803. Find the angle between the two curves $y=2 x^{2}$ and $y=x^{3}+$ $+2 x^{2}-1$ at their point of intersection.
Solution. Let's find the point of intersection of these curves. Solving the system of equations

$$
\left\{\begin{array}{l}
y=2 x^{2} \\
y=x^{3}+2 x^{2}-1
\end{array}\right.
$$

we get $x=1$ and $y=2$.

Thus, the point of intersection of the curves is $M_{0}(1 ; 2)$. Let's find the slopes of the tangents to the curves at $x=1$:

$$
\begin{gathered}
y^{\prime}=4 x, y^{\prime}(1)=4, \text { i.e., } k_{1}=4 \\
y^{\prime}=3 x^{2}+4 x, y^{\prime}(1)=7, \text { i.e., } k_{2}=7
\end{gathered}
$$

Using formula (3), we get

$$
\operatorname{tg} \varphi=\frac{7-4}{1+4 \cdot 7}=\frac{3}{29} \approx 0.1034
$$

from which

$$
\varphi \approx 5^{\circ} 54^{\prime}
$$