## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(t-\sin t) \\
y=3(1-\cos t)
\end{array}\right. \\
& \pi \leq t \leq 2 \pi
\end{aligned}
$$
## 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:

$$
\begin{aligned}
& x=3(t-\sin t) ; x_{t}^{\prime}=3(1-\cos t) \\
& y=3(1-\cos t) ; y_{t}^{\prime}=3 \sin t
\end{aligned}
$$

We get:

$$
\begin{aligned}
L & =\int_{\pi}^{2 \pi} \sqrt{9(1-\cos t)^{2}+9 \sin ^{2} t} d t= \\
& =3 \int_{\pi}^{2 \pi} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t= \\
& =3 \int_{\pi}^{2 \pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =3 \int_{\pi}^{2 \pi} \sqrt{2(1-\cos t)} d t=\left|\begin{array}{c}
\pi \leq t \leq 2 \pi ; \pi / 2 \leq t / 2 \leq \pi ; \sin (t / 2) \geq 0 \\
\sqrt{2(1-\cos t)}=\sqrt{4 \sin ^{2}(t / 2)}=2 \sin (t / 2)
\end{array}\right|= \\
& =3 \int_{\pi}^{2 \pi} 2 \sin (t / 2) d t=6 \int_{\pi}^{2 \pi} \sin (t / 2) d t=-\left.12 \cos (t / 2)\right|_{\pi} ^{2 \pi}=-12 \cdot(-1-0)=12
\end{aligned}
$$

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

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## Problem Kuznetsov Integrals 18-8

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