数学 - Python - JavaScript - 解析学 - 重積分 - 一般の集合の上の積分(双曲線、面積)

学習環境

解析入門(下) (松坂和夫数学入門シリーズ 6) (松坂和夫(著)、岩波書店)の第23章(重積分)、23.2(一般の集合の上の積分)、問題4.を取り組んでみる。

直線の方程式。

$x = \frac{x_{0}}{y_{0}} y$

双曲線の方程式。

$\begin{array}{l} x^{2} = (1 + \frac{y^{2}}{b^{2}}) a^{2} \\ x^{2} = \frac{(b^{2} + y^{2}) a^{2}}{b^{2}} \\ x = \frac{a}{b} \sqrt{b^{2} + y^{2}} \end{array}$

求める第1象限の部分についての面積。

$\begin{array}{l} \int_{0}^{y_{0}} (\frac{a}{b} \sqrt{b^{2} + y^{2}} - \frac{y_{0}}{x_{0}}) dy \\ = \frac{a}{b} \int_{0}^{y_{0}} \sqrt{b^{2} + y^{2}} dy - \frac{y_{0}}{y_{0}} \int_{0}^{y_{0}} 1 dy \end{array}$

1つ目の不積分について計算。

$\begin{array}{l} t = y + \sqrt{b^{2} + y^{2}} \\ t - y = \sqrt{b^{2} + y^{2}} \\ t^{2} - 2 t y + y^{2} = b^{2} + y^{2} \\ y = \frac{t^{2} - b^{2}}{2 t} = \frac{1}{2} (t - \frac{b^{2}}{t}) \\ \frac{dy}{dt} = \frac{1}{2} (1 + \frac{b^{2}}{t^{2}}) \\ \int \sqrt{b^{2} + y^{2}} dy \\ = \int \sqrt{b^{2} + \frac{1}{4} {(t - \frac{b^{2}}{t})}^{2}} \cdot \frac{1}{2} (1 + \frac{b^{2}}{t^{2}}) dt \\ = \frac{1}{4} \int \sqrt{4 b^{2} + {(t - \frac{b^{2}}{t})}^{2}} (1 + \frac{b^{2}}{t^{2}}) dt \\ = \frac{1}{4} \int \sqrt{4 b^{2} + t^{2} - 2 b^{2} + \frac{b^{4}}{t^{2}}} (1 + \frac{b^{2}}{t^{2}}) dt \\ = \frac{1}{4} \int \sqrt{t^{2} + 2 b^{2} + \frac{b^{4}}{t^{2}}} (1 + \frac{b^{2}}{t^{2}}) dt \\ = \frac{1}{4} \int (t + \frac{b^{2}}{t}) (1 + \frac{b^{2}}{t^{2}}) dt \\ = \frac{1}{4} \int (t + \frac{2 b^{2}}{t} + \frac{b^{4}}{t^{3}}) dt \\ = \frac{1}{4} (\frac{1}{2} t^{2} + 2 b^{2} \log t - \frac{b^{4}}{2 t^{2}}) \\ = \frac{1}{2} (\frac{1}{4} (t - \frac{b^{2}}{t}) (t + \frac{b^{2}}{t}) + b^{2} \log t) \end{array}$

1つ目の計算。

$\begin{array}{l} \frac{1}{4} (y + \sqrt{b^{2} + y^{2}} - \frac{b^{2}}{y + \sqrt{b^{2} + y^{2}}}) (y + \sqrt{b^{2} + y^{2}} + \frac{b^{2}}{y + \sqrt{b^{2} + y^{2}}}) \\ = \frac{1}{4} \frac{(2 y^{2} + 2 y \sqrt{b^{2} + y^{2}}) (2 y^{2} + 2 b^{2} + 2 y \sqrt{b^{2} + y^{2}})}{{(y + \sqrt{b^{2} + y^{2}})}^{2}} \\ = \frac{(y^{2} + y \sqrt{b^{2} + y^{2}}) (y^{2} + b^{2} + y \sqrt{b^{2} + y^{2}})}{{(y + \sqrt{b^{2} + y^{2}})}^{2}} \\ = \frac{y^{2} (y^{2} + b^{2}) + (y^{3} + y (y^{2} + b^{2})) \sqrt{b^{2} + y^{2}} + y^{2} (b^{2} + y^{2})}{{(y + \sqrt{b^{2} + y^{2}})}^{2}} \\ = \frac{y \sqrt{y^{2} + b^{2}} (y \sqrt{y^{2} + b^{2}} T^{2} Y^{2} + b^{2} + y \sqrt{b^{2} + y^{2}})}{{(y + \sqrt{b^{2} + y^{2}})}^{2}} \\ = y \sqrt{y^{2} + b^{2}} \end{array}$

よって、1つ目の定積分は、

$\begin{array}{l} \frac{a}{b} \frac{1}{2} {[y \sqrt{y^{2} + b^{2}} + b^{2} \log (y + \sqrt{b^{2} + y^{2}})]}_{0}^{y_{0}} \\ = \frac{a}{b} \frac{1}{2} (y_{0} \sqrt{y_{0}^{2} + b^{2}} + b^{2} \log (y_{0} + {\sqrt{b^{2} + y^{2}}}_{0}) - b^{2} \log b) \\ = \frac{a}{2 b} (y_{0} \sqrt{y_{0}^{2} + b^{2}} + b^{2} \log \frac{y_{0} + \sqrt{b^{2} + y_{0}^{2}}}{b}) \\ = \frac{a y_{0}}{2 b} \sqrt{y_{0}^{2} + b^{2}} + \frac{a b}{2} \log \frac{y_{0} + \sqrt{b^{2} + y_{0}^{2}}}{b} \end{array}$

