数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 面積 - 面積の公式(放物線、接線(直線)、弦、軸、三角形、図形の面積、比率)

数学読本〈5〉

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
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

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8.を取り組んでみる。

8. 1. 
      $\begin{array}{l}M\left(\frac{a+b}{2},\frac{a^2+b^2}{2}\right)\\ \frac{d}{dx}{x}^2=2x\end{array}$

      Aにおける放物線の接線の方程式。

      $\begin{array}{l}y=2a\left(x-a\right)+a^2\\ =2ax-a^2\end{array}$

      Bにおける放物線の接線の方程式。

      $\begin{array}{l}y=2b\left(x-b\right)+b^2\\ =2bx-b^2\end{array}$

      点Qを求める。

      $\begin{array}{l}2ax-a^2=2bx-b^2\\ x=\frac{a^2-b^2}{2\left(a-b\right)}=\frac{a+b}{2}\\ Q\left(\frac{a+b}{2},2a·\frac{a+b}{2}-a^2\right)\\ Q\left(\frac{a+b}{2},ab\right)\end{array}$

      よって、点MとQのx座標は等しいので、MとQを結ぶ直線MQは放物線の軸と平行である。

   2. 
      $\begin{array}{l}P\left(\frac{a+b}{2},{\left(\frac{a+b}{2}\right)}^2\right)\\ P\left(\frac{a+b}{2},\frac{a^2+2ab+b^2}{4}\right)\end{array}$

      線分MQの中点。

      $\begin{array}{l}\left(\frac{a+b}{2},\frac{\frac{a^2+b^2}{2}+ab}{2}\right)\\ =\left(\frac{a+b}{2},\frac{a^2+2ab+b^2}{4}\right)\end{array}$

      よって、直線MQと放物線の交点Pは線分MQの中点である。

   3. 
      

      点A、Bを通る直線の方程式。

      $\begin{array}{l}y=\frac{b^2-a^2}{b-a}\left(x-a\right)+a^2\\ =\left(a+b\right)\left(x-a\right)+a^2\\ =\left(a+b\right)x-ab\end{array}$

      放物線と弦ABで囲まれる図形の面積。

      $\begin{array}{l}{S}_1={\displaystyle {\int }_a^b\left(\left(\left(a+b\right)x-ab\right)-x^2\right)dx}\\ ={\left[\frac{a+b}{2}{x}^2-abx-\frac{1}{3}{x}^3\right]}_a^b\\ =\left(\frac{a+b}{2}{b}^2-a{b}^2-\frac{1}{3}{b}^3\right)-\left(\frac{a+b}{2}{a}^2-a^{2}b-\frac{1}{3}{a}^3\right)\\ =\frac{3a{b}^2+3b^3-6a{b}^2-2b^3-3a^3-3a^{2}b+6a^{2}b+2a^3}{6}\\ =\frac{b^3-3a{b}^2+3a^{2}b-a^3}{6}\\ =\frac{1}{6}{\left(b-a\right)}^3\end{array}$
   4. 
      

      直線AQの方程式。

      $\begin{array}{l}{S}_2={\displaystyle {\int }_a^{\frac{a+b}{2}}\left(x^2-\left(2ax-a^2\right)\right)dx}+{\displaystyle {\int }_{\frac{a+b}{2}}^b\left(x^2-\left(2bx-b^2\right)\right)dx}\\ ={\displaystyle {\int }_a^{\frac{a+b}{2}}\left(x^2-2ax+a^2\right)dx}+{\displaystyle {\int }_{\frac{a+b}{2}}^b\left(x^2-2bx+b^2\right)dx}\\ ={\displaystyle {\int }_a^{\frac{a+b}{2}}{\left(x-a\right)}^{2}dx}+{\displaystyle {\int }_{\frac{a+b}{2}}^b{\left(x-b\right)}^{2}dx}\\ ={\left[\frac{1}{3}{\left(x-a\right)}^3\right]}_a^{\frac{a+b}{2}}+{\left[\frac{1}{3}{\left(x-b\right)}^3\right]}_{\frac{a+b}{2}}^b\\ =\frac{1}{3}\left({\left[{\left(x-a\right)}^3\right]}_a^{\frac{a+b}{2}}+{\left[{\left(x-b\right)}^3\right]}_{\frac{a+b}{2}}^b\right)\\ =\frac{1}{3}\left({\left(\frac{a+b}{2}-a\right)}^3-{\left(\frac{a+b}{2}-b\right)}^3\right)\\ =\frac{1}{3}\left({\left(\frac{b-a}{2}\right)}^3-{\left(\frac{a-b}{2}\right)}^3\right)\\ =\frac{1}{3}\left({\left(\frac{b-a}{2}\right)}^3+{\left(\frac{b-a}{2}\right)}^3\right)\\ =\frac{1}{3}\left(2{\left(\frac{b-a}{2}\right)}^3\right)\\ =\frac{1}{3}\left(\frac{{\left(b-a\right)}^3}{4}\right)\\ =\frac{1}{12}{\left(b-a\right)}^3\end{array}$
   5. 
      

      点Pは線分MQの中点なので、△PABの面積は三角形QABの面積の半分。

      $\begin{array}{l}{S}_3=\frac{1}{2}\left(S_1+S_2\right)\\ =\frac{1}{2}\left(\frac{1}{6}{\left(b-a\right)}^3+\frac{1}{12}{\left(b-a\right)}^3\right)\\ =\frac{1}{2}·\frac{3}{12}{\left(b-a\right)}^3\\ =\frac{1}{8}{\left(b-a\right)}^3\end{array}$
   6. 
      $\begin{array}{l}{S}_1:S_2:S_3\\ =\frac{1}{6}{\left(b-a\right)}^3:\frac{1}{12}{\left(b-a\right)}^3:\frac{1}{8}{\left(b-a\right)}^3\\ =\frac{1}{6}:\frac{1}{12}:\frac{1}{8}\\ =4:2:3\end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Derivative, solve

print('8.')
x, a, b = symbols('x a b')
f = x ** 2
M = ((a + b) / 2, f.subs({x: (a + b) / 2}))
f1 = Derivative(f, x, 1)
fa = f1.subs({x: a}) * (x - a) + f.subs({x: a})
fb = f1.subs({x: b}) * (x - b) + f.subs({x: b})

Qx = solve(fa - fb, x)[0]
Q = (Qx, fa.subs({x: Qx}))

for t in [M, Q]:
    pprint(t[0].doit().factor())

入出力結果(Terminal, Jupyter(IPython))

$ ./sample8.py
8.
a + b
─────
  2  
a + b
─────
  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="-10">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="10">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-10">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="10">
<br>
<label for="a0">a = </label>
<input id="a0" type="number" value="-1">
<label for="b0">b = </label>
<input id="b0" 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="sample8.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_a0 = document.querySelector('#a0'),
    input_b0 = document.querySelector('#b0'),    
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_a0, input_b0],
    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,
    f1 = (x) => 2 * x,
    g = (x0) => (x) => f1(x0) * (x - x0) + f(x0);

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),
        a0 = parseFloat(input_a0.value),
        b0 = parseFloat(input_b0.value);
        

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        fa = g(a0),
        fb = g(b0),
        lines = [[a0, f(a0), b0, f(b0), 'blue'],
                 [a0, f(a0), 0, 0, 'blue'],
                 [0, 0, b0, f(b0), 'blue']],
        fns = [[f, 'green']],
        fns1 = [[fa, 'blue'],
                [fb, 'blue']],
        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 =
a = b =