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Oblivion

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- Thread starter Oblivion
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In summary: But, as I pointed out earlier, the problem does not involve speed and, in fact, doesn't even involve time. You will get the same result no matter what speed the spiders are moving...

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Oblivion

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BicycleTree

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I already know this one so I won't spoil it.

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DaveC426913

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Ans: [tex]7.85 feet? (1/4 of the circumference of a 10ft circle)

No. No. It's not a circular curved path, it's a logarythmic path.

No. No. It's not a circular curved path, it's a logarythmic path.

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T@P

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Oblivion

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It's a challenging problem, don't leave out speed in your figuring...

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<<<GUILLE>>>

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I've been thinking in it but I the best conclusion I have is never. is it correct?

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nnnnnnnn

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not sure what the answer is but they definitely continuously get closer to the center until they meet so its not never...<<<GUILLE>>> said:I've been thinking in it but I the best conclusion I have is never. is it correct?

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kleinwolf

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[tex] \frac{\dot{y}}{\dot{x}}=\frac{|x-y|}{|L-x-y|}[/tex]

the origin being here fixed at one corner.

(By symmetry to the next spider)..

Putting polar coordinates from the center of the square :

[tex] x=L/2+r*cos(\theta)[/tex]

[tex] y=L/2+r*sin(\theta) [/tex]

gives the equation :

[tex] \frac{cos(\theta)-sin(\theta)}{cos(\theta)+sin(\theta)}=\frac{\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}{\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)} [/tex]

which is easily solved for [tex] \dot{\theta}=-1 [/tex] and [tex] \dot{r}=-r [/tex]

This leads to [tex] r(t)=-Le^{-t} [/tex]

The problem is that the overall distance I get so is : [tex] D=\int_0^\infty\sqrt{\dot{x}^2+\dot{y}^2}dt=\sqrt{2}L[/tex]

This is not correct (the solution is on Mathworld for the general case)

Does somebody see the mistake ?

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xJuggleboy

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DaveC426913

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Oblivion said:It's a challenging problem, don't leave out speed in your figuring...

Hey, cut that out. Speed is a wild goose chase. Speed will have no factor in it.

If the spiders move at 1cm/sec, they will follow the same trajectory as if they were traveling at 100m/sec, thus, they will travel the same distance. Now, if you had asked how

In fact, it reduces to a pure geometry problem, and can be solved without any reference to time at all. (Or did you not notice that?)

Simply put, a line tangential to the curve at a given point, can be extended, and will intersect the equivalent point on the next curve. This is true for all points along the path.

eg. The point that is one foot distant from the corner will have a tangent that intersects the point on the next path (90 degrees away) that is one foot from its corner. Note that this is true at 1cm/sec and at 100m/sec.

(BTW, when you draw this same line for all four paths, you will have drawn a square - the 4 lines are at 90 degrees to each other. This results in always drawing squares of ever decreasing size. However, I don't know if this helps solve the problem.)

The problem of course, is that I never learned logs. I know the intuitive answer, but I don't know the rigors of the math.

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kleinwolf

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xJuggleboy said:

Yes..the spiders are "walking" an infinite amount of time, but the distance is finite. (But what we want is the distance, not the time)

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Moo Of Doom

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kleinwolf said:Yes..the spiders are "walking" an infinite amount of time, but the distance is finite. (But what we want is the distance, not the time)

How can any finite distance be traveled in an infinite amount of time if one moves at a constant speed?

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rygar

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Moo Of Doom said:How can any finite distance be traveled in an infinite amount of time if one moves at a constant speed?

because as you lower your speed, your travel time approaches infinity. you can lower you speed infinitely as well.

- #14

DaveC426913

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rygar said:because as you lower your speed, your travel time approaches infinity. you can lower you speed infinitely as well.

The spiders are walking at a

But, as I pointed out earlier, the problem does not involve speed and, in fact, doesn't even involve time. You will get the same result no matter what speed the spiders are moving at.

- #15

xJuggleboy

1. How can 2 things moving at right angels to each other ever meet?

2. If the squares they make are ever decreesing... How can they ever come to a point?

I still think the problem can't be done.

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DaveC426913

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xJuggleboy said:

1. How can 2 things moving at right angels to each other ever meet?

2. If the squares they make are ever decreesing... How can they ever come to a point?

I still think the problem can't be done.

Zeno's paradox.

Or convergent series.

Both explain how this happens.

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Jimmy Snyder

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At any given time after the spiders start out, but before they meet, the symmetry of their motion insures that the original problem is repeated in a smaller square. For an infinitesimal time after they start, they will travel approximately straight along the edge of the square. (the smaller the time interval, the better the approximation). At the end of that infinitesimal amount of time, the problem can be restated with a square that is infinitesimally close to the original square, in proportion to the distance already traveled. In other words, the entire trajectory should be as long as if they had traveled along the side of the original square.

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DaveC426913

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So, you're saying your answer is: 10 feet?jimmysnyder said:

At any given time after the spiders start out, but before they meet, the symmetry of their motion insures that the original problem is repeated in a smaller square. For an infinitesimal time after they start, they will travel approximately straight along the edge of the square. (the smaller the time interval, the better the approximation). At the end of that infinitesimal amount of time, the problem can be restated with a square that is infinitesimally close to the original square, in proportion to the distance already traveled. In other words, the entire trajectory should be as long as if they had traveled along the side of the original square.

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Jimmy Snyder

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DaveC426913 said:So, you're saying your answer is: ?

Yes, just that many whited out feet.

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kleinwolf

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DaveC426913 said:So, you're saying your answer is: 10 feet?

If your literal only reasoning would have been true, then it could apply to a triangle...however this is not the case. See : http://mathworld.wolfram.com/MiceProblem.html

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rygar

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DaveC426913 said:The spiders are walking at a. You're confusing the issue by misreading the original post.constant speed

But, as I pointed out earlier, the problem does not involve speed and, in fact, doesn't even involve time. You will get the same result no matter what speed the spiders are moving at.

i am not confusing the issue! they all walk at a constant speed--but that constant speed is arbitrary! it could be 17mph or 1cm/h!

i completely agree time has no effect as to the answer of this problem, i was just answering moo's question.

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BicycleTree

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Jimmy Snyder

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Thanks kleinwolf.

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kleinwolf

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Variation: put 3 spiders on a spherical triangle with 3 right angles (for simplification)

Four cannibalistic spiders refer to a fictional species of spider that is believed to feed on other spiders of the same species.

No, four cannibalistic spiders do not exist in reality. They are a concept that has been created for storytelling or entertainment purposes.

Since they are a fictional species, four cannibalistic spiders pose no threat to humans. However, other real species of spiders can be dangerous depending on the circumstances.

The characteristics of four cannibalistic spiders may vary depending on the source material. In general, they are described as being larger and more aggressive than other spider species, with a tendency to feed on their own kind.

No, there is no scientific evidence to support the existence of four cannibalistic spiders. They are purely a fictional concept and have not been observed or studied in the natural world.

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