Example 5. Find the radius of curvature of the cycloid $x=a(t-\sin t), y=$ $=a(1-\cos t)$ at any point of it.
Solution. From formulas (3.4) and (3.5), it follows that the radius of curvature of a curve given by parametric equations is determined by the formula

$$
R=\frac{\left(x^{\prime 2}+y^{\prime 2}\right)^{\frac{3}{2}}}{\left|x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}\right|}
$$

Let's find the first and second derivatives of the functions $x$ and $y$:

$$
x^{\prime}=a(1-\cos t) ; y^{\prime}=a \sin t ; \quad x^{\prime \prime}=a \sin t ; y^{\prime \prime}=a \cos t
$$

We will form the expressions $x^{\prime 2}+y^{\prime 2}$ and $\left(x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}\right)$:

$$
x^{\prime 2}+y^{\prime 2}=a^{2}(1-\cos t)^{2}+a^{2} \sin ^{2} t=a^{2}\left(1-2 \cos t+\cos ^{2} t+\right.
$$

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

Substituting the obtained expressions into the formula for the radius of curvature, we find

$$
R=\frac{\left[2 a^{2}(1-\cos t)\right]^{\frac{3}{2}}}{\left|-a^{2}(1-\cos t)\right|}=\frac{2 \sqrt{2} a^{3}(1-\cos t)^{\frac{3}{2}}}{a^{2}(1-\cos t)}=2 \sqrt{2} a \sqrt{1-\cos t}
$$

Since $\sqrt{1-\cos t}=\sqrt{2}\left|\sin \frac{t}{2}\right|$, then

$$
R=4 a\left|\sin \frac{t}{2}\right| \text {. }
$$

In particular, if $t=\pi$, then $R=4 a$; if $t=0$, then $R=0$.