数学 - Python - JavaScript - 解析学 - 積分 - 積分の応用 - 極座標による面積(三角関数(正弦)、2倍角)

学習環境

解析入門原書第3版 (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第3部(積分)、第13章(積分の応用)、2(極座標による面積)の練習問題2.を取り組んでみる。

$\begin{array}{l} \int \frac{1}{2} a^{2} \sin (2 θ) d θ \\ = \frac{1}{2} a^{2} (- \frac{1}{2} \cos (2 θ)) \\ = - \frac{1}{4} a^{2} \cos^{2} θ \end{array}$

よって求める曲線で囲まれる図形の面積は、

$\begin{array}{l} - \frac{1}{4} a^{2} {[\cos (2 θ)]}_{0}^{\frac{π}{2}} \\ = - \frac{1}{4} a^{2} (- 1 - 1) \\ = \frac{1}{2} a^{2} \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, Integral, sin, pi, sqrt, Rational
from sympy import solve

print('2.')
θ = symbols('θ')
a = symbols('a', positive=True)
f = sqrt(a ** 2 * sin(2 * θ))
I = Integral(Rational(1, 2) * f ** 2, (θ, 0, pi / 2))

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

入出力結果(Terminal, Jupyter(IPython))

$ ./sample2.py
2.
π               
─               
2               
⌠               
⎮  2            
⎮ a ⋅sin(2⋅θ)   
⎮ ─────────── dθ
⎮      2        
⌡               
0               

 2
a 
──
2 

$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="1">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.001" 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="sample2.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';

let a = 2,
    fr = theta => Math.sqrt(a ** 2 * Math.sin(2 * theta)),
    fx = theta => fr(theta) * Math.cos(theta),
    fy = theta => fr(theta) * Math.sin(theta),
    gr = theta => Math.sqrt(a ** 2 * Math.abs(Math.sin(2 * theta))),
    gx = theta => gr(theta) * Math.cos(theta),
    gy = theta => gr(theta) * Math.sin(theta),
    fns = [];

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 = [];

    for (let theta = 0; theta < Math.PI / 2; theta += dx) {
        points.push([fx(theta), fy(theta), 'red']);
    }
    for (let theta = Math.PI / 2; theta < Math.PI; theta += dx) {
        points.push([gx(theta), gy(theta), 'green']);
    }
    for (let theta = Math.PI; theta < 3 / 2 * Math.PI; theta += dx) {
        points.push([fx(theta), fy(theta), 'blue']);
    }
    for (let theta = 3 / 2 * Math.PI; theta < 2 * Math.PI; theta += dx) {
        points.push([gx(theta), gy(theta), 'orange']);
    }
    
    fns
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx) {
                let y = f(x);

                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('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].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

2018年11月19日月曜日

数学 - Python - JavaScript - 解析学 - 積分 - 積分の応用 - 極座標による面積(三角関数(正弦)、2倍角)

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