放物線の長さの計算
はじめに
- この記事で伝えたい事:放物線の長さ(曲線の長さの具体例計算)
- 想定読者：高校生
- この記事を読む為にあった方がいい予備知識：高校数学微積分
- 構成,アウトライン：補題1,補題2を使って命題3を導き、最後に具体例を挙げた
問題
$f(x)=ax^2$の$x=\alpha$から$\beta$までの長さを求めて下さい。
$$\int\frac{1}{cos^3\theta}d\theta=\frac{1}{4}log\bigl|\frac{sin\theta+1}{sin\theta-1}\bigr|+\frac{1}{2}\frac{sin\theta}{cos^2\theta}+C\ \ \ \ \ (Cは積分定数)$$
補題1proof:
$$\int\frac{1}{cos^3\theta}d\theta=\int\frac{cos\theta}{cos^4\theta}d\theta=\int\frac{cos\theta}{(1-sin^2\theta)^2}d\theta=\int\frac{(sin\theta)'}{(1-sin^2\theta)^2}d\theta \\ =\int\frac{1}{(1-t^2)^2}dt=\int\frac{1}{(t^2-1)^2}dt=\int\frac{1}{(t-1)^2(t+1)^2}dt \\=\int\frac{\frac{-1}{4}}{(t-1)}+\frac{\frac{1}{4}}{(t+1)}+\frac{\frac{1}{4}}{(t-1)^2}+\frac{\frac{1}{4}}{(t+1)^2}dt \\=\frac{1}{4}\int\frac{-1}{(t-1)}+\frac{1}{(t+1)}+\frac{1}{(t-1)^2}+\frac{1}{(t+1)^2}dt \\=\frac{1}{4}\bigl(-log|t-1|+log|t+1|-\frac{1}{t-1}-\frac{1}{t+1}\bigr)+C \\=\frac{1}{4}\bigl(-log|sin\theta-1|+log|sin\theta+1|-\frac{1}{sin\theta-1}-\frac{1}{sin\theta+1}\bigr)+C \\=\frac{1}{4}\bigl(-log|sin\theta-1|+log|sin\theta+1|-\bigl(\frac{1}{sin\theta-1}+\frac{1}{sin\theta+1}\bigr)\bigr)+C \\=\frac{1}{4}\bigl(-log\bigl|\frac{sin\theta+1}{sin\theta-1}\bigr|-\frac{2sin\theta}{sin^2\theta-1}\bigr)+C \\=\frac{1}{4}log\bigl|\frac{sin\theta+1}{sin\theta-1}\bigr|+\frac{1}{2}\frac{sin\theta}{cos^2\theta}+C\ \ \ \ \ \ \square $$
$$
\displaystyle sin(arctan(x))=\frac{x}{\sqrt{1+x^2}}
$$
※ arctanは$tan:\bigl[\frac{-\pi}{2},\frac{\pi}{2}\bigr]\rightarrow \mathbb{R}$の逆関数
補題2proof:
$\displaystyle sin(arctan(x))=\frac{x}{\sqrt{1+x^2}}$を示す。arctanはtanの逆関数なのだから
$tan(arctan(x))=x$と$\displaystyle 1+tan^2 \theta = \frac{1}{cos^2 \theta}$を用いて、
$$ 1+tan^2 (arctan(x)) = \frac{1}{cos^2 (arctan(x))} \Leftrightarrow \\
\displaystyle 1+x^2=\frac{1}{cos^2 (arctan(x))} $$
arctanの値域が$\displaystyle \bigl(\frac{-\pi}{2},\frac{\pi}{2}\bigr)$である事より$\displaystyle cos (arctan(x))=\frac{1}{\sqrt{1+x^2}}$
よって、$$
sin(arctan(x))=cos(arctan(x)) \cdot tan(arctan(x))\\
=\frac{1}{\sqrt{1+x^2}}\cdot x =\frac{x}{\sqrt{1+x^2}}\ \ \ \ \ \ \square
$$
$f(x)=ax^2$の$x=\alpha$から$\beta$までの長さは
$$\int_{\alpha}^{\beta}\sqrt{1+\bigl(\frac{d(ax^2)}{dx}\bigr)^2}dx\\
=\int_{\alpha}^{\beta}\sqrt{1+4a^2x^2}dx\\
=\frac{1}{2}log\bigl|\frac{2a\beta+\sqrt{1+4a^2\beta^2}}{2a\alpha+\sqrt{1+4a^2\alpha^2}}\bigr|+a\bigl(\beta\sqrt{1+4a^2\beta^2}-\alpha\sqrt{1+4a^2\alpha^2}\bigr)
$$
で与えられる。
命題3proof:
$$\int_{\alpha}^{\beta}\sqrt{1+\bigl(\frac{d(ax^2)}{dx}\bigr)^2}dx=\int_{\alpha}^{\beta}\sqrt{1+4a^2x^2}dx=(1)$$
$\displaystyle x=\frac{tan\theta}{2a};\theta\in \bigl[\frac{-\pi}{2},\frac{\pi}{2}\bigr]$とすると
$\displaystyle 1+4a^2x^2=1+tan^2\theta=\frac{1}{cos^2\theta},\\ \displaystyle dx=\frac{dx}{d\theta}d\theta=\frac{1}{2acos^2\theta}d\theta$
また$\displaystyle \alpha=\frac{tan\theta}{2a},\beta=\frac{tan\theta}{2a}$をそれぞれ解くと$\theta=arctan(2a\alpha),\theta=arctan(2a\beta)$より、
