411. Find the curvature of the curve: 1) $x=t^{2}, y=2 t^{3}$ at the point where $t=1$; 2) $y=\cos 2 x$ at the point where $x=\frac{\pi}{2}$.
Solution. 1) We find the derivatives $\dot{x}=2 t, \ddot{x}=2, \dot{y}=6 t^{2}$, $\ddot{y}=12 t$, and compute their values at the point where $t=1$:

$$
\dot{x}=2, \ddot{x}=2, \dot{y}=6, \ddot{y}=12
$$

and, substituting into formula (1), we get

$$
K=\frac{|\ddot{x} \ddot{y}-\ddot{y} x|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{\frac{3}{2}}}=\frac{2 \cdot 12-0.2}{\left(2^{2}+6^{2}\right)^{\frac{3}{2}}}=\frac{3}{20 \sqrt{10}}
$$

2) From the given equation, we find the first and second derivatives of $y$ with respect to $x$:

$$
y^{\prime}=-2 \sin 2 x, y^{\prime \prime}=-4 \cos 2 x
$$

we compute their values at the given point: $y^{\prime}\left(\frac{\pi}{2}\right)=0, y^{\prime \prime}\left(\frac{\pi}{2}\right)=4$ and, substituting into formula (1), we get $K=\frac{\left|y^{\prime \prime}\right|}{\left[1+\left(y^{\prime}\right)^{2}\right]^{\frac{3}{2}}}=4$.