In each step, the ant at point ($x_0$, $y_0$) chooses one of the lattice points ($x_1$, $y_1$) which satisfy $x_1 ≥ 0$ and $y_1 ≥ 1$ and goes straight to ($x_1$, $y_1$) at a constant velocity $v$. The value of $v$ depends on $y_0$ and $y_1$ as follows:
The left image is one of the possible paths for $d = 4$. First the ant goes from $A(0, 1)$ to $P_1(1, 3)$ at velocity $\frac{3 - 1}{\ln 3 - \ln 1} ≈ 1.8205$. Then the required time is $\frac{\sqrt{5}}{1.820} ≈ 1.2283$.
From $P_1(1, 3)$ to $P_2(3, 3)$ the ant travels at velocity 3 so the required time is $\frac{2}{3} ≈ 0.6667$. From $P_2(3, 3)$ to $B(4, 1)$ the ant travels at velocity $\frac{1 - 3}{\ln 1 - \ln 3} ≈ 1.8205$ so the required time is $\frac{\sqrt{5}}{1.8205} ≈ 1.2283$.
Thus the total required time is $1.2283 + 0.6667 + 1.2283 = 3.1233$.
The right image is another path. The total required time is calculated as $0.98026 + 1 + 0.98026 = 2.96052$. It can be shown that this is the quickest path for $d = 4$.
<imgclass="img-responsive center-block"alt="two possible paths for d = 4"src="https://cdn.freecodecamp.org/curriculum/project-euler/an-ant-on-the-move.jpg"style="background-color: white; padding: 10px;">
Let $F(d)$ be the total required time if the ant chooses the quickest path. For example, $F(4) ≈ 2.960\\,516\\,287$. We can verify that $F(10) ≈ 4.668\\,187\\,834$ and $F(100) ≈ 9.217\\,221\\,972$.
Find $F(10\\,000)$. Give your answer rounded to nine decimal places.