ベクトルの微分曲線の長さ、累乗、指数関数、置換積分法、双曲線関数(sinh、cosh)、逆関数、速度ベクトル、積分

続解析入門 (原書第2版)
楽天ブックス
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au PAY マーケット
原著: Calculus of Several Variables

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第2章(ベクトルの微分)、2(曲線の長さ)の練習問題3の解答を求めてみる。

$\frac{d}{dt} (t, 2 t, t^{2}) = (1, 2, 2 t)$
$\int \sqrt{1 + 4 + 4 t^{2}} dt = \int \sqrt{5 + 4 t^{2}} dt$
$t = \frac{\sqrt{5}}{2} u$
$1 = \frac{\sqrt{5}}{2} \frac{d u}{dt}$
$\int \sqrt{5 + 4 \cdot \frac{5}{4} u^{2}} \frac{\sqrt{5}}{2} d u = \frac{5}{2} \int \sqrt{1 + u^{2}} d u$
$u = \frac{e^{s} - e^{- s}}{2} = \sinh s$
$\frac{d u}{d s} = \frac{e^{s} + e^{- s}}{2}$
$1 + u^{2} = 1 + \sinh^{2} s = \cosh^{2} s = {(\frac{e^{s} + e^{- s}}{2})}^{2}$
$\frac{5}{2} \int \sqrt{1 + u^{2}} d u$
$= \frac{5}{2} \int \frac{e^{s} + e^{- s}}{2} \cdot \frac{e^{s} + e^{- s}}{2} d s$
$= \frac{5}{2^{3}} \int (e^{2 s} + e^{- 2 s} + 2) d s$
$= \frac{5}{2^{3}} (\frac{1}{2} e^{2 s} - \frac{1}{2} e^{- 2 s} + 2 s)$
$\begin{array}{l} t = 1, 3 \\ u = \frac{2}{\sqrt{5}}, \frac{6}{\sqrt{5}} \end{array}$
$\begin{array}{l} s = \sinh^{- 1} u \\ s = \sinh^{- 1} \frac{2}{\sqrt{5}}, \sinh^{- 1} \frac{6}{\sqrt{5}} \\ s = \log \sqrt{5}, \log (\frac{6}{\sqrt{5}} + \frac{\sqrt{41}}{\sqrt{5}}) \\ s = \log \sqrt{5}, \log \frac{6 + \sqrt{41}}{\sqrt{5}} \end{array}$
よって、求める曲線の長さは
$\frac{5}{2^{3}} (\frac{1}{2} ({(\frac{6 + \sqrt{41}}{\sqrt{5}})}^{2} - {(\sqrt{5})}^{2}) - \frac{1}{2} ({(\frac{6 + \sqrt{41}}{\sqrt{5}})}^{- 2} - {(\sqrt{5})}^{- 2}) + 2 (\log \frac{6 + \sqrt{41}}{\sqrt{5}} - \log \sqrt{5}))$
$= \frac{5}{2^{4}} (\frac{77 + 12 \sqrt{41}}{5} - 5) - \frac{5}{2^{4}} (\frac{5}{77 + 12 \sqrt{41}} - \frac{1}{5}) + \frac{5}{4} \log \frac{6 + \sqrt{41}}{\sqrt{5}} - \frac{5}{4} \log \sqrt{5}$
$= \frac{1}{2^{4}} (52 + 12 \sqrt{41}) - \frac{1}{2^{4}} (\frac{25}{77 + 12 \sqrt{41}} - 1) + \frac{5}{4} \log \frac{6 + \sqrt{41}}{\sqrt{5}} - \frac{5}{4} \log \sqrt{5}$
$= \frac{1}{4} (13 + 3 \sqrt{41}) + \frac{1}{2^{4}} \frac{52 + 12 \sqrt{41}}{77 + 12 \sqrt{41}} + \frac{5}{4} \log \frac{6 + \sqrt{41}}{\sqrt{5}} - \frac{5}{4} \log \sqrt{5}$
$= \frac{1}{4} (13 + 3 \sqrt{41}) + \frac{1}{4} \frac{13 + 3 \sqrt{41}}{77 + 12 \sqrt{41}} + \frac{5}{4} \log \frac{6 + \sqrt{41}}{\sqrt{5}} - \frac{5}{4} \log \sqrt{5}$
$\frac{d}{dt} (e^{3 t}, e^{- 3 t}, 3 \sqrt{2} t) = (3 e^{3 t}, - 3 e^{- 3 t}, 3 \sqrt{2})$
$\int \sqrt{9 e^{6 t} + 9 e^{6 t} + 9 \cdot 2} dt$
$= 3 \int \sqrt{2 e^{6 t} + 2} dt$
$= 3 \sqrt{2} \int \sqrt{e^{6 t} + 1} dt$
$u = e^{6 t}$
$\frac{d u}{dt} = 6 e^{6 t}$
$3 \sqrt{2} \int \sqrt{u} \frac{d u}{6 e^{6 t}}$
$= \frac{3 \sqrt{2}}{6} \int \frac{\sqrt{u}}{u} d u$
$= \frac{3 \sqrt{2}}{6} \int \frac{1}{\sqrt{u}} d u$
$= \frac{3 \sqrt{2}}{6} \cdot 2 \sqrt{u}$
$= \sqrt{2} \sqrt{u}$
$\begin{array}{l} t = 0, \frac{1}{3} \\ u = 6, 6 e^{2} \end{array}$
$\sqrt{2} (\sqrt{6 e^{2}} - \sqrt{6}) = 2 \sqrt{3} (e - 1)$