更上一层楼,数学是基础——渐近线和可分离变量微分方程

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渐近线

(同济)《高等数学》(第六版上)第一章总习题14
【(同济)《高等数学》(第六版上)第一章总习题14】

相关知识点

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点到直线的距离

d ( M , L ) = k x y + b 1 + k 2 d(M, L) = \frac{|kx - y +b|}{\sqrt{1 + k^2}}

洛必达法则

洛必达法则是在一定条件下通过分子分母分别求导,再求极限来确定未定式值的方法。
比如: lim t 0 e t 1 t = lim t 0 e t = 1 \displaystyle \lim_{t \to 0} \frac{e^t - 1}{t} = \displaystyle \lim_{t \to 0} {e^t} = 1

证明直线为曲线的渐近线的充分必要条件

d ( M , L ) = k x y + b 1 + k 2 lim x d ( M , L ) = lim x k x y + b 1 + k 2 = 0 ( y = f ( x ) ) lim x [ k x f ( x ) + b ] = 0 b = lim x [ f ( x ) k x ] . lim x [ f ( x ) x k ] = lim x 1 x [ f ( x ) k x ] = lim x 1 x b = 0 b = 0 k = lim x f ( x ) x .   k = lim x f ( x ) x b = lim x [ f ( x ) k x ] lim x d ( M , L ) = lim x k x y + b 1 + k 2 = 0 y = k x + b 线 线 . 证明:\because d(M, L) = \frac{|kx - y +b|}{\sqrt{1 + k^2}} \\ \therefore \displaystyle \lim_{x \to \infty} d(M, L) = \lim_{x \to \infty} \frac{|kx - y +b|}{\sqrt{1 + k^2}} = 0(y = f(x)),\\ 从而,得\displaystyle \lim_{x \to \infty} {[kx - f(x) + b] = 0},\\ 即 b = \displaystyle \lim_{x \to \infty} {[f(x) - kx]}. \\ 又 \displaystyle \lim_{x \to \infty}{[\frac{f(x)}{x} - k]} = \lim_{x \to \infty}{\frac{1}{x} \cdot [f(x) - kx]} = \lim_{x \to \infty}{\frac{1}{x} \cdot b} = 0 \cdot b = 0 \\ 得k = \displaystyle \lim_{x \to \infty} \frac{f(x)}{x}.【必要性证毕】\\ \space \\ 反之,如果: k = \displaystyle \lim_{x \to \infty} \frac{f(x)}{x} 和 b = \displaystyle \lim_{x \to \infty} {[f(x) - kx]} \\ 则\displaystyle \lim_{x \to \infty} d(M, L) = \lim_{x \to \infty} \frac{|kx - y +b|}{\sqrt{1 + k^2}} = 0,\\ 从而所得y = kx + b确为曲线之渐近线. 【充分性证毕】

求曲线 y = ( 2 x 1 ) e 1 x y = (2x - 1) e^{\frac {1}{x}} 的渐近线

k = lim x f ( x ) x = lim x ( 2 x 1 ) e 1 x x = lim x ( 2 x 1 ) x lim x e 1 x = 2 b = lim x [ f ( x ) k x ] = lim x [ ( 2 x 1 ) e 1 x 2 x ] = lim x 2 x ( e 1 x 1 ) lim x e 1 x = lim t 0 2 e t 1 t 1 t = 1 x = 2 1 = 1 k = \displaystyle \lim_{x \to \infty} \frac{f(x)}{x} \\ = \lim_{x \to \infty} {\frac{(2x - 1) e^{\frac {1}{x}}}{x}} =\lim_{x \to \infty} \frac{(2x - 1)}{x} \cdot \lim_{x \to \infty} e^{\frac {1}{x}} = 2\\ b = \lim_{x \to \infty} {[f(x) - kx]}\\ = \lim_{x \to \infty} {[(2x - 1) e^{\frac {1}{x}} - 2x]}\\ = \lim_{x \to \infty} 2x \cdot (e^{\frac {1}{x}} - 1) - \lim_{x \to \infty} e^{\frac {1}{x}}\\ = \lim_{t \to 0} 2 \cdot \frac {e^t - 1}{t} - 1【替换:t = \frac {1}{x}】\\ = 2 -1 = 1
所以曲线 y = ( 2 x 1 ) e 1 x y = (2x - 1) e^{\frac {1}{x}} 的渐近线为: y = 2 x + 1 y = 2x + 1

可分离变量微分方程

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基本形式 y = f ( y , x ) y^{'} = f(y, x)

【示例: y + 3 d y d x 5 = 0 y + 3 \cdot \frac {d_y}{d_x} - 5 = 0

d y d x = 5 y 3 d y 5 y = d x 3 l n 5 y = x 3 + c y = 5 ± 1 e ( x 3 + c )   5 ± 1 e ( x 3 + c ) + 3 ( 1 3 1 e ( x 3 + c ) ) 5 = 0 解:\frac {d_y}{d_x} = \frac {5 - y}{3} \\ \frac {d_y}{5 - y} = \frac {d_x}{3} \\ 两端分别积分得: \\ -ln⁡|5 - y| = \frac {x}{3} + c \\ 结果: \\ y = 5 ± \frac{1}{e^{(\frac {x}{3} + c)}} \\ \space \\ 验算:\\ 5 ± \frac {1}{e^{(\frac {x}{3} + c) }} + 3 \cdot (∓ \frac {1}{3} \cdot \frac {1}{e^{(\frac {x}{3} + c)} }) - 5 = 0

总结:

  • 数学能提供编程的思路,是程序员的内功。
  • 能将问题用数学语言表达清楚,基本就能写出代码了。

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