下册 7.5 多元函数微分的应用 第37题

\section*{解题过程：} （1）令 $\displaystyle F(x, y)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1$ ，则 $\displaystyle F_{x}=\frac{2 x}{a^{2}}, F_{y}=\frac{2 y}{b^{2}}$ ．函数点 $(x, y)$ 处的法向量为 $$ n= \pm\left(F_{x}, F_{y}\right)= \pm\left(\frac{2 x}{a^{2}}, \frac{2 y}{b^{2}}\right) $$ 所以函数在点 $\displaystyle \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ 处的内法向量为 $\displaystyle \boldsymbol{n}=-\left(\frac{\sqrt{2}}{a}, \frac{\sqrt{2}}{b}\right)$ ，单位内法向量为 $$ e_{n}=\left(-\frac{b}{\sqrt{a^{2}+b^{2}}},-\frac{a}{\sqrt{a^{2}+b^{2}}}\right)=(\cos \alpha, \cos \beta) . $$ 又因为 $\displaystyle \left.\frac{\partial f}{\partial x}\right|_{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}=-\left.\frac{2 x}{a^{2}}\right|_{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}=-\frac{\sqrt{2}}{a},\left.\frac{\partial f}{\partial y}\right|_{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}=-\left.\frac{2 y}{b^{2}}\right|_{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}=-\frac{\sqrt{2}}{b}$ ，所以 $$ \frac{\partial f}{\partial n}=\frac{\partial f}{\partial x} \cos \alpha+\frac{\partial f}{\partial y} \cos \beta=\frac{\sqrt{2}}{a} \cdot \frac{b}{\sqrt{a^{2}+b^{2}}}+\frac{\sqrt{2}}{b} \cdot \frac{a}{\sqrt{a^{2}+b^{2}}}=\frac{\sqrt{2}}{a b} \sqrt{a^{2}+b^{2}} . $$ （2）对可微函数，梯度方向是函数变化最快的方向． $$ f_{x}(x, y)=3 x^{2}-3 y, f_{y}(x, y)=3 y^{2}-3 x . $$ 设 $P\left(x_{0}, y_{0}\right)$ 为该曲线上的点，在该点的梯度为 $\left(3 x_{0}{ }^{2}-3 y_{0}, 3 y_{0}{ }^{2}-3 x_{0}\right)$ ，方向余弦为 $\displaystyle \left(\frac{x_{0}{ }^{2}-y_{0}}{r}, \frac{y_{0}{ }^{2}-x_{0}}{r}\right)$ ，其中 $r=\sqrt{\left(x_{0}{ }^{2}-y_{0}\right)^{2}+\left(y_{0}{ }^{2}-x_{0}\right)^{2}}$ 。函数在梯度方向的方向导数为 $$ \begin{aligned} \frac{\partial f}{\partial l} & =f_{x}\left(P_{0}\right) \cos \alpha+f_{y}\left(P_{0}\right) \cos \beta=3\left(x_{0}^{2}-y_{0}\right) \frac{x_{0}^{2}-y_{0}}{r}+3\left(y_{0}^{2}-x_{0}\right) \frac{y_{0}^{2}-x_{0}}{r} \\ & =3 \sqrt{\left(x_{0}^{2}-y_{0}\right)^{2}+\left(y_{0}^{2}-x_{0}\right)^{2}} . \end{aligned} $$ （3）求 $\displaystyle z=\arctan \frac{y}{x}$ 的一阶偏导数 $\displaystyle z_{x}=\frac{-y}{x^{2}+y^{2}}, z_{y}=\frac{x}{x^{2}+y^{2}}$ ． 圆 $x^{2}+y^{2}-2 x=0$ 在点 $\displaystyle P_{0}\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ 处的切线的斜率 $\displaystyle k=y_{x}\left(P_{0}\right)=\frac{1}{\sqrt{2}}$ ，切线的方向向量为 $\displaystyle \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ ．于是所求的方向导数为 $\displaystyle z_{l}=z_{x}\left(P_{0}\right) \frac{1}{\sqrt{2}}+z_{y}\left(P_{0}\right) \frac{1}{\sqrt{2}}=\frac{4}{3} \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right)=\frac{2-\sqrt{2}}{3}$ ． （4）由于 $\displaystyle \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x} \cos \alpha+\frac{\partial z}{\partial y} \sin \alpha=(2 x-y) \cos \alpha+(2 y-x) \sin \alpha$ ，所以 $$ \left.\frac{\partial z}{\partial v}\right|_{(1,1)}=\cos \alpha+\sin \alpha=\sin \left(\frac{\pi}{2}-\alpha\right)+\sin \alpha=2 \sin \frac{\pi}{4} \cos \left(\frac{\pi}{4}-\alpha\right) . $$ （I）当 $\displaystyle \alpha=\frac{\pi}{4}$ 时，沿 $\displaystyle \nu=\left(\cos \frac{\pi}{4}, \sin \frac{\pi}{4}\right)$ ，方向导数最大； （II）当 $\displaystyle \alpha=\frac{5 \pi}{4}$ 时，沿 $\displaystyle v=\left(\cos \frac{5 \pi}{4}, \sin \frac{5 \pi}{4}\right)$ ，方向导数最小。 （5）函数的一阶偏导数 $u_{x}=2 x, u_{y}=-2 z+2 y, u_{z}=-2 y$ ．于是函数在点 $M(-1,2,1)$ 处的梯度为 $(-2,2,-4)$ ． 对可微函数，梯度方向是函数变化最快的方向，故沿梯度方向的方向导数达到的最大值为 $|\operatorname{grad} \boldsymbol{u}|=2 \sqrt{6}$ ． （6）由于 $\displaystyle z_{x}=-\frac{F_{x}}{F_{z}}=-\frac{y z F_{1}+2 x F_{2}}{x y F_{1}+2 z F_{2}}, z_{y}=-\frac{F_{y}}{F_{z}}=-\frac{z x F_{1}+2 y F_{2}}{x y F_{1}+2 z F_{2}}$ ，所以 $\operatorname{grad} z=\left(z_{x}, z_{y}\right)$ ．