京都大学 情報学研究科 知能情報学専攻 2024年2月実施 基礎科目 F1-2

标签：

• Kyoto-University
• Continuity-and-Differentiability
• Multivariable-Differentiation
• Double-Integral

Author

itsuitsuki, 祭音Myyura

Description

Q.1

Consider the following functions $f_1(x)$ and $f_2(x)$.

$$f_1(x) = \begin{cases} x\sin\left(\frac{1}{x}\right) & (x \neq 0) \\ 0 & (x = 0) \end{cases}$$

$$f_2(x) = \begin{cases} x^2\sin\left(\frac{1}{x}\right) & (x \neq 0) \\ 0 & (x = 0) \end{cases}$$

(1) Evaluate whether $f_1(x)$ and $f_2(x)$ are differentiable, respectively.

(2) Evaluate whether their derivatives ${f_1}'(x)$ and ${f_2}'(x)$ are continuous at $x = 0$, respectively.

Q.2

Evaluate the directional differentiability and continuity of the following function.

$$f(x, y) = \begin{cases} \frac{xy^2}{x^2 + y^4} & (\text{Either } x \neq 0 \text{ or } y \neq 0) \\ 0 & (\text{Both } x = 0 \text{ and } y = 0) \end{cases}$$

Q.3

Compute the volume in the $xyz$-space for each of the following conditions.

(1) Under the surface $z = e^x \cos(y)$ and over the rectangle defined by $D = [0, 1] \times \left[0, \frac{\pi}{2}\right]$ on the $xy$-plane.

(2) Under the surface of the paraboloid $z = 1 - x^2 - y^2$ and over the $xy$-plane.

题目描述

1. 定义
   $$f_1(x)=\begin{cases}x\sin(1/x),&x\ne0\\0,&x=0,\end{cases} \qquad f_2(x)=\begin{cases}x^2\sin(1/x),&x\ne0\\0,&x=0.\end{cases}$$
   1. 分别判断 $f_1,f_2$ 是否可微；
   2. 分别判断导函数 $f_1',f_2'$ 在 $x=0$ 是否连续。
2. 对
   $$f(x,y)=\begin{cases} \dfrac{xy^2}{x^2+y^4},&(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0), \end{cases}$$
   判断方向可微性与连续性。
3. 求下列 $xyz$ 空间体积：
   1. $xy$ 平面矩形 $D=[0,1]\times[0,\pi/2]$ 上、曲面 $z=e^x\cos y$ 下方的体积；
   2. 抛物面 $z=1-x^2-y^2$ 与 $xy$ 平面之间的体积。

考点

• 连续性、可微性与导函数连续性：用差商和振荡项分别判断零点行为，区分可微与 $C^1$。
• 多元方向导数不蕴含连续：沿直线计算原点各方向导数，再用非线性路径检验二元极限。
• 二重积分求体积：在矩形域直接分离积分，在圆盘域用极坐标积分。

Kai

Q.1

(1)

$$\lim_{h\to 0}{f_1(h)-f_1(0)\over h}=\lim_{h\to 0}{h\sin(\frac1h)\over h}=\lim_{h\to 0}{\sin(\frac1h)}$$

does not exist. Hence $f_1$ is not differentiable.

$$\lim_{h\to 0}{f_2(h)-f_2(0)\over h}=\lim_{h\to 0}{h^2\sin(\frac1h)\over h}=\lim_{h\to 0}h\sin(\frac1h)=0$$

by Squeezing theorem. Hence $f_2$ is differentiable.

(2)

At $x=0$, $f_1'(x)$ does not exist, so it is not continuous.

From (1), $f'_2(0)=0$ and $\lim_{x\to 0}f_2'(x)=\lim_{x\to 0}(2x\sin{\frac1x}-\cos{\frac1x})$ does not exist, so $f'_2$ is also not continuous at $x=0$.

Q.2

We have $\lim_{h\to0}{f(h,0)-f(0,0)\over h}=0$ and $\lim_{h\to0}{f(0,h)-f(0,0)\over h}=0$. So $f$ is directionally differentiable at $x$ and $y$ axes.

By

$$\lim_{x=y^2,x\to 0}{xy^2\over x^2+y^4}=\frac12,$$

and

$$\lim_{x=0,y\to 0}{xy^2\over x^2+y^4}=0\ne \frac12,$$

limit of $f$ along different paths to $(0,0)$ is different, which means the limit does not exist. Hence $f$ is not continuous.

Q.3

(1)

$$\begin{aligned} V&=\int_0^1\int_0^{\pi/2}e^x\cos (y) dy dx \\ &=\int_0^1e^x dx\int_0^{\pi/2}\cos y \\ &=e-1. \end{aligned}$$

(2)

The partial surface to be integrated is

$$S=\{(x,y)|x^2+y^2\le 1\}.$$

Let $x=r\cos \theta,y=r\sin\theta$, $r\in[0,1]$ and $\theta\in[0,2\pi)$.

$$\begin{aligned} V&=\iint_{(x,y)\in S}zd xd y\\ &=\iint_{(r,\theta)\in S}(1-r^2)|J|d rd \theta \\&=\int_0^{2\pi}\int_0^1(1-r^2)rd rd \theta \\&=2\pi \int_0^1(r-r^3)d r \\&=2\pi (\frac12-\frac14)={\pi\over 2}. \end{aligned}$$