京都大学 情報学研究科 通信情報システム専攻 2022年8月実施 専門基礎A [A-1]

标签：

• 京都大学
• 局部极值
• 二重积分
• 弧长
• 特征值与特征向量
• 内积与正交性

Author

SUN, 祭音Myyura

Description

大学公表の原題 Answer all the following questions.

(1)

Find all the local maxima and minima, and corresponding $x$ and $y$ with respect to function $f(x,y)$. Let $x$ and $y$ be real numbers.

$$f(x,y)=x^3-x^2y+xy^2-x$$

(2)

Let $D$ be a domain bounded by $y=\sqrt{x}$ and $y=x$, where $x\ge 0$. Compute the following integral $I$.

$$I=\iint_D e^{-y}\,dx\,dy$$

(3)

Find the length of the curve given as follows.

$$y=x^{3/2}\quad \left(0\le x\le \frac{4}{3}\right)$$

(4)

Find the eigenvectors of matrix $A$, and show the conditions on which they become orthogonal to each other. Let $x$ be a real number.

$$A=\begin{pmatrix} x & a\\ b & x \end{pmatrix}$$

题目描述

回答全部问题。

1. 对实数 $x,y$，求函数

   $$f(x,y)=x^3-x^2y+xy^2-x$$

   的所有局部极大值、局部极小值及其对应的 $(x,y)$。

2. 设 $D$ 为 $x\ge0$ 时曲线 $y=\sqrt x$ 与 $y=x$ 围成的区域，计算

   $$I=\iint_D e^{-y}\,dx\,dy.$$

3. 求曲线

   $$y=x^{3/2},\qquad 0\le x\le\frac43$$

   的弧长。

4. 对矩阵

   $$A=\begin{pmatrix}x&a\\b&x\end{pmatrix},$$

   求其特征向量，并给出这些特征向量彼此正交的条件，其中 $x$ 为实数。

Kai

(1)

To find critical points, set partial derivatives to zero:

$$\begin{cases} f_x = 3x^2 - 2yx + y^2 - 1 = 0 \\ f_y = -x^2 + 2xy = x(2y - x) = 0 \end{cases}$$

From $f_y = 0$, we have $x = 0$ or $x = 2y$.

• If $x = 0$, then $y^2 - 1 = 0 \Rightarrow y = \pm 1$. Points: $(0, 1), (0, -1)$.
• If $x = 2y$, then $3(2y)^2 - 2y(2y) + y^2 - 1 = 0 \Rightarrow 9y^2 = 1 \Rightarrow y = \pm \frac{1}{3}$. Points: $(\frac{2}{3}, \frac{1}{3}), (-\frac{2}{3}, -\frac{1}{3})$.

Second-order partial derivatives:

$$f_{xx} = 6x - 2y, \quad f_{xy} = -2x + 2y, \quad f_{yy} = 2x$$

Hessian determinant $H = f_{xx}f_{yy} - (f_{xy})^2$:

• For $(0, 1): H = ( -2)(0) - (2)^2 = -4 < 0$ (Saddle point)
• For $(0, -1): H = (2)(0) - (-2)^2 = -4 < 0$ (Saddle point)
• For $(\frac{2}{3}, \frac{1}{3}): H = (\frac{10}{3})(\frac{4}{3}) - (-\frac{2}{3})^2 = 4 > 0$. Since $f_{xx} = \frac{10}{3} > 0$, it is a local minimum:

$$f(\frac{2}{3}, \frac{1}{3}) = (\frac{2}{3})^3 - (\frac{2}{3})^2(\frac{1}{3}) + (\frac{2}{3})(\frac{1}{3})^2 - \frac{2}{3} = -\frac{4}{9}$$

• For $(-\frac{2}{3}, -\frac{1}{3}): H = (-\frac{10}{3})(-\frac{4}{3}) - (\frac{2}{3})^2 = 4 > 0$. Since $f_{xx} = -\frac{10}{3} < 0$, it is a local maximum:

$$\begin{aligned} f(-\frac{2}{3}, -\frac{1}{3}) &= (-\frac{2}{3})^3 - (-\frac{2}{3})^2(-\frac{1}{3}) \\ &+ (-\frac{2}{3})(-\frac{1}{3})^2 - (-\frac{2}{3}) \\ &= \frac{4}{9} \end{aligned}$$

(2)

