Marsh Crossing

Frodo and Sam need to travel 100 leagues due East from point A to point B. On normal terrain, they can cover 10 leagues per day, and so the journey would take 10 days. However, their path is crossed by a long marsh which runs exactly South-West to North-East, and walking through the marsh will slow them down. The marsh is 50 leagues wide at all points, and the mid-point of AB is located in the middle of the marsh. A map of the region is shown in the diagram below:

The marsh consists of 5 distinct regions, each 10 leagues across, as shown by the shading in the map. The strip closest to point A is relatively light marsh, and can be crossed at a speed of 9 leagues per day. However, each strip becomes progressively harder to navigate, the speeds going down to 8, 7, 6 and finally 5 leagues per day for the final region of marsh, before it ends and the terrain becomes easier again, with the speed going back to 10 leagues per day.

If Frodo and Sam were to head directly East for point B, they would travel exactly 100 leagues, and the journey would take approximately 13.4738 days. However, this time can be shortened if they deviate from the direct path.

Find the shortest possible time required to travel from point A to B, and give your answer in days, rounded to 10 decimal places.

To approach this, we need to discover how to optimize Frodo’s and Sam’s path via employing calculus. We begin by defining the problem with some math.

Let’s say Frodo and Sam deviate from the East direction by an angle θ, their travel path would then become a straight diagonal across the marsh. This path length would be larger than the 50 leagues they would travel going straight East through the marsh, but it could potentially provide a shorter traveling time because the longer path could mean less time spent in the tougher parts of the marsh.

The additional distance ∆D they would need to cover due to this deviation can be given by ∆D = 50*tan(θ).

Also, the time they would spend in each strip of the marsh would equal the width of the strip divided by their velocity in the strip. For the ith strip, this would be given by:

Timei = 10/(10-i) * sec(θ), where sec(θ)=1/cos(θ)

The total journey time would then be the sum of the times for all strips plus the time for the ∆D journey at their velocity on normal terrain:

T_total = ∆D/10 + sum(Timei for i=1 to 5)

To find the minimum of T_total, we would take the derivative of T_total with respect to θ and equate it to zero. Then, solve for θ.

The first term, ∆D/10 differentiates to 5*sec^2(θ) while the second term, the sum(Timei for i=1 to 5), differentiates to sum((10/(10-i)) * tan(θ) * sec(θ) for i=1 to 5)

Equate this derivative of T_total to zero and solve for θ, the angle that gives the minimum T_total. This θ giving the minimum time needs a numerical approach for solution like the Newton-Raphson method.

Finally, substitute this θ into the equation of T_total to get the minimum time. Additionally, the accuracy of the final answer would depend on the iterative method chosen. Without the actual computation, it’s hard to provide the final answer. This would involve programming computation for efficiency.

In a simpler form, the problem necessitate determining the right angle θ that would minimize the total travel time from point A to point B. Although they travel more, the path they take reduces traveling through the toughest parts of the marsh thereby cutting down the overall travel time.

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