## Problem Statement

Calculate the lengths of the arcs of the curves given by the parametric equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2(t-\sin t) \\
y=2(1-\cos t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$
## Solution

The length of the arc of a curve defined by parametric equations is determined by the formula

$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$

Let's find the derivatives with respect to $t$ for the given curve:

$$
\begin{aligned}
& x=2(t-\sin t) ; x_{t}^{\prime}=2(1-\cos t) \\
& y=2(1-\cos t) ; y_{t}^{\prime}=2 \sin t
\end{aligned}
$$

We get:

$$
\begin{aligned}
L & =\int_{0}^{\pi / 2} \sqrt{4(1-\cos t)^{2}+4 \sin ^{2} t} d t= \\
& =2 \int_{0}^{\pi / 2} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t= \\
& =2 \int_{0}^{\pi / 2} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =2 \int_{0}^{\pi / 2} \sqrt{2(1-\cos t)} d t=\mid 0 \leq t \leq \pi / 2 ; 0 \leq t / 2 \leq \pi / 4 \\
& =2 \int_{0}^{\pi / 2} 2 \sin (t / 2) d t=4 \int_{0}^{\pi / 2} \sin (t / 2) d t= \\
& =-\left.8 \cos (t / 2)\right|_{0} ^{\pi / 2}=-8 \cdot(\cos (\pi / 4)-\cos (0))=-8 \cdot\left(\frac{\sqrt{2}}{2}-1\right)=4(2-\sqrt{2})
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2%D0%B0_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%BD%D0%B8%D0%BA_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_18-25»

Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals

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## Problem Kuznetsov Integrals 18-26

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