7. Two ants, one starting at $(-1,1)$, the other at $(1,1)$, walk to the right along the parabola $y=x^{2}$ such that their midpoint moves along the line $y=1$ with constant speed 1. When the left ant first hits the line $y=\frac{1}{2}$, what is its speed?
Solution: $\quad 3 \sqrt{3}-3$
When the left ant first hits the line $y=\frac{1}{2}$, the right ant hits the line $y=\frac{3}{2}$. The left ant is then at $\left(-\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$, and the right ant is at $\left(\frac{\sqrt{6}}{2}, \frac{3}{2}\right)$. Let the left ant have velocity with components $v_{x}$ and $v_{y}$, the right ant velocity with components $w_{x}$ and $w_{y}$. Since $\frac{d y}{d x}=2 x, \frac{v_{y}}{v_{x}}=-\sqrt{2}$ and $\frac{w_{y}}{w_{x}}=\sqrt{6}$. Since the midpoint of the ants moves at speed 1 along the line $y=1, \frac{1}{2}\left(v_{x}+w_{x}\right)=1$ and $\frac{1}{2}\left(v_{y}+w_{y}\right)=0$. Therefore, $\sqrt{2} v_{x}=-v_{y}=w_{y}=\sqrt{6} w_{x}=\sqrt{6}\left(2-v_{x}\right)$. Solving for $v_{x}$ gives $\frac{2 \sqrt{6}}{\sqrt{6}+\sqrt{2}}=3-\sqrt{3}$. Then the speed of the left ant is
$$
\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{v_{x}^{2}+\left(-\sqrt{2} v_{x}\right)^{2}}=\sqrt{3}\left|v_{x}\right|=3 \sqrt{3}-3 .
$$