11.4. In a glass, the axial section of which is the graph of the function $y=x^{4}$, a "cherry" - a ball of radius $R$ - is placed. For which values of $R$ will the ball touch the bottom of the glass (the point $(0,0)$)?
## Solution.

We examine the indicated axial section. Since the axial section of the ball is a circle of radius $R$, then this circle must be tangent to the $O x$ axis at the point $O=(0 ; 0)$. Therefore, the center of the circle is on the $O y$ axis at the point $O_{1}=(0 ; R)$.

We write the condition that the graph of the circumference of radius $R$ with the center at the point $O_{1}=(0 ; R)$ and the graph of the function $y=x^{4}$ have common points (i.e., they either intersect or are tangent).

$$
\begin{aligned}
& \left\{\begin{array}{c}
x^{2}+(y-R)^{2}=R^{2} \\
y=x^{4}
\end{array}\right. \\
& x^{2}+\left(x^{4}-R\right)^{2}=R^{2} \\
& x^{2}\left(1+x^{6}-2 R x^{2}\right)=0
\end{aligned}
$$

The value $x_{1}=0$ gives the point of tangency of the graphs at the origin. Therefore, other points of tangency (or intersection) of the graphs are given by the condition

$$
1+x_{0}^{6}=2 R x_{0}^{2}, \quad x_{0} \neq 0
$$

Let $\quad x_{0}^{2}=t>0$. Then we have

$$
1+t^{3}=2 R t, \quad t>0
$$

This equation has a solution if and only if the line $y=2 R t$ (which passes through the point $(0,0)$) is tangent to or intersects the graph $G_{g}$ of the function $g(t)=1+t^{3}$ (for $t>0$).

We find the condition for the tangency of the graphs of the functions. The tangent to the graph $G_{g}$ has the equation

$$
y-g\left(t_{0}\right)=g^{\prime}\left(t_{0}\right) \cdot\left(t-t_{0}\right)
$$

Then

$$
y=1+t_{0}^{3}+3 t_{0}^{2} \cdot\left(t-t_{0}\right)=3 t_{0}^{2} \cdot t+\left(1-2 t_{0}^{3}\right)
$$

On the other hand $y=2 R t$. Therefore

$$
\left\{\begin{array} { c } 
{ 3 t _ { 0 } ^ { 2 } = 2 R } \\
{ 1 - 2 t _ { 0 } ^ { 3 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{c}
R_{0}=\frac{3 \sqrt[3]{2}}{4} \\
x_{0}= \pm \sqrt{t_{0}}= \pm \frac{1}{\sqrt[6]{2}}
\end{array}\right.\right.
$$

Thus:

1) For $0<R<R_{0}=\frac{3 \sqrt[3]{2}}{4}$, if the "cherry" is tangent to the bottom of the cup, it remains within the limits of the cup's walls. For this, it is sufficient to show that there exists $x_{0} \in[-R, R]$ such that

$$
f\left(x_{0}\right)=R-\sqrt{R^{2}-x_{0}^{2}}-x_{0}^{4}<0
$$

2) For $R>R_{0}=\frac{3 \sqrt[3]{2}}{4}$, if the "cherry" is tangent to the bottom of the cup, it exits the limits of the cup's walls. For this, it is sufficient to show that there exists $x_{0} \in[-R, R]$ such that

$$
f\left(x_{0}\right)=R-\sqrt{R^{2}-x_{0}^{2}}-x_{0}^{4}>0
$$

$$
R-\frac{1}{\sqrt[3]{4}} < R_{0}=\frac{3 \sqrt[3]{2}}{4}
$$

Solutions