曲線の長さを求める問題

■問題

曲線 $3$\displaystyle{ y=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2} }$ $3$\displaystyle{ $ 0\leq x\leq 1 $ }$ の長さを求めよ．

■解説動画

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■答

$3$\displaystyle{ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log $ 1+\sqrt{2} $ }$

ヒント

曲線の長さを参考にする

■解説

図の赤の実線部分の長さを求める．

$3$\displaystyle{ y=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2} }$ ， $3$\displaystyle{ {y}^{\prime }=x }$

よって，曲線の長さ $3$L$ は公式より

$3$\displaystyle{ L=\displaystyle{ \int _{0}^{1} \sqrt{ 1+{x}^{2} }dx } }$

となる． $3$\displaystyle{ x=\tan t }$ とおく置換積分にによって値を求める．

$3$\displaystyle{ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}} }$ →　 $3$\displaystyle{ dx=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}}dt }$

$3$\displaystyle{x}$	$3$\displaystyle{ 0\rightarrow 1 }$
$3$\displaystyle{t}$	$3$\displaystyle{ 0\rightarrow \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }$

よって

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \sqrt{ 1+{ \tan }^{2}t } }\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}}dt }$

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \sqrt{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}} } }\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}}dt }$

$3$\displaystyle{ 0< t< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }$ では， $3$\displaystyle{ \cos t> 0 }$ ．よって

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos t \hspace{2}} }\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}t \hspace{2}}dt }$

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{3}t \hspace{2}}dt } }$

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \frac{\hspace{2} \cos t \hspace{2}}{\hspace{2} { \cos }^{4}t \hspace{2}} }dt }$

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} } \frac{\hspace{2} \cos t \hspace{2}}{\hspace{2} { $1- { \sin }^{2}t$ }^{2} \hspace{2}} }dt }$

$3$\displaystyle{ u=\sin t }$ とおく．

$3$\displaystyle{ \frac{\hspace{2} du \hspace{2}}{\hspace{2} dt \hspace{2}}=\cos t }$ →　 $3$\displaystyle{ \cos tdt=du }$

$3$\displaystyle{t}$	$3$\displaystyle{0\rightarrow \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}}$
$3$\displaystyle{u}$	$3$\displaystyle{0\rightarrow \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}}}$

よって

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $1- {u}^{2}$ }^{2} \hspace{2}} }du }$

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $1- u$ }^{2}{ $1+u$ }^{2} \hspace{2}} }du }$

ここで，被積分関数を部分分数に分解する．

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 1- u $ }^{2}{ $ 1+u $ }^{2} \hspace{2}}=\frac{\hspace{2}a\hspace{2}}{\hspace{2} { $ 1- u $ }^{2} \hspace{2}}+\frac{\hspace{2}b\hspace{2}}{\hspace{2} 1- u \hspace{2}}+\frac{\hspace{2}c\hspace{2}}{\hspace{2} { $ 1+u $ }^{2} \hspace{2}}+\frac{\hspace{2}d\hspace{2}}{\hspace{2} 1+u \hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2} a+b $ 1- u $ \hspace{2}}{\hspace{2} { $ 1- u $ }^{2} \hspace{2}}+\frac{\hspace{2} c+d $ 1+u $ \hspace{2}}{\hspace{2} { $ 1+u $ }^{2} \hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2} $ a+b $ - bu \hspace{2}}{\hspace{2} { $ 1- u $ }^{2} \hspace{2}}+\frac{\hspace{2} $ c+d $ +du \hspace{2}}{\hspace{2} { $ 1+u $ }^{2} \hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2} \{ $ a+b $ - bu \} { $ 1+u $ }^{2}+ \{ $ c+d $ +du \} { $ 1- u $ }^{2} \hspace{2}}{\hspace{2} { $ 1- u $ }^{2}{ $ 1+u $ }^{2} \hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2} \{ $ a+b $ - bu \} $ 1+2u+{u}^{2} $ + \{ $ c+d $ +du \} $ 1- 2u+{u}^{2} $ \hspace{2}}{\hspace{2} { $ 1- u $ }^{2}{ $ 1+u $ }^{2} \hspace{2}} }$

・分子の計算

$3$\displaystyle{ a+b+2 $ a+b $ u+ $ a+b $ {u}^{2} }$ $3$\displaystyle{ - bu- 2b{u}^{2}- b{u}^{3} }$ $3$\displaystyle{ c+d- 2 $ c+d $ u+ $ c+d $ {u}^{2} }$ $3$\displaystyle{ du- 2d{u}^{2}+d{u}^{3} }$

$3$\displaystyle{= a+b+c+d }$ $3$\displaystyle{ + $ 2a+b- 2c- d $ t }$ $3$\displaystyle{ + $ a- b+c- d $ {t}^{2} }$ $3$\displaystyle{ + $ - b+d $ {t}^{3} }$

ここで

$3$\displaystyle{ a+b+c+d=1 }$ ･･････(1)

$3$\displaystyle{ 2a+b- 2c- d=0 }$ ･･････(2)

$3$\displaystyle{ a- b+c- d=0 }$ ･･････(3)

