珍珠湾ART

标题: 小狗过河 [打印本页]

作者: ob 时间: 2005-2-24 23:45
标题: 小狗过河

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？

作者: IG 时间: 2005-2-25 02:19
标题: 回复：小狗过河

about 500km?

作者: IG 时间: 2005-2-26 00:41
标题: 回复：小狗过河

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

作者: fzy 时间: 2005-2-26 01:51
标题: There must be a better way

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.

作者: 独木桥 时间: 2005-2-26 02:17
标题: 回复：There must be a better way

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.

作者: fzy 时间: 2005-2-26 02:21
标题: 回复：回复：There must be a better way

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

作者: 独木桥 时间: 2005-2-26 02:25
标题: 回复：回复：回复：There must be a better way

Yes, I get it.

作者: fzy 时间: 2005-3-2 01:37
标题: 回复：小狗过河

What is the answer to the problem?

欢迎光临珍珠湾ART (http://www.zzwav.com/)