6.84. Find the curvature, radius of curvature, and coordinates of the center of curvature of the curve $x=\alpha(t)=t-\sin t, y=\beta(t)=1-\cos t$ at the point $t=\pi / 2$.
Solution.

$$
\begin{aligned}
& K=\frac{\left|\frac{d^{2} \beta}{d t^{2}} \cdot \frac{d \alpha}{d t}-\frac{d \beta}{d t} \cdot \frac{d^{2} \alpha}{d t^{2}}\right|}{\left(\left(\frac{d \alpha}{d t}\right)^{2}+\left(\frac{d \beta}{d t}\right)^{2}\right)^{3 / 2}}=\frac{\left|\cos t(1-\cos t)-\sin ^{2} t\right|}{\left((1-\cos t)^{2}+\sin ^{2} t\right)^{3 / 2}} \\
& K\left(\frac{\pi}{2}\right)=\frac{1}{2 \sqrt{2}} \\
& R(t)=\frac{\left(\left(\frac{d \alpha}{d t}\right)^{2}+\left(\frac{d \beta}{d t}\right)^{2}\right)^{3 / 2}}{\left|\frac{d^{2} \beta}{d t^{2}} \cdot \frac{d \alpha}{d t}-\frac{d \beta}{d t} \cdot \frac{d^{2} \alpha}{d t^{2}}\right|}=\frac{\left((1-\cos t)^{2}+\sin ^{2} t\right)^{3 / 2}}{\left|\cos t(1-\cos t)-\sin ^{2} t\right|} \\
& \quad R\left(\frac{\pi}{2}\right)=2 \sqrt{2} \\
& X=x-\frac{\left((1+\cos t)^{2}+\sin ^{2} t\right) \sin t}{\cos t(1-\cos t)-\sin ^{2} t} \\
& \quad Y=y+\frac{\left((1-\cos t)^{2}+\sin ^{2} t\right)(1-\cos t)}{\cos t(1-\cos t)-\sin ^{2} t} \\
& X\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+1, \quad Y\left(\frac{\pi}{2}\right)=-1
\end{aligned}
$$