The domain $D$ is bounded by $y = \sqrt{x}$ and $y = x$, which intersect at $(0, 0)$ and $(1, 1)$. For $0 \le y \le 1$, the range of $x$ is $y^2 \le x \le y$.

$$I = \iint_D e^{-y} dA = \int_0^1 \int_{y^2}^y e^{-y} dx dy = \int_0^1 (y - y^2) e^{-y} dy$$

Using integration by parts:

$$\begin{aligned} \int (y - y^2) e^{-y} dy &= -(y - y^2)e^{-y} + \int (1 - 2y) e^{-y} dy\\ &= (y^2 - y)e^{-y} - (1 - 2y)e^{-y} + \int (-2) e^{-y} dy\\ &= (y^2 - y - 1 + 2y + 2)e^{-y} \\ &= (y^2 + y + 1)e^{-y} \end{aligned}$$

Evaluating from 0 to 1:

$$I = \left[ (y^2 + y + 1)e^{-y} \right]_0^1 = 3e^{-1} - 1$$

(3)

The arc length $L$ for $y = x^{3/2}$ from $x=0$ to $x=4/3$:

$$y' = \frac{3}{2}x^{1/2} \Rightarrow 1 + (y')^2 = 1 + \frac{9}{4}x$$

$$L = \int_0^{4/3} \sqrt{1 + \frac{9}{4}x} dx$$

Let $u = 1 + \frac{9}{4}x$, then $du = \frac{9}{4}dx$. When $x=0, u=1$; when $x=4/3, u=4$.

$$\begin{aligned} L &= \int_1^4 \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \left[ \frac{2}{3}u^{3/2} \right]_1^4 \\ &= \frac{8}{27}(4^{3/2} - 1^{3/2}) = \frac{8}{27}(8 - 1) \\ &= \frac{56}{27} \end{aligned}$$

(4)

The characteristic equation is

$$\det(A-\lambda I) =\begin{vmatrix}x-\lambda&a\\b&x-\lambda\end{vmatrix} =(x-\lambda)^2-ab=0.$$

If $ab\ne0$, put $s=\sqrt{ab}$ (over $\mathbb C$). Then

$$\lambda_\pm=x\pm s.$$

For $\lambda_+=x+s$,

$$\begin{pmatrix}-s&a\\b&-s\end{pmatrix} \begin{pmatrix}v_1\\v_2\end{pmatrix}=0, \qquad -sv_1+av_2=0,$$

and for $\lambda_-=x-s$,

$$\begin{pmatrix}s&a\\b&s\end{pmatrix} \begin{pmatrix}v_1\\v_2\end{pmatrix}=0, \qquad sv_1+av_2=0.$$

Therefore

$$E_{\lambda_+}=\operatorname{span}\left\{\begin{bmatrix}a\\s\end{bmatrix}\right\}, \qquad E_{\lambda_-}=\operatorname{span}\left\{\begin{bmatrix}a\\-s\end{bmatrix}\right\}.$$

If $a=b=0$, the eigenspace for $\lambda=x$ is $\mathbb R^2$. If $ab=0$ but $(a,b)\ne(0,0)$, the only eigenvalue is $x$ and its eigenspace is $\operatorname{span}\{(1,0)^T\}$ for $a\ne0$, or $\operatorname{span}\{(0,1)^T\}$ for $b\ne0$.

For two real eigenvectors we need $ab>0$. Their inner product is

$$\begin{bmatrix}a\\s\end{bmatrix}^{\!T} \begin{bmatrix}a\\-s\end{bmatrix} =a^2-ab=a(a-b),$$

so they are orthogonal exactly when $a=b\ne0$. When $a=b=0$, any orthogonal basis is an eigenbasis. Hence $A$ has an orthogonal real eigenbasis if and only if $a=b$.

If orthogonality is taken over $\mathbb C$ with the Hermitian inner product, the condition is instead $|a|=|b|$. Indeed, for $ab\ne0$ the displayed eigenvectors satisfy

$$\begin{bmatrix}a\\s\end{bmatrix}^{\!*} \begin{bmatrix}a\\-s\end{bmatrix} =|a|^2-|s|^2=|a|^2-|ab|,$$

which vanishes exactly when $|a|=|b|$. This also includes the scalar case $a=b=0$; if exactly one is zero, there is no eigenbasis. In particular, for real $a,b$ with $a=-b\ne0$, the eigenvalues are nonreal but their eigenvectors are Hermitian-orthogonal. The condition $a=b$ above is specifically for a real orthogonal eigenbasis.