$$(1)=\int_{arctan(2a\alpha)}^{arctan(2a\beta)}\sqrt{\frac{1}{cos^2\theta}}\frac{1}{2acos^2\theta}d\theta=\frac{1}{2a}\int_{arctan(2a\alpha)}^{arctan(2a\beta)}\frac{1}{cos^3\theta}d\theta=(2)$$
補題1より、$$(2)=\bigl[\frac{1}{4}log\bigl|\frac{sin\theta+1}{sin\theta-1}\bigr|+\frac{1}{2}\frac{sin\theta}{cos^2\theta}\bigr]_{arctan(2a\alpha)}^{arctan(2a\beta)}
\\
=\frac{1}{4}log\bigl|\frac{sin(arctan(2a \beta))+1}{sin(arctan(2a \beta))-1}\bigr|+\frac{1}{2}\frac{sin(arctan(2a \beta))}{cos^2 (arctan(2a \beta))}-\frac{1}{4}log\bigl|\frac{sin (arctan(2a \alpha))+1}{sin (arctan(2a \alpha))-1}\bigr|-\frac{1}{2}\frac{sin (arctan(2a \alpha))}{cos^2 (arctan(2a \alpha))}
\\
=\frac{1}{4}log\bigl|\frac{sin(arctan(2a \beta))+1}{sin(arctan(2a \beta))-1}\bigr|+\frac{1}{2}\frac{tan^2(arctan(2a\beta))}{sin(arctan(2a \beta))}-\frac{1}{4}log\bigl|\frac{sin (arctan(2a \alpha))+1}{sin (arctan(2a \alpha))-1}\bigr|-\frac{1}{2}\frac{tan^2(arctan(2a \alpha))}{sin (arctan(2a \alpha))}
\\
=(3)
$$補題2より
$$(3)=
\frac{1}{4}log\bigl|\frac{\frac{2a\beta}{\sqrt{1+4a^2\beta^2}}+1}{\frac{2a\beta}{\sqrt{1+4a^2\beta^2}}-1}\bigr|+\frac{1}{2}\frac{4a^2\beta^2}{\frac{2a\beta}{\sqrt{1+4a^2\beta^2}}}-\frac{1}{4}log\bigl|\frac{\frac{2a\alpha}{\sqrt{1+4a^2\alpha^2}}+1}{\frac{2a\alpha}{\sqrt{1+4a^2\alpha^2}}-1}\bigr|-\frac{1}{2}\frac{4a^2\alpha^2}{\frac{2a\alpha}{\sqrt{1+4a^2\alpha^2}}}
\\=
\frac{1}{4}log\bigl|\frac{2a\beta+\sqrt{1+4a^2\beta^2}}{2a\beta-\sqrt{1+4a^2\beta^2}}\bigr|+(\sqrt{1+4a^2\beta^2})a\beta-\frac{1}{4}log\bigl|\frac{2a\alpha+\sqrt{1+4a^2\alpha^2}}{2a\alpha-\sqrt{1+4a^2\alpha^2}}\bigr|-(\sqrt{1+4a^2\alpha^2})a\alpha
\\
=
\frac{1}{4}log\bigl|\frac{\bigl(2a\beta+\sqrt{1+4a^2\beta^2}\bigr)^2}{(-1)}\bigr|-\frac{1}{4}log\bigl|\frac{\bigl(2a\alpha+\sqrt{1+4a^2\alpha^2}\bigr)^2}{(-1)}\bigr|+a\bigl(\beta\sqrt{1+4a^2\beta^2}-\alpha\sqrt{1+4a^2\alpha^2}\bigr)
\\
=
\frac{1}{4}log\bigl|\frac{\bigl(2a\beta+\sqrt{1+4a^2\beta^2}\bigr)^2}{\bigl(2a\alpha+\sqrt{1+4a^2\alpha^2}\bigr)^2}\bigr|+a\bigl(\beta\sqrt{1+4a^2\beta^2}-\alpha\sqrt{1+4a^2\alpha^2}\bigr)
\\
=
\frac{1}{2}log\bigl|\frac{2a\beta+\sqrt{1+4a^2\beta^2}}{2a\alpha+\sqrt{1+4a^2\alpha^2}}\bigr|+a\bigl(\beta\sqrt{1+4a^2\beta^2}-\alpha\sqrt{1+4a^2\alpha^2}\bigr)\ \ \ \ \ \ \square
$$
具体的に$\displaystyle a=\frac{1}{2},\alpha=0,\beta=1$を適用する。
曲線の長さ：
$$ \int_{0}^{1}\sqrt{1+\bigl(\frac{d}{dx}\bigl(\frac{x^2}{2}\bigr)\bigr)^2}dx=\int_{0}^{1}\sqrt{1+x^2}dx \\=\frac{1}{2}log\bigl|\frac{1+\sqrt{2}}{1}\bigr|+\frac{1}{2}\cdot\sqrt{2}=\frac{1}{2}log(1+\sqrt{2})+\frac{1}{\sqrt{2}} =1.147\cdots $$
最短距離(直線)：
$(0,0)$から$\displaystyle \bigl(1,\frac{1}{2}\bigr)$であるから、$$ \sqrt{1^2+\bigl(\frac{1}{2}\bigr)^2}=\sqrt{\frac{5}{4}}=\frac{\sqrt5}{2}=\frac{2.2362\cdots}{2}=1.118\cdots$$
となっており、確かに最短距離である直線よりも曲線の長さの方が少々長い。
