距離空間の世界 R^nにおける曲線対数関数、放物線の長さ、パラメーター表示に変換、置換積分法

学習環境

解析入門(中) (松坂和夫数学入門シリーズ 5) (松坂和夫 (著)、岩波書店)の第12章(距離空間の世界)、12.4(R^nにおける曲線)、問題5の解答を求めてみる。

\begin{array}{l} x = t \\ y = \log t \\ 1 \leq t \leq 3 \end{array}

\begin{array}{l} \frac{dx}{dt} = 1 \\ \frac{dy}{dt} = \frac{1}{t} \end{array}

\int \sqrt{1 + \frac{1}{t^{2}}} dt

= \int \sqrt{\frac{1 + t^{2}}{t^{2}}} dt

= \int \frac{\sqrt{1 + t^{2}}}{t} dt

置換積分法。

\begin{array}{l} \sqrt{1 + t^{2}} = u \\ \frac{t}{\sqrt{1 + t^{2}}} = \frac{d u}{dt} \\ 1 + t^{2} = u^{2} \\ t^{2} = u^{2} - 1 \\ \sqrt{2} \leq u \leq \sqrt{10} \end{array}

\int \frac{u}{t} \cdot \frac{\sqrt{1 + t^{2}}}{t} d u

= \int \frac{u^{2}}{u^{2} - 1} d u

= \int \frac{u^{2} - 1 + 1}{u^{2} - 1} d u

= \int (1 + \frac{1}{(u + 1) (u - 1)}) d u

= u + \frac{1}{2} \int (\frac{1}{u - 1} - \frac{1}{u + 1}) d u

= u + \frac{1}{2} (\log (u - 1) - \log (u + 1))

= u + \frac{1}{2} \log \frac{u - 1}{u + 1}

よって、求める曲線の長さは

\sqrt{10} - \sqrt{2} + \frac{1}{2} (\log \frac{\sqrt{10} - 1}{\sqrt{10} + 1} - \log \frac{\sqrt{2} - 1}{\sqrt{2} + 1})

= \sqrt{10} - \sqrt{2} + \frac{1}{2} \log \frac{{(\sqrt{10} - 1)}^{2}}{9} - \log {(\sqrt{2} - 1)}^{2}

= \sqrt{10} - \sqrt{2} + \frac{1}{2} \log {(\frac{\sqrt{10} - 1}{3 (\sqrt{2} - 1)})}^{2}

= \sqrt{10} - \sqrt{2} + \log \frac{\sqrt{10} - 1}{3 (\sqrt{2} - 1)}

Wolfram

Integrate[Sqrt[1 + 1 / t^2], {t, 1, 3}]

N[%,5]

Sqrt[10] - Sqrt[2] + Log[(Sqrt[10] - 1) / (3 * (Sqrt[2] - 1))]

N[%,5]

2.3020

Plot[Log[x]]

Plot::argr: Plot called with 1 argument; 2 arguments are expected.

Plot[Log[x], {x, 0, 10}]

Plot[Log[x], {x, 1, 3}]