与えられた曲線の長さを求める問題です。 (1) $y = \sqrt{1-x^2}$ ($0 \leq x \leq \frac{1}{2}$) (2) $y = 2x\sqrt{x}$ ($0 \leq x \leq 1$) (3) $y = x^2$ ($0 \leq x \leq 2$)

1. 問題の内容

与えられた曲線の長さを求める問題です。

(1) $y = \sqrt{1-x^2}$ ($0 \leq x \leq \frac{1}{2}$)

(2) $y = 2x\sqrt{x}$ ($0 \leq x \leq 1$)

(3) $y = x^2$ ($0 \leq x \leq 2$)

2. 解き方の手順

曲線の長さLは一般に、次の式で求められます。

$L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx$

(1) $y = \sqrt{1-x^2}$ の場合

$\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$

$(\frac{dy}{dx})^2 = \frac{x^2}{1-x^2}$

$1 + (\frac{dy}{dx})^2 = 1 + \frac{x^2}{1-x^2} = \frac{1}{1-x^2}$

$L = \int_{0}^{1/2} \sqrt{\frac{1}{1-x^2}} dx = \int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} dx = [\arcsin(x)]_{0}^{1/2} = \arcsin(\frac{1}{2}) - \arcsin(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}$

(2) $y = 2x\sqrt{x} = 2x^{3/2}$ の場合

$\frac{dy}{dx} = 3\sqrt{x}$

$(\frac{dy}{dx})^2 = 9x$

$1 + (\frac{dy}{dx})^2 = 1 + 9x$

$L = \int_{0}^{1} \sqrt{1 + 9x} dx$

$u = 1 + 9x$ とおくと、$du = 9 dx$ より $dx = \frac{1}{9} du$.

$x=0$ のとき $u=1$, $x=1$ のとき $u=10$.

$L = \int_{1}^{10} \frac{1}{9} \sqrt{u} du = \frac{1}{9} \int_{1}^{10} u^{1/2} du = \frac{1}{9} [\frac{2}{3} u^{3/2}]_{1}^{10} = \frac{2}{27} [u^{3/2}]_{1}^{10} = \frac{2}{27} (10^{3/2} - 1^{3/2}) = \frac{2}{27} (10\sqrt{10} - 1)$

(3) $y = x^2$ の場合

$\frac{dy}{dx} = 2x$

$(\frac{dy}{dx})^2 = 4x^2$

$1 + (\frac{dy}{dx})^2 = 1 + 4x^2$

$L = \int_{0}^{2} \sqrt{1 + 4x^2} dx = \int_{0}^{2} \sqrt{(2x)^2 + 1} dx$

ここで与えられた積分公式 $\int \sqrt{x^2 + A} dx = \frac{1}{2} x\sqrt{x^2 + A} + \frac{A}{2} \log|x + \sqrt{x^2 + A}| + C$ を用いる。

$2x = t$とおくと、$x=t/2$ より $dx = \frac{1}{2}dt$

$\int \sqrt{t^2 + 1} \frac{1}{2} dt = \frac{1}{2}\int \sqrt{t^2 + 1} dt = \frac{1}{2} (\frac{1}{2}t\sqrt{t^2 + 1} + \frac{1}{2} \log|t + \sqrt{t^2 + 1}|)$

$=\frac{1}{4}t\sqrt{t^2 + 1} + \frac{1}{4} \log|t + \sqrt{t^2 + 1}| = \frac{1}{4} 2x\sqrt{4x^2 + 1} + \frac{1}{4} \log|2x + \sqrt{4x^2 + 1}|$

$= \frac{1}{2} x\sqrt{4x^2 + 1} + \frac{1}{4} \log|2x + \sqrt{4x^2 + 1}|$

$L = [\frac{1}{2} x\sqrt{4x^2 + 1} + \frac{1}{4} \log|2x + \sqrt{4x^2 + 1}|]_{0}^{2} = \frac{1}{2} (2)\sqrt{4(2^2) + 1} + \frac{1}{4} \log|2(2) + \sqrt{4(2^2) + 1}| - 0 - \frac{1}{4} \log|1|$

$= \sqrt{17} + \frac{1}{4} \log(4 + \sqrt{17})$

3. 最終的な答え

(1) $\frac{\pi}{6}$

(2) $\frac{2}{27} (10\sqrt{10} - 1)$

(3) $\sqrt{17} + \frac{1}{4} \log(4 + \sqrt{17})$