So, finish the diagram. If the man enters the water at point C, x meters upstream from A, and lands at point D, y meters downstream from B, then we have distances of
land: (d^2+x^2)
water: √(x+y)^2+d^2)
land: √(y^2+(2d)^2)
The time it takes to cross the river is shortest when he swims directly north, taking d/(v/3) = 3d/v seconds. In that time, he drifts downstream at speed v, making y = 3d meters. So, now we see that the distances are
√(d^2+x^2) + √(d^2+16x^2) + √(9x^2+4d^2)
Now, divide each distance by its corresponding speed, and we have
1/(3v) (√(d^2+x^2) + 9√(d^2+16x^2) + √(9x^2+4d^2))
That's all very messy, but the v does not affect anything at this point. So, let us express x in terms of d. That is, consider the ratio x/d, rather than an absolute distance. If we let u = x/d, then we have the time
t = d/(3v) (√(1+u^2) + 9√(1+16u^2) + √(4+9u^2))
To minimize the time, we must have dt/du=0. It's a bit messy, but we find that dt/du=0 when u=0. That is, the man runs directly from A to the river, swims right across, then runs from where he lands on to B.
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Let's say he swims from A directly north, being taken downstream d/(v/3)*v meters to point C. Then his distances are
d on land from A to river
d north in water
√((3d)^2 + (2d)^2) on land to B
time: d/3v + d/(v/3) + d√13/3v
= d/v (1/3 + 3 + √13/3)
= dv/3 (10+√13)
A river flows at the speed of v from west to east. How should a man who is at a point A of the southern bank of the river, which is d meters away from the river, and wants to reach a point B on the northern bank of the river, which is 2d meters away from the river, choose his path in order to reach the destination at the shortest time? Assume the width of the river to be d, the man to run at a speed of 3v and swim at the speed of v/3 in still water, and the points A and B on the same meridian.
1 answer
1 answer