数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 簡単な微分方程式 - 二三の応用(球形、水、容量、流出量、水面の降下速度、容量の減少速度)

学習環境

Surface 3 (4G LTE)、Surface 3 タイプカバー、Surface ペン(端末)
Windows 10 Pro (OS)
数式入力ソフト(TeX, MathML): MathType
MathML対応ブラウザ: Firefox、Safari
MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
参考書籍

数学読本〈5〉微分法の応用/積分法/積分法の応用/行列と行列式(松坂和夫(著)、岩波書店)の第20章(面積、体積、長さ - 積分法の応用)、20.4(簡単な微分方程式)、二三の応用、問45.を取り組んでみる。

1. 球形の中心を通る断面を考える。
  $\begin{array}{l} s^{2} + u^{2} = R^{2} \\ u^{2} = R^{2} - s^{2} \\ V = \int_{- R}^{x - R} u^{2} π d s \\ = \int_{- R}^{x - R} (R^{2} - s^{2}) π d s \\ = π \int_{- R}^{x - R} (R^{2} - s^{2}) d s \\ = π {[R^{2} s - \frac{1}{3} s^{3}]}_{- R}^{x - R} \\ = π ((R^{2} (x - R) - \frac{1}{3} {(x - R)}^{3}) - (- R^{3} + \frac{1}{3} R^{3})) \\ = π (R^{2} x - R^{3} - \frac{1}{3} (x^{3} - 3 x^{2} R + 3 x R^{2} - R^{3}) + R^{3} - \frac{1}{3} R^{3}) \\ = π (R^{2} x - \frac{1}{3} x^{3} + x^{2} R - x R^{2} + \frac{1}{3} R^{3} - \frac{1}{3} R^{3}) \\ = π (- \frac{1}{3} x^{3} + x^{2} R) \\ = π (R x^{2} - \frac{1}{3} x^{3}) \end{array}$
2. $\begin{array}{l} \frac{d V}{d t} = - k \sqrt{x} \\ \frac{d V}{d x} \cdot \frac{d x}{d t} = - k \sqrt{x} \\ \frac{d}{d x} (π (R x^{2} - \frac{1}{3} x^{3})) \frac{d x}{d t} = - k \sqrt{x} \\ π (2 R x - x^{2}) \frac{d x}{d t} = - k \sqrt{x} \\ - \frac{d x}{d t} = \frac{k \sqrt{x}}{π (2 R x - x^{2})} \\ - \frac{d x}{d t} = \frac{k \sqrt{x}}{π x (2 R - x)} \\ - \frac{d x}{d t} = \frac{k}{π \sqrt{x} (2 R - x)} \end{array}$
3. 微分方程式の一般解。
  $\begin{array}{l} - \frac{d x}{d t} = \frac{k}{π \sqrt{x} (2 R - x)} \\ - π \sqrt{x} (2 R - x) d x = k d t \\ \int (- π \sqrt{x} (2 R - x)) d x = \int k d t \\ - π \int (2 R \sqrt{x} - x^{\frac{3}{2}}) d x = \int k d t \\ - π (2 R \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}}) = k t + C \\ - π x^{\frac{3}{2}} (2 R \frac{2}{3} - \frac{2}{5} x) = k t + C \\ - 2 π x^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} x) = k t + C \end{array}$
  0秒のとき、水深は2Rなので。初期条件t = 0、x=2RとしてCを求める。
  $\begin{array}{l} - 2 π {(2 R)}^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} \cdot 2 R) = k 0 + C \\ C = - 2 π {(2 R)}^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} \cdot 2 R) \end{array}$
  よって、求める関係式。
  $- 2 π x^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} x) = k t - 2 π {(2 R)}^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} \cdot 2 R)$
4. 前問の関係式で x = Rとする。
  $\begin{array}{l} - 2 π R^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} R) = k t - 2 π {(2 R)}^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} \cdot 2 R) \\ t = \frac{2 π {(2 R)}^{\frac{3}{2}} (\frac{2}{3} R - \frac{2}{5} R) - 2 π R^{\frac{3}{2}} (\frac{2}{3} R - \frac{1}{5} R)}{k} \\ = \frac{2 π R^{\frac{3}{2}} (2^{\frac{3}{2}} (\frac{2}{3} R - \frac{2}{5} R) - (\frac{2}{3} R - \frac{1}{5} R))}{k} \\ = \frac{2 π R^{\frac{3}{2}} (2^{\frac{5}{2}} (\frac{1}{3} R - \frac{1}{5} R) - (\frac{2}{3} R - \frac{1}{5} R))}{k} \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, pi, Rational

print('45.')

R, k = symbols('R k')
t = 2 * pi * R ** Rational(3, 2) * (2 ** Rational(5, 2)
                                    * (R / 3 - R / 5) - (2 * R / 3 - R / 5)) / k

for t0 in [t, t.expand(), t.factor()]:
    pprint(t0)
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample45.py
45.
     3/2 ⎛  7⋅R   8⋅√2⋅R⎞
2⋅π⋅R   ⋅⎜- ─── + ──────⎟
         ⎝   15     15  ⎠
─────────────────────────
            k            

        5/2            5/2
  14⋅π⋅R      16⋅√2⋅π⋅R   
- ───────── + ────────────
     15⋅k         15⋅k    

     5/2            
2⋅π⋅R   ⋅(-7 + 8⋅√2)
────────────────────
        15⋅k        

$

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="0.001">
<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="r1">R = </label>
<input id="r1" type="number" min="0" value="2">

<label for="x0">x0 = </label>
<input id="x0" type="number" value="1">

<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="sample45.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'),
    input_r1 = document.querySelector('#r1'),
    input_x0 = document.querySelector('#x0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_r1, input_x0],
    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 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),
        r1 = parseFloat(input_r1.value),
        x0 = parseFloat(input_x0.value);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        lines = [[x0, y1, x0, y2, 'blue']],
        f1 = (x) => Math.sqrt(r1 ** 2 - x ** 2),
        f2 = (x) => -f1(x),
        fns = [[f1, 'green'],
               [f2, 'green']],
        fns1 = [],
        fns2 = [];

    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]);
            }
        }
    });
    fns1.forEach((o) => {
        let [fn, color] = o;
        
        lines.push([x1, fn(x1), x2, fn(x2), color]);
    });
    fns2.forEach((o) => {
        let [fn, color] = o;

        for (let x = x1; x <= x2; x += dx0) {
            let g = fn(x);
            
            lines.push([x1, g(x1), x2, g(x2), 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('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.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.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

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

    [fns, fns1, fns2].forEach((fs) => p(fs.join('\n')));
};

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

r = dx =
x1 = x2 =
y1 = y2 =
R = x0 =

Kamimura's blog

2017年10月31日火曜日

数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 簡単な微分方程式 - 二三の応用(球形、水、容量、流出量、水面の降下速度、容量の減少速度)

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