《高等数学》习题6.5选做
Elegia
2021-08-08 00:02:28
在下面的习题中,出现的函数 $\displaystyle f( x,y)$ 或 $\displaystyle F( u)$ 一律假定有连续的一阶偏导数或导数。
1. 求下列复合函数的偏导数或导数:
(1) $\displaystyle z=\sqrt{u^{2} +v^{2}} ,u=xy,v=y^{2}$。
解:
$$
\begin{aligned}
\mathrm{d} z & =\frac{u\mathrm{d} u+v\mathrm{d} v}{\sqrt{u^{2} +v^{2}}}\\
& =\frac{u( y\mathrm{d} x+x\mathrm{d} y) +2vy\mathrm{d} y}{\sqrt{u^{2} +v^{2}}}\\
& =\frac{xy( y\mathrm{d} x+x\mathrm{d} y) +2y^{3}\mathrm{d} y}{| y| \sqrt{x^{2} +y^{2}}}\\
& =\frac{xy^{2}\mathrm{d} x+\left( x^{2} +2y^{2}\right) y\mathrm{d} y}{| y| \sqrt{x^{2} +y^{2}}}\\
& =\frac{x| y| }{\sqrt{x^{2} +y^{2}}}\mathrm{d} x+\frac{\left( x^{2} +2y^{2}\right)\operatorname{sgn} y}{\sqrt{x^{2} +y^{2}}}\mathrm{d} y
\end{aligned}
$$
(2) $\displaystyle z=\frac{u^{2}}{v}$,其中 $\displaystyle u=y\mathrm{e}^{x} ,v=x\ln y$
解:
$$
\begin{aligned}
\mathrm{d} z & =\frac{2uv\mathrm{d} u-u^{2}\mathrm{d} v}{v^{2}}\\
& =\frac{2uv\left(\mathrm{e}^{x}\mathrm{d} y+y\mathrm{e}^{x}\mathrm{d} x\right) -u^{2}\left(\ln y\mathrm{d} x+\frac{x}{y}\mathrm{d} y\right)}{v^{2}}\\
& =\frac{( 2x-1) y^{2}\mathrm{e}^{2x}\mathrm{d} x+( 2x-y) y\mathrm{e}^{2x}\mathrm{d} x}{x^{2}\ln y}
\end{aligned}
$$
(5) $\displaystyle z=f\left( x^{2} -y^{2} ,\mathrm{e}^{xy}\right)$
解:
$$
\begin{aligned}
\mathrm{d} z & =f_{x} \cdot \mathrm{d}\left( x^{2} -y^{2}\right) +f_{y} \cdot \mathrm{d}\left(\mathrm{e}^{xy}\right)\\
& =2f_{x}( x\mathrm{d} x-y\mathrm{d} y) +f_{y} \cdot \mathrm{e}^{xy}( y\mathrm{d} x+x\mathrm{d} y)\\
& =\left( 2xf_{x} +y\mathrm{e}^{xy} f_{y}\right)\mathrm{d} x+\left( -2yf_{x} +x\mathrm{e}^{xy} f_{y}\right)\mathrm{d} y
\end{aligned}
$$
2. 设 $\displaystyle u=f\left( x+y+z,x^{2} +y^{2} +z^{2}\right)$,求 $\displaystyle \Delta u$
解:设 $\displaystyle \alpha =x+y+z,\beta =x^{2} +y^{2} +z^{2}$,那么
$$
\begin{aligned}
\frac{\partial ^{2} u}{\partial x^{2}} & =\frac{\partial }{\partial x}( f_{\alpha }( \alpha ,\beta ) +f_{\beta }( \alpha ,\beta ) \cdot 2x)\\
& =f_{\alpha \alpha }( \alpha ,\beta ) +( f_{\alpha \beta }( \alpha ,\beta ) +f_{\beta \alpha }) \cdot ( 1+x) +f_{\beta \beta }( \alpha ,\beta ) \cdot 4x^{2} +2f_{\beta }( \alpha ,\beta )
\end{aligned}
$$
因此 $\displaystyle \Delta u=3f_{\alpha \alpha } +( f_{\alpha \beta } +f_{\beta \alpha }) \cdot ( 3+x+y+z) +4f_{\beta \beta } \cdot \left( x^{2} +y^{2} +z^{2}\right) +6f_{\beta }$。
