A man launches his boat from point A on a bank of a straight river, 4 km wide, and wants to reach point B, 4 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between B and C and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.)
2007-11-08 07:14:00 · 2 answers · asked by syem64 2 in Science & Mathematics ➔ Mathematics
This is a straightfoward calculus problem. Let x = distance from D to B. Then the total time is (1/8)x + (1/6)√(16+(4-x)²)). Differentiating this, setting the expression to 0 and solving for x, we find that x = -0.54 km, or a spot just beyond B. The reason why this gives the minimum time is because after the man reaches this spot -0.54 km beyond B, he is able to turn back the clock running back to B by -0.54/8 = -0.0675 h, for a mininum time of 0.94096 hours. Of course, if he just rowed directly to B, the time would have been 0.94281 hours, but that's no fun.
2007-11-08 07:36:44 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋
Time=[(sqrt(16+x^2))/6] + [(4-x)/8]
Here is the equation thats the hardest part. Next find the derivative, and set it equal to zero. Find x, that is how far the rower lands from point B.
2007-11-08 15:39:46 · answer #2 · answered by gang$tahtooth 5 · 0⤊ 0⤋