# 曲線の長さ

$s={\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$$={\int }_{\alpha }^{\beta }\sqrt{{\left\{{u}^{\prime }\left(t\right)\right\}}^{2}+{\left\{{v}^{\prime }\left(t\right)\right\}}^{2}}dt$

$s={\int }_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$$={\int }_{a}^{b}\sqrt{1+{\left\{{f}^{\prime }\left(x\right)\right\}}^{2}}dx$

となる．ただし，$a=u\left(\alpha \right)$$b=u\left(\beta \right)$ である．

## ■導出

$\Delta {s}_{i}$$=\sqrt{{\left(\Delta {x}_{i}\right)}^{2}+{\left(\Delta {y}_{i}\right)}^{2}}$

$=\sqrt{{\left(\frac{\Delta {x}_{i}}{\Delta {t}_{i}}\right)}^{2}+{\left(\frac{\Delta {y}_{i}}{\Delta {t}_{i}}\right)}^{2}}\Delta {t}_{i}$

$\underset{n\to \infty }{\mathrm{lim}}\sum _{i=1}^{n}\sqrt{{\left(\frac{\Delta {x}_{i}}{\Delta {t}_{i}}\right)}^{2}+{\left(\frac{\Delta {y}_{i}}{\Delta {t}_{i}}\right)}^{2}}\Delta {t}_{i}$ $={\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$

$={\int }_{\alpha }^{\beta }\sqrt{{\left\{{u}^{\prime }\left(t\right)\right\}}^{2}+{\left\{{v}^{\prime }\left(t\right)\right\}}^{2}}dt$

となる． 一方

$\Delta {s}_{i}$$=\sqrt{{\left(\Delta {x}_{i}\right)}^{2}+{\left(\Delta {y}_{i}\right)}^{2}}$$=\sqrt{1+{\left(\frac{\Delta {y}_{i}}{\Delta {x}_{i}}\right)}^{2}}\Delta {x}_{i}$

と考えると，曲線$\text{AB}$  $\left(a\leqq x\leqq b\right)$ の長さは

$\underset{n\to \infty }{\mathrm{lim}}\sum _{i=1}^{n}\sqrt{1+{\left(\frac{\Delta {y}_{i}}{\Delta {x}_{i}}\right)}^{2}}\Delta {x}_{i}$$={\int }_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$

$={\int }_{a}^{b}\sqrt{1+{\left\{{f}^{\prime }\left(x\right)\right\}}^{2}}dx$

となりる．

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