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来源: 馨尹 于 06-10-19 17:57:15
设有一条绳子(很好的橡皮绳)长1米,而且每秒均匀拉伸10厘米,
一只虫子(当然是理想的虫子啦,永远不死的虫子,)从绳子的一端爬向
另一端,每秒爬1厘米,
问,虫子能爬到另一端否?
Deuss: This is the continuous version of the problem. Other people already presented solutions. Here we use the idea of measuring the distance in the original rope, i.e. at any moment, we can imagine to scale back the rope to original, and count the bug position/distance there. At time t, the bug crawls dt cm in time dt, which is corresponding to distance100/(100+10t)dt=10/(10+t)dt in original rope.
So, after time T, the distane crawled in original rope is Int_0^T(10/(10+t)dt)=10(ln(T+10)-ln(10)),
set it equal to 100, we get T=10*exp(10)-10=220254.7, that's the time the bug reaches the other end.
The discrete version of the problem is stated as follows:
There is a rubber band with one end tied to a post, and the other end tied to the tail of a kangaroo. A flea is initially at the post, and the rubber band is 100 feet long. Now the flea tries to jump to the kangaroo at 1 feet per jump. After each flea jump, the kangaroo gets scared and leap forward 10 feet.
Question is, can the flea reach the kangaroo ?
Answer is yes, in roughly 240000 seconds:
At time t, the far end of the rope is at 100+10t.
From time (t-1) to t, the bug crawls from x_{t-1}=x to 1+x*(100+10t)/(100+10(t-1)) = 1+x+x/(10+(t-1))=x_t
We can also view this way, let the rope scale back to original, then, the distance pulled by rope becomes 0, and the 1cm is scaled back to 100/(100+10t)=10/(10+t). So, the distance in the ORIGINAL rope at time t is
10(1/10+1/11+1/12+....1/(10+t)) this goes to infinity.
The t first making it >=100 would be the answer, roughly 240000 by solving ln(10+t)+0.577215665-(1+1/2+...+1/9)=10
设有一条绳子(很好的橡皮绳)长1米,而且每秒均匀拉伸10厘米,
一只虫子(当然是理想的虫子啦,永远不死的虫子,)从绳子的一端爬向
另一端,每秒爬1厘米,
问,虫子能爬到另一端否?
Deuss: This is the continuous version of the problem. Other people already presented solutions. Here we use the idea of measuring the distance in the original rope, i.e. at any moment, we can imagine to scale back the rope to original, and count the bug position/distance there. At time t, the bug crawls dt cm in time dt, which is corresponding to distance100/(100+10t)dt=10/(10+t)dt in original rope.
So, after time T, the distane crawled in original rope is Int_0^T(10/(10+t)dt)=10(ln(T+10)-ln(10)),
set it equal to 100, we get T=10*exp(10)-10=220254.7, that's the time the bug reaches the other end.
The discrete version of the problem is stated as follows:
There is a rubber band with one end tied to a post, and the other end tied to the tail of a kangaroo. A flea is initially at the post, and the rubber band is 100 feet long. Now the flea tries to jump to the kangaroo at 1 feet per jump. After each flea jump, the kangaroo gets scared and leap forward 10 feet.
Question is, can the flea reach the kangaroo ?
Answer is yes, in roughly 240000 seconds:
At time t, the far end of the rope is at 100+10t.
From time (t-1) to t, the bug crawls from x_{t-1}=x to 1+x*(100+10t)/(100+10(t-1)) = 1+x+x/(10+(t-1))=x_t
We can also view this way, let the rope scale back to original, then, the distance pulled by rope becomes 0, and the 1cm is scaled back to 100/(100+10t)=10/(10+t). So, the distance in the ORIGINAL rope at time t is
10(1/10+1/11+1/12+....1/(10+t)) this goes to infinity.
The t first making it >=100 would be the answer, roughly 240000 by solving ln(10+t)+0.577215665-(1+1/2+...+1/9)=10
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