数学 - Python - JavaScript - 細分による加法 - 積分法 – 定積分の性質と計算 - 定積分の置換積分法(平方根、三角関数(正弦、余弦、2倍角の公式))

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19章(細分による加法 - 積分法)、19.3(定積分の性質と計算)、定積分の置換積分法、問27.を取り組んでみる。

1. $\begin{array}{l} x = 3 \sin θ \\ \frac{d x}{d θ} = 3 \cos θ \\ x = - 3 \\ - 3 = 3 \sin θ \\ \sin θ = - 1 \\ θ = - \frac{π}{2} \\ x = 3 \\ 3 = 3 \sin θ \\ \sin θ = 1 \\ θ = \frac{π}{2} \\ \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{9 - 3^{2} \sin^{2} θ} 3 \cos θ d θ \\ = \int_{- \frac{π}{2}}^{\frac{π}{2}} 3 \sqrt{1 - \sin^{2} θ} 3 \cos θ d θ \\ = 9 \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{\cos^{2} θ} \cos θ d θ \\ = 9 \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ d θ \\ = 9 \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos 2 θ + 1}{2} d θ \\ = 9 {[\frac{1}{4} \sin 2 θ + \frac{1}{2} θ]}_{- \frac{π}{2}}^{\frac{π}{2}} \\ = \frac{9}{2} (\frac{π}{2} + \frac{π}{2}) \\ = \frac{9}{2} π \end{array}$
2. $\begin{array}{l} x = a \sin θ \\ \frac{d x}{d θ} = a \cos θ \\ x = 0 \\ 0 = a \sin θ \\ θ = 0 \\ x = \frac{a}{2} \\ \frac{a}{2} = a \sin θ \\ \sin θ = \frac{1}{2} \\ θ = \frac{π}{6} \\ \int_{0}^{\frac{π}{6}} \sqrt{a^{2} - a^{2} \sin^{2} θ} a \cos θ d θ \\ = \int_{0}^{\frac{π}{6}} a \sqrt{1 - \sin^{2} θ} a \cos θ d θ \\ = a^{2} \int_{0}^{\frac{π}{6}} \sqrt{\cos^{2} θ} \cos θ d θ \\ = a^{2} \int_{0}^{\frac{π}{6}} \cos^{2} θ d θ \\ = a^{2} \int_{0}^{\frac{π}{6}} \frac{\cos 2 θ + 1}{2} d θ \\ = a^{2} {[\frac{1}{4} \sin 2 θ + \frac{1}{2} θ]}_{0}^{\frac{π}{6}} \\ = a^{2} (\frac{1}{4} \sin \frac{π}{3} + \frac{1}{2} \cdot \frac{π}{6}) \\ = a^{2} (\frac{\sqrt{3}}{8} + \frac{π}{12}) \end{array}$
3. $\begin{array}{l} x = a \sin θ \\ \frac{d x}{d θ} = a \cos θ \\ x = 0 \\ 0 = a \sin θ \\ θ = 0 \\ x = \frac{a}{2} \\ \frac{a}{2} = a \sin θ \\ \sin θ = \frac{1}{2} \\ θ = \frac{π}{6} \\ \int_{0}^{\frac{π}{6}} \frac{1}{\sqrt{a^{2} - a^{2} \sin^{2} θ}} a \cos θ d θ \\ = \int_{0}^{\frac{π}{6}} \frac{1}{a \sqrt{1 - \sin^{2} θ}} a \cos θ d θ \\ = \int_{0}^{\frac{π}{6}} \frac{1}{\sqrt{\cos^{2} θ}} \cos θ d θ \\ = \int_{0}^{\frac{π}{6}} 1 d θ \\ = {[θ]}_{0}^{\frac{π}{6}} \\ = \frac{π}{6} \end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Integral, plot, sqrt, sin, cos, exp, pi

print('27.')
x = symbols('x')
a = symbols('a', positive=True)
fs = [(sqrt(9 - x ** 2), (-3, 3)),
      (sqrt(a ** 2 - x ** 2), (0, a / 2)),
      (1 / sqrt(a ** 2 - x ** 2), (0, a / 2))]


for i, (f, (x1, x2)) in enumerate(fs, 1):
    try:
        print(f'({i})')
        I = Integral(f, (x, x1, x2))
        for g in [I, I.doit()]:
            pprint(g)
        p = plot(f.subs({a: 2}), show=False, legend=True)
        p.save(f'sample27_{i}.svg')
    except Exception as err:
        print(type(err), err)
    print()

入出力結果(Terminal, IPython)

$ ./sample27.py
27.
(1)
3                  
⌠                  
⎮     __________   
⎮    ╱    2        
⎮  ╲╱  - x  + 9  dx
⌡                  
-3                 
9⋅π
───
 2 

(2)
a                
─                
2                
⌠                
⎮    _________   
⎮   ╱  2    2    
⎮ ╲╱  a  - x   dx
⌡                
0                
    2      2
√3⋅a    π⋅a 
───── + ────
  8      12 

(3)
a                
─                
2                
⌠                
⎮      1         
⎮ ──────────── dx
⎮    _________   
⎮   ╱  2    2    
⎮ ╲╱  a  - x     
⌡                
0                
π
─
6

$

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="a0">a = </label>
<input id="a0" 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="sample27.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'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_a0],
    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 f1 = (x) => Math.sqrt(9 - x ** 2);

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);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        f2 = (x) => Math.sqrt(a0 ** 2 - x ** 2),
        f3 = (x) => 1 / Math.sqrt(a0 ** 2 - x ** 2),
        lines = [[-3, y1, -3, y2, 'red'],
                 [3, y1, 3, y2, 'red'],
                 [0, y1, 0, y2, 'purple'],
                 [a0 / 2, y1, a0 / 2, y2, 'purple']],
        fns = [[f1, 'green'],
               [f2, 'blue'],
               [f3, 'orange']],
        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 =
a =

Kamimura's blog

2017年9月1日金曜日

数学 - Python - JavaScript - 細分による加法 - 積分法 – 定積分の性質と計算 - 定積分の置換積分法(平方根、三角関数(正弦、余弦、2倍角の公式))

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