## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=5(t-\sin t) \\
y=5(1-\cos t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$
## Solution

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

$$
\begin{aligned}
& x_{t}^{\prime}=5(1-\cos t) ; \quad y_{t}^{\prime}=5 \sin t \\
& L=5 \int_{0}^{\pi} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t=5 \int_{0}^{\pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =5 \int_{0}^{\pi} \sqrt{2-2 \cos t} d t=5 \int_{0}^{\pi} \sqrt{4 \sin ^{2}} \frac{t}{2} d t=10 \int_{0}^{\pi} \sin \frac{t}{2} d t=-\left.20 \cos \frac{t}{2}\right|_{0} ^{\pi}=20
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_18-1"

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

- Last edited on this page: 20:47, 9 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 18-2

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