小狗过河

ob · 发表于 2005-2-24 23:45:13

一只狗和它的主人在岸边隔河正相对，河水以常速流动。
狗以在静水中每小时2千米的游速向对岸游去，在全部时间里它的游向始终朝着它的主人。
此人注意到，狗游到河宽三分之二的地方才停止向下游漂流，并且游过河的时间比在静水中要多花五分钟。
问：小河有多宽？

IG · 发表于 2005-2-25 02:19:45

about 500km?

IG · 发表于 2005-2-26 00:41:49

I mean it's about 500m, not 500km.

fzy · 发表于 2005-2-26 01:51:48

I have to solve a differential equation to get the path of the dog. It is

x = C(tan(theta/2))^a

where C is the width of the river, a the ratio of the speeds of the dog and the river, and theta the angle between dog-person and the river bank (theta goes down from 90 to 0). From this we can figure out a and C using the conditions. But I do not want to do it manually, and a computer simulation gives C = 146 meter.

The problem does not look so bad. So I must have made a mistake some where or there is a short cut.

独木桥 · 发表于 2005-2-26 02:17:08

I got an equation different from yours by a factor 1/sin(theta):

r = C(tan(theta/2))^a/sin(theta),

where the notations are the same as yours. If you check what will happen

as theta tends to 0 in the cases when a>1 and a<1, you find your equation

can not be right.

fzy · 发表于 2005-2-26 02:21:59

they are the same. I have x on the LHS, not r.

独木桥 · 发表于 2005-2-26 02:25:15

Yes, I get it.

fzy · 发表于 2005-3-2 01:37:09

What is the answer to the problem?

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