よって、求める双曲線の弧で囲まれた図形の面積は、

$\begin{array}{l} \frac{a y_{0}}{2 b} \sqrt{y_{0}^{2} + b^{2}} + \frac{a b}{2} \log \frac{y_{0} + {\sqrt{b^{2} + y^{2}}}_{0}}{b} - \frac{x_{0}}{y_{0}} {[\frac{1}{2} y^{2}]}_{0}^{y_{0}} \\ = \frac{a y_{0}}{2 b} \sqrt{y_{0}^{2} + b^{2}} + \frac{a b}{2} \log \frac{y_{0} + \sqrt{b^{2} + y_{0}^{2}}}{b} - \frac{x_{0} y_{0}}{2} \\ \frac{x_{0}^{2}}{a^{2}} - \frac{y_{0}^{2}}{b^{2}} = 1 \\ \frac{x_{0}^{2} b^{2}}{a^{2}} = b^{2} + y_{0}^{2} \\ {\sqrt{b^{2} + y^{2}}}_{0} = \frac{x_{0} b}{a} \\ \frac{y_{0} + \sqrt{b^{2} + y_{0}^{2}}}{b} \\ = \frac{y_{0}}{b} + \sqrt{y + \frac{y_{0}^{2}}{b^{2}}} \\ = \frac{y_{0}}{b} + \frac{x_{0}}{a} \end{array}$

ゆえに、

$\frac{a b}{2} \log (\frac{x_{0}}{a} + \frac{y_{0}}{b})$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, sqrt, Integral

print('4.')

a, b, x, y, x0, y0 = symbols('a, b, x, y, x0, y0', positive=True)
f = a / b * sqrt(b ** 2 + y ** 2) - y0 / x0
I = Integral(f, (y, 0, y0))

for t in [f, I, I.doit()]:
    pprint(t.simplify())
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample4.py
4.
     _________     
    ╱  2    2      
a⋅╲╱  b  + y     y₀
────────────── - ──
      b          x₀

y₀                         
⌠                          
⎮  ⎛     _________     ⎞   
⎮  ⎜    ╱  2    2      ⎟   
⎮  ⎜a⋅╲╱  b  + y     y₀⎟   
⎮  ⎜────────────── - ──⎟ dy
⎮  ⎝      b          x₀⎠   
⌡                          
0                          

         ⎛y₀⎞           __________      
a⋅b⋅asinh⎜──⎟          ╱  2     2      2
         ⎝b ⎠   a⋅y₀⋅╲╱  b  + y₀     y₀ 
───────────── + ────────────────── - ───
      2                2⋅b            x₀

$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.0001" value="0.005">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-5">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-5">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="5">
<br>
<label for="n0">n = </label>
<input id="n0" type="number" value="2">

<button id="draw0">draw</button>
<button id="clear0">clear</button>

<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>

<script src="sample4.js"></script>

JavaScript

let div0 = document.querySelector('#graph0'),
    pre0 = document.querySelector('#output0'),
    width = 600,
    height = 600,
    padding = 50,
    btn0 = document.querySelector('#draw0'),
    btn1 = document.querySelector('#clear0'),
    input_r = document.querySelector('#r0'),
    input_dx = document.querySelector('#dx'),
    input_x1 = document.querySelector('#x1'),
    input_x2 = document.querySelector('#x2'),
    input_y1 = document.querySelector('#y1'),
    input_y2 = document.querySelector('#y2'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2],
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let a = 2,
    b = 3,
    fns = [[x => Math.sqrt((x ** 2 / a ** 2 - 1) * b ** 2), 'red']];

let draw = () => {
    pre0.textContent = '';

    let r = parseFloat(input_r.value),
        dx = parseFloat(input_dx.value),
        x1 = parseFloat(input_x1.value),
        x2 = parseFloat(input_x2.value),
        y1 = parseFloat(input_y1.value),
        y2 = parseFloat(input_y2.value);
            
    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        lines = [];
    
    fns
        .forEach((o) => {
            let [fn, color] = o;
            
            for (let x = x1; x <= x2; x += dx) {
                let y = fn(x);
                
                if (Math.abs(y) < Infinity) {
                    points.push([x, y, color]);
                }
            }
        });
    
    let xscale = d3.scaleLinear()
        .domain([x1, x2])
        .range([padding, width - padding]);

    let yscale = d3.scaleLinear()
        .domain([y1, y2])
        .range([height - padding, padding]);

    let xaxis = d3.axisBottom().scale(xscale);
    let yaxis = d3.axisLeft().scale(yscale);
    div0.innerHTML = '';
    let svg = d3.select('#graph0')
        .append('svg')
        .attr('width', width)
        .attr('height', height);

    svg.selectAll('circle')
        .data(points)
        .enter()
        .append('circle')
        .attr('cx', (d) => xscale(d[0]))
        .attr('cy', (d) => yscale(d[1]))
        .attr('r', r)
        .attr('fill', (d) => d[2] || 'green');

    svg.selectAll('line')
        .data([[x1, 0, x2, 0], [0, y1, 0, y2]].concat(lines))
        .enter()
        .append('line')
        .attr('x1', (d) => xscale(d[0]))
        .attr('y1', (d) => yscale(d[1]))
        .attr('x2', (d) => xscale(d[2]))
        .attr('y2', (d) => yscale(d[3]))
        .attr('stroke', (d) => d[4] || 'black');
    
    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

    svg.append('g')
        .attr('transform', `translate(${padding}, 0)`)
        .call(yaxis);
    p(fns.join('\n'));
};

inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw();

r = dx =
x1 = x2 =
y1 = y2 =
n =

Kamimura's blog

2018年11月13日火曜日

数学 - Python - JavaScript - 解析学 - 重積分 - 一般の集合の上の積分(双曲線、面積)

0 コメント:

コメントを投稿