【高校数学Ⅲ】6-1-3 曲線の長さ｜問題集

\(P\)の座標\(x\)は
\(\displaystyle x=2+\int_0^3 (4-2t)dt\)
\(\displaystyle =2+[4t-t^2]_0^3\)
\(=5\)

\(\displaystyle s=\int_0^\pi |\sin2t|dt\)
\(\displaystyle =\int_0^\frac{\pi}{2} \sin2tdt-\int_\frac{\pi}{2}^\pi \sin2tdt\)
\(\displaystyle =\left[-\frac{1}{2}\cos2t\right]_0^\frac{\pi}{2}+\left[\frac{1}{2}\cos2t\right]_\frac{\pi}{2}^\pi\)
\(\displaystyle =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=2\)

\(\displaystyle \frac{dx}{dt}=-e^{-t}\cos\pi t-e^{-t}\pi\sin\pi t\)
\(\displaystyle =-e^{-t}(\cos\pi t+\pi\sin\pi t)\)
\(\displaystyle \frac{dy}{dt}=-e^{-t}\sin\pi t+e^{-t}\pi\cos\pi t\)
\(\displaystyle =-e^{-t}(\sin\pi t-\pi\cos\pi t)\)
道のり\(s\)は
\(\displaystyle s=\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\)
\(\displaystyle =\int_0^2 \sqrt{e^{-2t}(1+\pi^2)}dt\)
\(\displaystyle =\sqrt{1+\pi^2}\int_0^2 e^{-t} dt\)
\(\displaystyle =\sqrt{1+\pi^2}[-e^{-t}]_0^2\)
\(\displaystyle =\sqrt{1+\pi^2}\left(1-\frac{1}{e^2}\right)\)

\(\displaystyle \frac{dx}{d\theta}=1-\cos\theta\)
\(\displaystyle \frac{dy}{d\theta}=\sin\theta\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{(1-\cos\theta)^2+(\sin\theta)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{2(1-\cos\theta)}d\theta\)
\(\displaystyle =\int_0^{2\pi} 2\sin\frac{\theta}{2} d\theta\)
\(\displaystyle =2\left[-2\cos\frac{\theta}{2}\right]_0^{2\pi}\)
\(=-4(-1-1)\)
\(=8\)

\(\displaystyle \frac{dx}{d\theta}=2(1-\cos\theta)\)
\(\displaystyle \frac{dy}{d\theta}=2\sin\theta\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{2^2・(1-\cos\theta)^2+4\sin^2\theta}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{8(1-\cos\theta)}d\theta\)
\(\displaystyle =\sqrt{8}\int_0^{2\pi} \sqrt{2}\sin\frac{\theta}{2} d\theta\)
\(\displaystyle =-8\left[\cos\frac{\theta}{2}\right]_0^{2\pi}\)
\(=-8(-1-1)\)
\(=16\)

\(\displaystyle \frac{dy}{dx}=\frac{3}{2}x^{\frac{1}{2}}\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^5 \sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}dx\)
\(\displaystyle =\int_0^5 \sqrt{1+\frac{9}{4}x}dx\)
\(\displaystyle =\frac{2}{3}・\frac{4}{9}\left[\left(1+\frac{9}{4}x\right)^{\frac{3}{2}}\right]_0^5\)
\(\displaystyle =\frac{8}{27}\left(\left(\frac{49}{4}\right)^{\frac{3}{2}}-1\right)\)
\(\displaystyle =\frac{335}{27}\)

\(\displaystyle \frac{dy}{dx}=\frac{2x+(x-3)}{6\sqrt{x}}=\frac{x-1}{2\sqrt{x}}\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_1^4 \sqrt{1+\left(\frac{x-1}{2\sqrt{x}}\right)^2}dx\)
\(\displaystyle =\int_1^4 \left(\frac{1}{2}x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{1}{2}}\right)dx\)
\(\displaystyle =\frac{1}{2}\left[\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}\right]_1^4\)
\(\displaystyle =\frac{1}{2}\left(\left(\frac{2}{3}・8+2・2\right)-\left(\frac{2}{3}+2\right)\right)\)
\(\displaystyle =\frac{10}{3}\)