$3$\displaystyle{ - b+d=0 }$ ･･････(4)

である．(4)より

$3$\displaystyle{ d=b }$ ･･････(5)

(5)を(2)に代入

$3$\displaystyle{ 2a+b- 2c- b=0 }$

$3$\displaystyle{ a- c=0 }$

$3$\displaystyle{ a=c }$ ･･････(6)

(5)を(3)に代入

$3$\displaystyle{ a- b+c- b=0 }$

$3$\displaystyle{ a+c- 2b=0 }$

これに(6)を代入

$3$\displaystyle{ a+a- 2b=0 }$

$3$\displaystyle{ 2a- 2b=0 }$

$3$\displaystyle{ a=b }$ ･･････(7)

(7)と(5)より

$3$\displaystyle{ d=a }$ ･･････(8)

(6)，(7)，(8)を(1)に代入

$3$\displaystyle{ a+a+a+a=1 }$

$3$\displaystyle{ 4a=1 }$

$3$\displaystyle{ a=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$

よって

$3$\displaystyle{ a=b=c=d=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$

となる．したがって

$3$\displaystyle{ =\displaystyle{ \int _{0}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 1- u $ }^{2}{ $ 1+u $ }^{2} \hspace{2}} } }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\displaystyle{ \int _{0}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} } \{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 1- u $ }^{2} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- u \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 1+u $ }^{2} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+u \hspace{2}} \} dx } }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \[ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- u \hspace{2}}- \log \| 1- u \| - \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+u \hspace{2}}+\log \| 1+u \| \] _{0}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} } }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \hspace{2}}- \log \| 1- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \| - \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \hspace{2}}+\log \| 1+\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \| - 1+1 $ }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2} \sqrt{2}- 1 \hspace{2}}- \log $ \frac{\hspace{2} \sqrt{2}- 1 \hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} $ - \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2} \sqrt{2}+1 \hspace{2}}+\log $ \frac{\hspace{2} \sqrt{2}+1 \hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} $ \} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ \frac{\hspace{2} \sqrt{2} $ \sqrt{2}+1 $ - \sqrt{2} $ \sqrt{2}- 1 $ \hspace{2}}{\hspace{2} $ \sqrt{2}- 1 $ $ \sqrt{2}+1 $ \hspace{2}}+\log \frac{\hspace{2} \sqrt{2}+1 \hspace{2}}{\hspace{2} \sqrt{2}- 1 \hspace{2}} \} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ \frac{\hspace{2} 2+\sqrt{2}- 2+\sqrt{2} \hspace{2}}{\hspace{2} 2- 1 \hspace{2}}+\log \frac{\hspace{2} 2+2\sqrt{2}+1 \hspace{2}}{\hspace{2} 2- 1 \hspace{2}} \} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ 2\sqrt{2}+\log $ 3+2\sqrt{2} $ \} }$

$3$\displaystyle{ =\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\log $ 3+2\sqrt{2} $ }$

$3$\displaystyle{ =\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\log { \left({ \sqrt{2}+1 }\right) }^{2} }$

$3$\displaystyle{ =\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log $ \sqrt{2}+1 $ }$

■別解

$3$\displaystyle{ L=\displaystyle{ \int _{0}^{1} \sqrt{ 1+{x}^{2} }dx } }$

部分積分法を用いて計算をする．

$3$\displaystyle{ =\displaystyle{ \int _{0}^{1} { $x$ }^{\prime }\sqrt{ 1+{x}^{2} }dx } }$

$3$\displaystyle{ = \[ x\sqrt{ 1+{x}^{2} } \] _{0}^{1}- \displaystyle{ \int _{0}^{1} x\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\frac{\hspace{2} 2x \hspace{2}}{\hspace{2} \sqrt{ 1+{x}^{2} } \hspace{2}} }dx }$

$3$\displaystyle{ =\sqrt{2}- \displaystyle{ \int _{0}^{1} \frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} \sqrt{ 1+{x}^{2} } \hspace{2}}dx } }$

$3$\displaystyle{ =\sqrt{2}- \displaystyle{ \int _{0}^{1} \frac{\hspace{2} 1+{x}^{2}- 1 \hspace{2}}{\hspace{2} \sqrt{ 1+{x}^{2} } \hspace{2}}dx } }$

$3$\displaystyle{ =\sqrt{2}- \displaystyle{ \int _{0}^{1} \sqrt{ 1+{x}^{2} }dx+\displaystyle{ \int _{0}^{1} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 1+{x}^{2} } \hspace{2}}dx } } }$

$3$\displaystyle{ =\sqrt{2}- L+ \[ \log \| x+\sqrt{ 1+{x}^{2} } \| \] _{0}^{1} }$

$3$\displaystyle{ =\sqrt{2}- L+\log $ 1+\sqrt{2} $ }$

よって

$3$\displaystyle{ 2L=\sqrt{2}+\log $ 1+\sqrt{2} $ }$

$3$\displaystyle{ L=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log $ 1+\sqrt{2} $ }$

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最終更新日：2025年6月9日