## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2.5(t-\sin t) \\
y=2.5(1-\cos t)
\end{array}\right. \\
& \frac{\pi}{2} \leq t \leq \pi
\end{aligned}
$$
## Solution

The length 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:

$$
\begin{aligned}
& x=2.5(t-\sin t) ; x_{t}^{\prime}=2.5(1-\cos t) \\
& y=2.5(1-\cos t) ; y_{t}^{\prime}=2.5 \sin t
\end{aligned}
$$

We get:

$$
\begin{aligned}
L & =\int_{\pi / 2}^{\pi} \sqrt{2.5^{2}(1-\cos t)^{2}+2.5^{2} \sin ^{2} t} d t= \\
& =2.5 \int_{\pi / 2}^{\pi} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t= \\
& =2.5 \int_{\pi / 2}^{\pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =2.5 \int_{\pi / 2}^{\pi} \sqrt{2(1-\cos t)} d t=\left\lvert\, \begin{array}{l}
\pi / 2 \leq t \leq \pi \\
\sin (t / 2) \geq 0
\end{array}\right. \\
& =2.5 \int_{\pi / 2}^{\pi} 2 \sin (t / 2) d t=5 \int_{\pi / 2}^{\pi} \sin (t / 2) d t= \\
& =-\left.10 \cos (t / 2)\right|_{\pi / 2} ^{\pi}=-10 \cdot(\cos (\pi / 2)-\cos (\pi / 4))=-10 \cdot\left(0-\frac{\sqrt{2}}{2}\right)=5 \sqrt{2}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+18-13$ ” Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals

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## Problem Kuznetsov Integrals 18-14

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