物理应用

2022.6.30

一元函数微分学应用

物理应用
$\begin{matrix} v = \frac{d s}{d t} \\ a = \frac{d v}{d t} \end{matrix}$
相关变化率
$a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = \frac{d v}{d x} \cdot v$
几何应用
- 曲率
  $\begin{matrix} k = \frac{| y^{″} |}{(1 + (y^{'})^{2})^{\frac{3}{2}}} \\ R = f r a c 1 k \end{matrix}$

形心
$\begin{matrix} \overset{―}{x} = \frac{\iint x d x}{\iint d σ} = \frac{\int_{a}^{b} d x \int_{y_{1}}^{y_{2}} x d y}{\int_{a}^{b} d x \int_{y_{1}}^{y_{2}} d y} = \frac{\int_{a}^{b} x (y_{1} - y_{2}) d x}{\int_{a}^{b} y_{1} - y_{2} d x} \\ \overset{―}{y} = \frac{\iint y d x}{\iint d σ} = \frac{\int_{a}^{b} d x \int_{y_{1}}^{y_{2}} y d y}{\int_{a}^{b} d x \int_{y_{1}}^{y_{2}} d y} = \frac{\int_{a}^{b} 0.5 (y_{1}^{2} - y_{2}^{2}) d x}{\int_{a}^{b} y_{1} - y_{2} d x} \end{matrix}$
弧长
$\begin{matrix} d s = \sqrt{d x^{2} + d y^{2}} & , 微元 \\ s = \int_{a}^{b} \sqrt{1 + y^{' 2}} d x & , 直角坐标系 \\ s = \int_{α}^{β} \sqrt{(x_{t}^{'})^{2} + (y_{t}^{'})^{2}} d t & , 参数方程 \\ s = \int_{α}^{β} \sqrt{(r)^{2} + (r_{θ}^{'})^{2}} d θ & , 极坐标系 \end{matrix}$
表面积
$\begin{matrix} d S = 2 π R \cdot d s & , 微元 \\ d S = \int_{a}^{b} 2 π | y (x) | \sqrt{1 + y^{' 2}} d x & , 直角坐标 \\ d S = \int_{a}^{b} 2 π | y (t) | \sqrt{x_{t}^{' 2} + y_{t}^{' 2}} d t & , 参数方程 \end{matrix}$

\begin{matrix} x^{2} \frac{d^{2} y}{d x^{2}} + A x \frac{d y}{d x} + B y = f (x) \\ {\begin{cases} x = e^{t} & , x > 0 \\ x = - e^{t} & , x < 0 \end{cases} \\ {\begin{cases} x = e^{t} & , x > 0 \\ x = - e^{t} & , x < 0 \end{cases} \\ {\begin{cases} \frac{d^{2} y}{d t^{2}} + (A - 1) \frac{d y}{d t} + B y = f (e^{t}), x > 0 \\ \frac{d^{2} y}{d t^{2}} + (A - 1) \frac{d y}{d t} + B y = f (- e^{t}), x < 0 \end{cases} \end{matrix}

标准形式
$\begin{matrix} f (x) \approx S (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos \frac{n π}{l} x + b_{n} \sin \frac{n π}{l} x \\ a_{n} = \frac{1}{l} \int_{- l}^{l} f (x) \cos \frac{n π}{l} x d x \\ b_{n} = \frac{1}{l} \int_{- l}^{l} f (x) \cos \frac{n π}{l} x d x \end{matrix}$
f(x)是奇函数~正弦级数
$\begin{matrix} a_{n} = 0 \\ b_{n} = \frac{2}{l} \int_{0}^{l} f (x) \sin \frac{n π}{l} x d x \\ f (x) = \sum_{n = 1}^{\infty} b_{n} \sin \frac{n π}{l} x \end{matrix}$

f(x)是偶函数~余弦函数
$\begin{matrix} a_{n} = \frac{2}{l} \int_{0}^{l} f (x) \cos \frac{n π}{l} x d x \\ b_{n} = 0 \\ f (x) = a_{0} + \sum_{n = 1}^{\infty} a_{n} \cos \frac{n π}{l} x \end{matrix}$
奇延拓：展开成正弦级数
偶延拓：展开成余弦级数
狄利克雷收敛定理
$\begin{matrix} S (x) = {\begin{cases} f (x) & , x 为连续点 \\ \frac{f (x - 0) + f (x + 0)}{2} & , x 为间断点 \\ \frac{f (- l + 0) + f (l - 0)}{2} & , x = \pm l \end{cases} \end{matrix}$