Example 3. Find the length of the arc of one arch of the cycloid $x=a(t-\sin t)$, $y=a(1-\cos t)$.
Solution. A moving point describes one arch of a cycloid (Fig. 5.11) when $t$ changes from 0 to $2 \pi$. Therefore, in formula (5.19), $t_{1}=0, t_{2}=2 \pi$.

Let's find the expression for the integrand in formula (5.19). Differentiating the parametric equations of the cycloid, we get:

$$
x_{t}^{\prime}=a(1-\cos t) ; y_{t}^{\prime}=a \sin t
$$

Thus,

$$
\begin{gathered}
x_{t}^{\prime 2}+y_{t}^{\prime 2}=a^{2}\left(1-2 \cos t+\cos ^{2} t\right)+a^{2} \sin ^{2} t= \\
=a^{2}(1-2 \cos t+1)=2 a^{2}(1-\cos t)=4 a^{2} \sin ^{2} \frac{t}{2} \\
\sqrt{x^{\prime 2}+y^{\prime 2}}=\sqrt{4 a^{2} \sin ^{2} \frac{t}{2}}=2 a \sin \frac{t}{2}
\end{gathered}
$$

Substituting this expression into formula (5.19), we find

$$
\begin{aligned}
l=\int_{0}^{2 \pi} \sqrt{x^{\prime 2}+y^{\prime 2}} d t & =\int_{0}^{2 \pi} 2 a \sin \frac{t}{2} d t=2 a \int_{0}^{2 \pi} \sin \frac{t}{2} d t= \\
=4 a \int_{0}^{2 \pi} \sin \frac{t}{2} d\left(\frac{t}{2}\right) & =-\left.4 a \cos \frac{t}{2}\right|_{0} ^{2 \pi}=-4 a(\cos \pi-\cos 0)= \\
& =-4 a(-1-1)=8 a
\end{aligned}
$$

Thus, $l=8 a$. For example, when $a=\frac{1}{2} l=4$, and when $a=1 \quad l=8$.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-216.jpg?height=588&width=474&top_left_y=571&top_left_x=28)

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-216.jpg?height=396&width=451&top_left_y=669&top_left_x=611)

Fig. 5.15

## Fig. 5.14