曲線の長さを求める問題

■問題

曲線 ${\displaystyle y=\frac{1}{2}{x}^2}$　${\displaystyle \left(0\leqq x\leqq 1\right)}$ の長さを求めよ．

■答

${\displaystyle \frac{\sqrt{2}}{2}+\frac{1}{2}\log \left(1+\sqrt{2}\right)}$

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．

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

よって，曲線の長さ$L$は公式より

${\displaystyle L={\displaystyle {\int }_0^1\sqrt{1+x^2}dx}}$

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

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

よって

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\sqrt{1+{\tan }^{2}t}}\cdot \frac{1}{{\cos }^{2}t}dt}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\sqrt{\frac{1}{{\cos }^{2}t}}}\cdot \frac{1}{{\cos }^{2}t}dt}$

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

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{1}{\cos t}}\cdot \frac{1}{{\cos }^{2}t}dt}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{3}t}dt}}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{\cos t}{{\cos }^{4}t}}dt}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{\cos t}{{(1-{\sin }^{2}t)}^2}}dt}$

${\displaystyle u=\sin t}$とおく．

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

よって

${\displaystyle ={\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{{(1-u^2)}^2}}du}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{{(1-u)}^2{(1+u)}^2}}du}$

ここで，被積分関数を部分分数に分解する．

・分子の計算

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

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

ここで

${\displaystyle a+b+c+d=1}$　･･････(1)

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

${\displaystyle a-b+c-d=0}$　･･････(3)

${\displaystyle -b+d=0}$　･･････(4)

である．(4)より

${\displaystyle d=b}$　･･････(5)

(5)を(2)に代入

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

${\displaystyle a-c=0}$

${\displaystyle a=c}$　･･････(6)

(5)を(3)に代入

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

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

これに(6)を代入

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

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

${\displaystyle a=b}$ 　･･････(7)

(7)と(5)より

${\displaystyle d=a}$ 　･･････(8)

(6)，(7)，(8)を(1)に代入

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

${\displaystyle 4a=1}$

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

よって

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

となる．したがって

${\displaystyle ={\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{{\left(1-u\right)}^2{\left(1+u\right)}^2}}}$

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

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

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

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

 

■別解

${\displaystyle L={\displaystyle {\int }_0^1\sqrt{1+x^2}dx}}$

部分積分法を用いて計算をする．

${\displaystyle ={\displaystyle {\int }_0^1{\left(x\right)}^{\text{'}}\sqrt{1+x^2}dx}}$

${\displaystyle ={\left[x\sqrt{1+x^2}\right]}_0^1-{\displaystyle {\int }_0^{1}x\frac{1}{2}\frac{2x}{\sqrt{1+x^2}}}dx}$

${\displaystyle =\sqrt{2}-{\displaystyle {\int }_0^1\frac{x^2}{\sqrt{1+x^2}}dx}}$

${\displaystyle =\sqrt{2}-{\displaystyle {\int }_0^1\frac{1+x^2-1}{\sqrt{1+x^2}}dx}}$

${\displaystyle =\sqrt{2}-L+{\left[\log \left|x+\sqrt{1+x^2}\right|\right]}_0^1}$

${\displaystyle =\sqrt{2}-L+\log \left(1+\sqrt{2}\right)}$

よって

${\displaystyle 2L=\sqrt{2}+\log \left(1+\sqrt{2}\right)}$

${\displaystyle L=\frac{\sqrt{2}}{2}+\frac{1}{2}\log \left(1+\sqrt{2}\right)}$

 

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最終更新日：2024年5月13日