1.数直線上を運動する点\(P\)の速度が時刻\(t\)の関数として\(v=4-2t\)で与えられている。\(t=0\)における\(P\)の座標が\(2\)のとき、\(t=3\)のときの\(P\)の座標を求めなさい。
\(P\)の座標\(x\)は
\(\displaystyle x=2+\int_0^3 (4-2t)dt\)
\(\displaystyle =2+[4t-t^2]_0^3\)
\(\displaystyle =5\)
2.数直線上を運動する点\(P\)があり、時刻\(t\)における点\(P\)の速さは\(v=\sin2t\)であるとする。\(t=0\)から\(t=\pi\)までに\(P\)が通過する道のり\(s\)を求めなさい。
\(\displaystyle s=\int_0^\pi |\sin2t|dt\)
\(\displaystyle =\int_0^\frac{\pi}{2} \sin2tdt-\int_\frac{\pi}{2}^\pi \sin2tdt\)
\(\displaystyle =\left[-\frac{1}{2}\cos2t\right]_0^\frac{\pi}{2}+\left[\frac{1}{2}\cos2t\right]_\frac{\pi}{2}^\pi\)
\(\displaystyle =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(\displaystyle =2\)
3.座標平面上を運動する点\(P\)の時刻\(t\)における座標\((x,y)\)が\(x=e^{-t}\cos\pi t,y=e^{-t}\sin\pi t\)で表されるとき、\(t=0\)から\(t=2\)までに\(P\)が通過する道のり\(s\)を求めなさい。
\(\displaystyle \frac{dx}{dt}=-e^{-t}\cos\pi t-e^{-t}\pi\sin\pi t\)
\(\displaystyle =-e^{-t}(\cos\pi t+\pi\sin\pi t)\)
\(\displaystyle \frac{dy}{dt}=-e^{-t}\sin\pi t+e^{-t}\pi\cos\pi t\)
\(\displaystyle =-e^{-t}(\sin\pi t-\pi\cos\pi t)\)
道のり\(s\)は
\(\displaystyle s=\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\)
\(\displaystyle =\int_0^2 \sqrt{(-e^{-2t}(\cos\pi t+\pi\sin\pi t)^2+(-e^{-2t}(\sin\pi t-\pi\cos\pi t)^2}dt\)
\(\displaystyle =\int_0^2 \sqrt{e^{-2t}(1+\pi^2)}dt\)
\(\displaystyle =\sqrt{1+\pi^2}\int_0^2 e^{-t} dt\)
\(\displaystyle =\sqrt{1+\pi^2}[-e^{-t}]_0^2\)
\(\displaystyle =\sqrt{1+\pi^2}\left(1-\frac{1}{e^2}\right)\)
4.次の曲線の長さ\(s\)を求めなさい。
(1)サイクロイド\(\displaystyle x=\theta-\sin\theta,y=1-\cos\theta(0\leqq\theta\leqq2\pi)\)
\(\displaystyle \frac{dx}{d\theta}=1-\cos\theta\)
\(\displaystyle \frac{dy}{d\theta}=\sin\theta\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{(1-\cos\theta)^2+(\sin\theta)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{2(1-\cos\theta)}d\theta\)
\(\displaystyle =\int_0^{2\pi} 2\sin\frac{\theta}{2} d\theta\)
\(\displaystyle =2\left[-2\cos\frac{\theta}{2}\right]_0^{2\pi}\)
\(\displaystyle =-4(-1-1)\)
\(\displaystyle =8\)
(2)サイクロイド\(\displaystyle x=2(\theta-\sin\theta),y=2(\theta-\cos\theta)(0\leqq\theta\leqq2\pi)\)
\(\displaystyle \frac{dx}{d\theta}=2(1-\cos\theta)\)
\(\displaystyle \frac{dy}{d\theta}=2\sin\theta\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{2^2・(1-\cos\theta)^2+4\sin^2\theta}d\theta\)
\(\displaystyle =\int_0^{2\pi} \sqrt{8(1-\cos\theta)}d\theta\)
\(\displaystyle =\sqrt{8}\int_0^{2\pi} \sqrt{2}\sin\frac{\theta}{2} d\theta\)
\(\displaystyle =-8\left[\cos\frac{\theta}{2}\right]_0^{2\pi}\)
\(\displaystyle =-8(-1-1)\)
\(\displaystyle =16\)
(3)曲線\(y=x\sqrt{x}(0\leqq x\leqq 5)\)
\(\displaystyle \frac{dy}{dx}=\frac{3}{2}x^{\frac{1}{2}}\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_0^5 \sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}dx\)
\(\displaystyle =\int_0^5 \sqrt{1+\frac{9}{4}x}dx\)
\(\displaystyle =\frac{2}{3}・\frac{4}{9}\left[\left(1+\frac{9}{4}x\right)^{\frac{3}{2}}\right]_0^5\)
\(\displaystyle =\frac{8}{27}\left(\left(\frac{49}{4}\right)^{\frac{3}{2}}-1\right)\)
\(\displaystyle =\frac{335}{27}\)
(4)曲線\(\displaystyle y=\frac{1}{3}(x-3)\sqrt{x}(1\leqq x\leqq 4)\)
\(\displaystyle \frac{dy}{dx}=\frac{2x+(x-3)}{6\sqrt{x}}=\frac{x-1}{2\sqrt{x}}\)
曲線の長さ\(s\)は
\(\displaystyle s=\int_1^4 \sqrt{1+\left(\frac{x-1}{2\sqrt{x}}\right)^2}dx\)
\(\displaystyle =\int_1^4 \left(\frac{1}{2}x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{1}{2}}\right)dx\)
\(\displaystyle =\frac{1}{2}\left[\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}\right]_1^4\)
\(\displaystyle =\frac{1}{2}\left(\left(\frac{2}{3}・8+2・2\right)-\left(\frac{2}{3}+2\right)\right)\)
\(\displaystyle =\frac{10}{3}\)