数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 面積 - 面積の公式(2つの放物線で囲まれる図形の面積)

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)、20.1(面積)、面積の公式、問10.を取り組んでみる。

1つ目の放物線の焦点。
$(0, \frac{1}{4})$
2つ目の放物線を変形。
$\begin{array}{l} a x = {(y + \frac{1}{2} b)}^{2} - \frac{1}{4} b^{2} \\ a x = {(y + \frac{1}{2} b)}^{2} - \frac{1}{4} b^{2} \end{array}$
この放物線の焦点。
$(\frac{a}{4} - \frac{1}{4 a} b^{2}, - \frac{1}{2} b)$
求めるa、bの値。
$\begin{array}{l} (\frac{a}{4} - \frac{1}{4 a} b^{2}, - \frac{1}{2} b) \\ \frac{a}{4} - \frac{b^{2}}{4 a} = 0 \\ - \frac{1}{2} b = \frac{1}{4} \\ a^{2} - b^{2} = 0 \\ b = - \frac{1}{2} \\ a^{2} = b^{2} \\ a^{2} = \frac{1}{4} \\ a > 0 \\ a = \frac{1}{2} \end{array}$
2つの放物線で囲まれる図形の面積。
$\begin{array}{l} y = x^{2} \\ \frac{1}{2} x = y^{2} - \frac{1}{2} y \\ x = 2 y^{2} - y \\ = y (2 y - 1) \\ y = 0, \frac{1}{2} \\ {(y - \frac{1}{4})}^{2} - \frac{1}{16} = \frac{1}{2} x \\ {(y - \frac{1}{4})}^{2} = \frac{1}{2} x + \frac{1}{16} \\ y - \frac{1}{4} = \pm \sqrt{\frac{1}{2} x + \frac{1}{16}} \\ y = \frac{1}{4} \pm \sqrt{\frac{1}{2} x + \frac{1}{16}} \\ x = 2 x^{4} - x^{2} \\ 2 x^{4} - x^{2} - x = 0 \\ x (2 x^{3} - x - 1) = 0 \\ x (x - 1) (2 x^{2} + 2 x + 1) = 0 \\ x = 0, 1 \\ \int_{0}^{1} ((\frac{1}{4} + \sqrt{\frac{1}{2} x + \frac{1}{16}}) - x^{2}) d x + \int_{0}^{\frac{1}{2}} (0 - (2 y^{2} - y)) d x \\ = \int_{0}^{1} ((\frac{1}{4} + \sqrt{\frac{8}{16} x + \frac{1}{16}}) - x^{2}) d x + \int_{0}^{\frac{1}{2}} (0 - (2 y^{2} - y)) d x \\ = \int_{0}^{1} ((\frac{1}{4} + \frac{1}{4} \sqrt{8 x + 1}) - x^{2}) d x + \int_{0}^{\frac{1}{2}} (0 - (2 y^{2} - y)) d x \\ = \frac{1}{4} \int_{0}^{1} (1 + \sqrt{8 x + 1} - 4 x^{2}) d x - \int_{0}^{\frac{1}{2}} (2 y^{2} - y) d x \\ = \frac{1}{4} {[x + \frac{2}{3} \cdot \frac{1}{8} {(8 x + 1)}^{\frac{3}{2}} - \frac{4}{3} x^{3}]}_{0}^{1} - {[\frac{2}{3} y^{3} - \frac{1}{2} y^{2}]}_{0}^{\frac{1}{2}} \\ = \frac{1}{4} {[x + \frac{1}{12} {(8 x + 1)}^{\frac{3}{2}} - \frac{4}{3} x^{3}]}_{0}^{1} - {[\frac{2}{3} y^{3} - \frac{1}{2} y^{2}]}_{0}^{\frac{1}{2}} \\ = \frac{1}{4} ((1 + \frac{1}{12} 9^{\frac{3}{2}} - \frac{4}{3}) - (\frac{1}{12})) - (\frac{2}{3} \cdot \frac{1}{2^{3}} - \frac{1}{2} \cdot \frac{1}{2^{2}}) \\ = \frac{1}{4} (1 + \frac{3^{3}}{3 \cdot 4} - \frac{4}{3} - \frac{1}{12}) - (\frac{1}{12} - \frac{1}{8}) \\ = \frac{1}{4} (1 + \frac{9}{4} - \frac{4}{3} - \frac{1}{12}) - \frac{1}{12} + \frac{1}{8} \\ = \frac{1}{4} (1 + \frac{9}{4} - \frac{4}{3} - \frac{1}{12} - \frac{1}{3} + \frac{1}{2}) \\ = \frac{1}{4} \cdot \frac{12 + 27 - 16 - 1 - 4 + 6}{12} \\ = \frac{1}{4} \cdot \frac{24}{12} \\ = \frac{1}{4} \cdot 2 \\ = \frac{1}{2} \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Rational, solve, plot, Integral, sqrt

print('10.')
x, y = symbols('x y')
a = Rational(1, 2)
b = -a
f = y - x ** 2
g = a * x - (y ** 2 + b * y)

s1 = solve(f, y)
s2 = solve(g, y)

pprint(s1)
pprint(s2)

p = plot(*s1, *s2, (x, -2, 2), show=False, legend=True)
p.save('sample10.svg')

pprint(solve(s1[0] - s2[1], x))
pprint(solve(s1[0] - s2[0], x))
pprint(solve(y ** 2 + b * y))

I = Integral(Rational(1, 4) + sqrt(Rational(1, 2) * x + Rational(1, 16)) -
             x**2, (x, 0, 1)) + Integral(0 - (2 * y ** 2 - y), (y, 0, Rational(1, 2)))

pprint(I)
pprint(I.doit())

入出力結果(Terminal, Jupyter(IPython))

$ ./sample10.py
10.
⎡ 2⎤
⎣x ⎦
⎡    _________        _________    ⎤
⎢  ╲╱ 8⋅x + 1    1  ╲╱ 8⋅x + 1    1⎥
⎢- ─────────── + ─, ─────────── + ─⎥
⎣       4        4       4        4⎦
[1]
⎡     1   ⅈ    1   ⅈ⎤
⎢0, - ─ - ─, - ─ + ─⎥
⎣     2   2    2   2⎦
[0, 1/2]
                      1                             
1/2                   ⌠                             
 ⌠                    ⎮ ⎛           ________    ⎞   
 ⎮  ⎛     2    ⎞      ⎮ ⎜   2      ╱ x   1     1⎟   
 ⎮  ⎝- 2⋅y  + y⎠ dy + ⎮ ⎜- x  +   ╱  ─ + ──  + ─⎟ dx
 ⌡                    ⎮ ⎝       ╲╱   2   16    4⎠   
 0                    ⌡                             
                      0                             
1/2
$

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">

<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="sample10.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 f = (x) => x ** 2,
    g1 = (x) => - Math.sqrt(8 * x + 1) / 4 + 1 / 4,
    g2 = (x) => Math.sqrt(8 * x + 1) / 4 + 1 / 4;

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 = [[f, 'red'],
               [g1, 'green'],
               [g2, 'blue']],
        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 =

Kamimura's blog

2017年9月25日月曜日

数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 面積 - 面積の公式(2つの放物線で囲まれる図形の面積)

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