# Calculus/Arc length/Solutions

1. Find the length of the curve $y=x^{3/2}$ from $x=0$ to $x=1$.
\begin{align}s&=\int_{0}^{1}\sqrt{1+\left(\frac{d\left(x^{3/2}\right)}{dx}\right)^{2}}dx\\ &=\int_{0}^{1}\sqrt{1+\left(\frac{3}{2}\sqrt{x}\right)^{2}}dx\\ &=\int_{0}^{1}\sqrt{1+\frac{9}{4}x}dx\end{align}

Let

$u=1+\frac{9}{4}x;\qquad du=\frac{9}{4}dx;\qquad dx=\frac{4}{9}du$

Then

\begin{align}s&=\int_{u(0)}^{u(1)}\sqrt{u}\frac{4}{9}du\\ &=\frac{4}{9}\frac{2}{3}u^{3/2}\biggr|_{u(0)}^{u(1)}\\ &=\frac{8}{27}\left(1+\frac{9}{4}x\right){}^{3/2}\biggr|_{0}^{1}\\ &=\frac{8}{27}\left(\left(1+\frac{9}{4}\right){}^{3/2}-1\right)\\ &=\frac{8}{27}\left(\left(\frac{13}{4}\right){}^{3/2}-1\right)\\ &=\frac{8}{27}\left(\left(\frac{13}{2^{2}}\right){}^{3/2}-1\right)\\ &=\frac{8}{27}\left(\frac{13^{3/2}}{2^{3}}-1\right)\\ &=\frac{8}{27}\left(\frac{13^{3/2}}{8}-1\right)\\ &=\mathbf{\frac{13^{3/2}-8}{27}}\end{align}
2. Find the length of the curve $y=\frac{e^x +e^{-x}}{2}$ from $x=0$ to $x=1$.
\begin{align}s&=\int_{0}^{1}\sqrt{1+\left(\frac{d\left(\frac{e^{x}+e^{-x}}{2}\right)}{dx}\right)^{2}}dx\\ &=\int_{0}^{1}\sqrt{1+\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}}dx\\ &=\int_{0}^{1}\sqrt{1+\frac{e^{2x}-2+e^{-2x}}{4}}dx\\ &=\int_{0}^{1}\sqrt{\frac{e^{2x}+2+e^{-2x}}{4}}dx\\ &=\int_{0}^{1}\sqrt{\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}}dx\\ &=\int_{0}^{1}\frac{e^{x}+e^{-x}}{2}dx\\ &=\frac{e^{x}-e^{-x}}{2}\biggr|_{0}^{1}\\ &=\mathbf{\frac{e-\frac{1}{e}}{2}}\end{align}
3. Find the circumference of the circle given by the parametric equations $x(t)=R\cos(t)$, $y(t)=R\sin(t)$, with $t$ running from $0$ to $2\pi$.
\begin{align}s&=\int_{0}^{2\pi}\sqrt{\left(\frac{d(R\cos(t))}{dt}\right)^{2}+\left(\frac{d(R\sin(t))}{dt}\right)^{2}}dt\\ &=\int_{0}^{2\pi}\sqrt{\left(-R\sin(t)\right)^{2}+\left(R\cos(t)\right)^{2}}dt\\ &=\int_{0}^{2\pi}\sqrt{R^{2}(\sin^{2}(t)+\cos^{2}(t))}dt\\ &=\int_{0}^{2\pi}Rdt\\ &=Rt\bigr|_{0}^{2\pi}\\ &=\mathbf{2\pi R}\end{align}
4. Find the length of one arch of the cycloid given by the parametric equations $x(t)=R(t-\sin(t))$, $y(t)=R(1-\cos(t))$, with $t$ running from $0$ to $2\pi$.
\begin{align}s&=\int_{0}^{2\pi}\sqrt{\left(\frac{d(R(t-\sin(t)))}{dt}\right)^{2}+\left(\frac{d(R(1-\cos(t)))}{dt}\right)^{2}}dt\\ &=\int_{0}^{2\pi}\sqrt{\left(R(1-\cos(t))\right)^{2}+\left(R\sin(t)\right)^{2}}dt\\ &=\int_{0}^{2\pi}R\sqrt{\left(1-\cos(t)\right)^{2}+\sin^{2}(t)}dt\\ &=\int_{0}^{2\pi}R\sqrt{1-2\cos(t)+\cos^{2}(t)+\sin^{2}(t)}dt\\ &=\int_{0}^{2\pi}R\sqrt{2-2\cos(t)}dt\end{align}

Using the trigonometric identity

$\sin^{2}\left(\frac{t}{2}\right)=\frac{1-\cos\left(t\right)}{2}$

, we have

\begin{align}s&=\int_{0}^{2\pi}R\sqrt{4\sin^{2}\left(\frac{t}{2}\right)}dt\\ &=\int_{0}^{2\pi}2R\sin\left(\frac{t}{2}\right)dt\\ &=-4R\cos\left(\frac{t}{2}\right)\biggr|_{0}^{2\pi}\\ &=-4R(-1-1)\\ &=\mathbf{8R}\end{align}