Moving circular disk between two parallel sinusoidal curves Find the largest radius of the circle that can be "rolled" between the curves $y = sin(x)$ and $y = sin(x)+1$.
After two weeks of research, I finally give up.

 A: The diameter $d$ of the largest circle is given by the minimum distance between the curves, that is
$$d=\min_{x,y}\ \sqrt{ (y-x)^2 + (1+\sin(y)-\sin(x))^2} = \sqrt{\min_{u,v}\ 4u^2 + (1+2\sin(u)\cos(v))^2},$$
where $u=\frac{y-x}{2}$, $v=\frac{y+x}{2}$. Notice that any pair of $u,v$ uniquely defines $x,y$ (and vice verse) and thus we can minimize over $u,v$ instead of $x,y$. So, we have
$$d^2 = \min_{u,v}\ 4u^2 + (1+2\sin(u)\cos(v))^2 = \min_u\ \left(4u^2 + \min_v\ (1+2\sin(u)\cos(v))^2\right).$$
Without loss of generality, we assume that $\sin(u)\geq 0$. We split minimization over $u$ into two cases:
$$\min_{u} \ldots = \min\left\{ \min_{u,\ \sin(u)\geq \frac{1}{2}} \ldots, \min_{u,\ \sin(u)< \frac{1}{2}} \ldots\right\}.$$
Case 1: $\sin(u)\geq \frac{1}{2}$. Then by taking an appropriate $v$, we can make $(1+2\sin(u)\cos(v))^2=0$. Furthermore, $\sin(u)\geq \frac{1}{2}$ implies $|u|\geq \frac{\pi}{6}$. Thus, the minimum value for $d$ in this case is $\frac{\pi}{3}$.
Case 2: $\sin(u)<\frac{1}{2}$. To minimize the $(1+2\sin(u)\cos(v))^2$ in this case, we need to take $v$ such that $\cos(v)=-1$. Therefore, we need to compute
$$\min_u 4u^2+(1-2\sin(u))^2.$$
It is clear that this minimum does not exceed 1 and we can restrict $u$ to the interval $[0,\frac{1}{2}]$. The minimum is attained at the zero of the derivative $8u-4(1-2\sin(u))\cos(u)$, which is $u\approx 0.24737$, and gives the diameter $d \approx 0.71075$.
Clearly, case 2 delivers a smaller value and thus the largest diameter is $d \approx 0.71075$.