3. 设 $\displaystyle u=f( \xi ,\eta )$,其中 $\displaystyle \xi =\mathrm{e}^{x}\cos y,\eta =\mathrm{e}^{x}\sin y$,求 $\displaystyle \frac{\partial ^{2} u}{\partial x^{2}}$ 与 $\displaystyle \frac{\partial ^{2} u}{\partial y^{2}}$。
解:$\displaystyle \mathrm{d} u=f_{x} \cdot \mathrm{d} \xi +f_{y} \cdot \mathrm{d} \eta =f_{x} \cdot ( \xi \mathrm{d} x-\eta \mathrm{d} y) +f_{y} \cdot ( \eta \mathrm{d} x+\xi \mathrm{d} y)$。所以 $\displaystyle \frac{\partial ^{2} u}{\partial x^{2}} =( f_{x} \cdot \xi +f_{y} \cdot \eta ) '=\left( f_{xx} \cdot \xi ^{2} +( f_{xy} +f_{yx}) \cdot \xi \eta +f_{yy} \cdot \eta ^{2}\right) +f_{x} \cdot \xi +f_{y} \cdot \eta $,而 $\displaystyle \frac{\partial ^{2} u}{\partial y^{2}} =( -f_{x} \cdot \eta +f_{y} \cdot \xi ) '=\left( f_{xx} \cdot \eta ^{2} -( f_{xy} +f_{yx}) \xi \eta +f_{yy} \cdot \xi ^{2}\right) -f_{x} \cdot \xi -f_{y} \cdot \eta $。
6. 设函数 $\displaystyle u( x,y)$ 有二阶连续偏导且满足 Laplace 方程:$\displaystyle \frac{\partial ^{2} u}{\partial x^{2}} +\frac{\partial ^{2} u}{\partial y^{2}} =0$,证明:做换元 $\displaystyle x=\mathrm{e}^{s}\cos t,y=\mathrm{e}^{s}\sin t$ 后,满足关于 $\displaystyle s,t$ 的 Laplace 方程 $\displaystyle \frac{\partial ^{2} u}{\partial s^{2}} +\frac{\partial ^{2} u}{\partial t^{2}} =0$。
解:利用例题 3 中的结果,相加得到
$$
\begin{aligned}
\frac{\partial ^{2} u}{\partial s^{2}} +\frac{\partial ^{2} u}{\partial t^{2}} & =f_{xx} \cdot \left( \xi ^{2} +\eta ^{2}\right) +f_{yy} \cdot \left( \eta ^{2} +\xi ^{2}\right)\\
& =\mathrm{e}^{2s}( f_{xx} +f_{yy})\\
& =\mathrm{e}^{2s} \cdot 0\\
& =0
\end{aligned}
$$
9. 设 $\displaystyle z=f( x,y)$ 在一个平面区域 $\displaystyle D$ 内有定义。假定 $\displaystyle D$ 对于任意一点 $\displaystyle ( x_{0} ,y_{0})$,直线 $\displaystyle y=y_{0}$ 与 $\displaystyle D$ 的交是一个开区间,且 $\displaystyle z$ 在区域 $\displaystyle D$ 中有连续的一阶偏导数。若对 $\displaystyle x$ 的偏导数恒为 $\displaystyle 0$,证明存在 $\displaystyle F( y)$ 使得 $\displaystyle f( x,y) =F( y)$。
解:考虑 $\displaystyle y=y_{0}$ 交 $\displaystyle D$ 于 $\displaystyle ( x_{l} ,x_{r}) \times \{y_{0}\}$,取 $\displaystyle x_{m} =\frac{x_{l} +x_{r}}{2}$,那么根据中值定理有 $\displaystyle f( x,y_{0}) =f( x_{m} ,y_{0}) +f_{x}( \xi ,y_{0})( x-x_{m}) =f( x_{m} ,y_{0})$,因此对每个 $\displaystyle y_{0}$,取 $\displaystyle F( y_{0}) =f( x_{m} ,y_{0})$ 即为所求。