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Sine River 2 (Posted on 2024-06-17) Difficulty: 4 of 5
The Sine River is located between y = sin(x) and y = 1 + sin(x) (x in radians, not degrees).
Your starting point is on the shore, at location (0,0) and your destination is location (10,2).
Your running speed is 1 unit per minute. Your swimming speed is 1/2 units per minute.

Assuming there is no current in the river (and no wind), what is the minimum travel time to the destination, and by what pathway?

Note that in the earlier problem, Sine River, the destination was (5,1) and the solution was limited to straight segments. For this problem, a portion of the fastest route might be a curved path.

No Solution Yet Submitted by Larry    
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Solution solution | Comment 1 of 2
If we traced beams of light at various initial angles as in the original Sine River, we'd never hit the target. The largest angles would miss the second encounter with the riveer, and those that are not high enough to do that would be refracted downward (southward) twice.

The best we could do is to come out of the river as in the first occasion and run in a direction that makes the line with that slope tangent to the next encounter with the north bank of the river. Then run along that bank until another tangent line would include the destination point.

We want the refracted angle at which we exit the river (the first and only time) to just be tangent to the north bank of the river on its next north crest, which means it's to have only one intersection point, rather than two.

The vpasolve function includes an initial guess as to a solution, so it's to be hoped that as far as the intersecton of the line coming out of the water with the top (northern) sine curve will be different if we give the function an initial guess of 7.3 vs an initial guess of 8 so long as there are indeed two solutions.

Indeed two different initial guesses given to the vpasolve function do produce the two different solutions when two are present. The closest I can get to an initial starting angle is 10.1291541007574° north of east, at which the refracted path grazes the top sine curve at (7.591855, 1.965841).

Clearly the tangent that finally leads to (10,2) is horizontal, starting at the second crest of the top sine curve, at (pi*5/2, 2). What's missing is the length of the arc of the  raised sine curve from (7.591855, 1.965841) to (pi*5/2, 2). Numerical integration gives 0.2650582 as this length.


                                                                     minutes
Starts                       0, 0
                    2.69061 running                                  2.69061
Enters Water                 2.648681340915, 0.473192636927 
                    0.745212 swimming                                1.490424
Leaves Water                 3.290847372588, 0.851298821260
                    4.44307 running                                  4.44307
Arrives at river bank        7.591855, 1.965841
                    0.265058 running along bank                      0.265058
Leaves river bank            5*pi/2, 2
                    2.14602 running                                  2.14602
Arrives at destination       10, 2                                   
                                                                   -----------
                                                                   
                                                                    11.035182     

  Posted by Charlie on 2024-06-17 15:08:21
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