サイクロイドの微分・面積・体積・弧長

\(\dfrac{dy}{dt}=a\sin t\)
\(\dfrac{dx}{dt}=a(1-\cos t)\)
よって
\(\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\color{red}{\dfrac{\sin t}{1-\cos t}}\)

---

\(\dfrac{d^2y}{dx^2}\)
\(=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\)
\(=\dfrac{dt}{dx}\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\)
\(=\dfrac{1}{\dfrac{dx}{dt}}\dfrac{d}{dt}\left(\dfrac{\sin t}{1-\cos t}\right)\)
\(=\dfrac{1}{a(1-\cos t)}\dfrac{\cos t(1-\cos t) -\sin^2 t}{(1-\cos t)^2}\)
\(=\dfrac{\cos t -1}{a(1-\cos t)^3}\)
\(=\color{red}{-\dfrac{1}{a(1-\cos t)^2}}\)

\(\displaystyle S=\int_{0}^{2\pi a} y\,dx\)

ここで,
\(\begin{array}{l|l}
x & 0\rightarrow 2\pi a \\
\hline
t & 0\rightarrow 2\pi \\
\end{array}
\)

\(y=a(1-\cos t)\)
\(dx=\dfrac{dx}{dt}dt=a(1-\cos t)dt\)
より,
\(S\)
\(\displaystyle =\int_{0}^{2\pi}a^2 (1-\cos t)^2 dt\)
\(\displaystyle =a^2 \int_{0}^{2\pi}(1-2\cos t + \cos^2 t)dt\)
\(\displaystyle =a^2 \int_{0}^{2\pi}(1-2\cos t + \dfrac{1+\cos 2t}{2})dt\)
\(=a^2\left[\dfrac{3t}{2}-2\sin t+\dfrac{1}{4} \sin 2t\right]_{0}^{2\pi}\)
\(=a^2\cdot\dfrac{3}{2}\cdot 2\pi=\color{red}{3\pi a^2}\)

\(\displaystyle V=\pi\int_{0}^{2\pi a} y^2\,dx\)

ここで,
\(\begin{array}{l|l}
x & 0\rightarrow 2\pi a \\
\hline
t & 0\rightarrow 2\pi \\
\end{array}
\)

\(y=a(1-\cos t)\)
\(dx=\dfrac{dx}{dt}dt=a(1-\cos t)dt\)
より,
\(V\)
\(\displaystyle =\pi\int_{0}^{2\pi}a^3 (1-\cos t)^3 dt\)
\(\displaystyle =\pi\int_{0}^{2\pi}a^3(1-3\cos t +3\cos^2 t -\cos^3 t )dt\)
\(\displaystyle =\pi a^3\int_{0}^{2\pi}\left(1-3\cos t + 3\cdot\dfrac{1+\cos 2t}{2} -\dfrac{\cos 3t + 3\cos t}{4}\right) dt\)
\(=\pi a^3 \left[\color{blue}{t}-3\sin t +\color{blue}{\dfrac{3t}{2}}+\dfrac{3\sin 2t}{4}-\dfrac{\sin 3t}{12} -\dfrac{3\sin t}{4}\right]_{0}^{2\pi}\)
\(=\pi a^3\color{blue}{\dfrac{5}{2}}\cdot 2\pi=\color{red}{5{\pi}^2 a^3}\)

\(L\)
\(=\displaystyle \int_{0}^{2\pi}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,\,dt\)
\(=\displaystyle \int_{0}^{2\pi}\sqrt{a^2(1-\cos t)^2+ a^2\sin^2 t}\,\,dt\)
\(=\displaystyle a \int_{0}^{2\pi}\sqrt{1-2\cos t +\cos^2 t +\sin^2 t}\,\,dt\)
\(=\displaystyle a \int_{0}^{2\pi}\sqrt{2(1-\cos t)}\,\,dt\)
\(=\displaystyle a \int_{0}^{2\pi}\sqrt{2\cdot 2\sin^2 \dfrac{t}{2}}\,\,dt\)
\(=\displaystyle 2a \int_{0}^{2\pi}\sin \dfrac{t}{2}\,\,dt\)
\(=2a\left[-2\cos\dfrac{t}{2}\right]_0^{2\pi}\)
\(=2a\cdot(-2)\cdot(-1-1)=\color{red}{8a}\)