \xi ,\eta ) ,y( \xi ,\eta )) ,v=v( x( \xi ,\eta ) ,y( \xi ,\eta ))$ 有定义。证明：$\displaystyle \frac{\operatorname{D}( u,v)}{\operatorname{D}( \xi ,\eta )} =\frac{\operatorname{D}( u,v)}{\operatorname{D}( x,y)} \cdotp \frac{\operatorname{D}( x,y)}{\operatorname{D}( \xi ,\eta )}$。 解：由 $\displaystyle \mathrm{d} u=u_{x}\mathrm{d} x+u_{y}\mathrm{d} y=u_{x}( x_{\xi }\mathrm{d} \xi +x_{\eta }\mathrm{d} \eta ) +u_{y}( y_{\xi }\mathrm{d} \xi +y_{\eta }\mathrm{d} \eta )$……可以得到 $$ \begin{bmatrix} \frac{\partial u}{\partial \xi } & \frac{\partial u}{\partial \eta }\\ \frac{\partial v}{\partial \xi } & \frac{\partial v}{\partial \eta } \end{bmatrix} =\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} \cdotp \begin{bmatrix} \frac{\partial x}{\partial \xi } & \frac{\partial x}{\partial \eta }\\ \frac{\partial y}{\partial \xi } & \frac{\partial y}{\partial \eta } \end{bmatrix} $$ 可以两边取行列式，既得 $\displaystyle \frac{\operatorname{D}( u,v)}{\operatorname{D}( \xi ,\eta )} =\frac{\operatorname{D}( u,v)}{\operatorname{D}( x,y)} \cdotp \frac{\operatorname{D}( x,y)}{\operatorname{D}( \xi ,\eta )}$。