積分法パラメーター曲線、弧の長さ、定積分

学習環境

微分積分学 (ちくま学芸文庫) (吉田洋一(著)、筑摩書房)のⅣ.(積分法)、演習問題22.の解答を求めてみる。

\begin{array}{l} 3 t - t^{3} = 0 \\ t (3 - t^{2}) = 0 \\ t = \pm \sqrt{3} \end{array}

求める弧の長さ。

\int_{0}^{\sqrt{3}} \sqrt{{(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2}} dt

= \int_{0}^{\sqrt{3}} \sqrt{{(6 t)}^{2} + {(3 - 3 t^{2})}^{2}} dt

= 3 \int_{0}^{\sqrt{3}} \sqrt{4 t^{2} + 1 - 2 t^{2} + t^{4}}

= 3 \int_{0}^{\sqrt{3}} \sqrt{t^{4} + 2 t^{2} + 1} dt

= 3 \int_{0}^{\sqrt{3}} \sqrt{{(t^{2} + 1)}^{2}} dt

= 3 \int_{0}^{\sqrt{3}} (t^{2} + 1) dt

= 3 {[\frac{1}{3} t^{3} + t]}_{0}^{\sqrt{3}}

= 3 (\sqrt{3} + \sqrt{3})

= 6 \sqrt{3}

コード

#!/usr/bin/env python3
from unittest import TestCase, main
from sympy import sqrt, Integral, Derivative, solve
from sympy.plotting import plot_parametric
from sympy.abc import t

print('22.')

x = 3 * t ** 2
y = 3 * t - t ** 3
x1, y1 = [Derivative(o, t, 1).doit() for o in [x, y]]


class Test(TestCase):
    def test(self):
        self.assertEqual(
            float(Integral(sqrt(x1 ** 2 + y1 ** 2), (t, 0, sqrt(3))).doit()),
            float(6 * sqrt(3))
        )
        self.assertEqual(
            float(Integral(sqrt(x1 ** 2 + y1 ** 2), (t, -sqrt(3), 0)).doit()),
            float(6 * sqrt(3))
        )


p = plot_parametric(
    (x, y, (t, -2.5, -sqrt(3))),
    (x, y, (t, -sqrt(3), 0)),
    (x, y, (t, 0, sqrt(3))),
    (x, y, (t, sqrt(3), 2.5)),
    legend=True,
    show=False
)
colors = ['red', 'green', 'blue', 'brown', 'orange',
          'purple', 'pink', 'gray', 'skyblue', 'yellow']

for o, color in zip(p, colors):
    o.line_color = color
p.save('sample22.png')
p.show()
if __name__ == "__main__":
    main()

入出力結果

% ./sample22.py -v
22.
test (__main__.Test) ... ok

----------------------------------------------------------------------
Ran 1 test in 3.504s

OK
%