On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.

In each step, the ant at point (x₀, y₀) chooses one of the lattice points (x₁, y₁) which satisfy x₁ ≥ 0 and y₁ ≥ 1 and goes straight to (x₁, y₁) at a constant velocity v. The value of v depends on y₀ and y₁ as follows:

• If y₀ = y₁, the value of v equals y₀.
• If y₀ ≠ y₁, the value of v equals (y₁ - y₀) / (ln(y₁) - ln(y₀)).

The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P₁(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
From P₁(1, 3) to P₂(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P₂(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is 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.

Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960516287.
We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.

Find F(10000). Give your answer rounded to nine decimal places.