Easy Questions?

[Swatopluk]

The proof that the worm will reach the end in the end looks relatively easy.
If it does not crawl, its relative position on the rod does not change. If it crawls even at the lowest speed it improves its relative position and thereby its speed toward the end.
The latter fact should prevent the function from becoming asymptotic.
When it reaches the position on the rod where its own speed and the positional expansion speed add up to the total expansion speed its distance from the desired end reaches its maximum, then begins to shrink.
I haven't done the calculation yet how long it actually takes but I remember that the numbers get huge.

[Sibling Lambicus the Toluous]

I couln't be bothered doing the math, and frankly I'm not sure if I remember how to do the math, but I'm pretty sure the answer is, it depends on the starting position.

[...]

Thus if v3>v1 or if (v2+v1(x0/L)>v1, then the worm will get to the end, otherwise it won't make it.

The starting point's given: it's the fixed end of the rod.

And the worm will make it, since it's given that v₂ > v₁.  Therefore, v₃ > v₁ as well.

Wait... hang on.  I think I was making the problem more difficult than it had to be.  Is the time taken just L/v₂?

I haven't done the math, but I think if you look at the worm's movement as a percentage of the instantaneous length of the rod, you'll find that the change in percentage is constant with time; all the terms with v₁ in them should drop out of the equation.

Edit: I tried the math and it got messy, but I think it works in concept: at any point in time, the additional speed the worm receives due to the expansion of the rod is equal to the expansion of the rod itself (as it has to be, the way the problem is set up).  Therefore, the faster the rod expands, the more extra speed the worm will receive. 

I'm fairly certain that over the entire length of the rod, this extra "push" will result in a "bonus worm distance" (i.e. the extra distance travelled by the worm above and beyond what it would have done at speed v₂) that will work out to the final length of the rod minus the initial length.

[Swatopluk]

The solution I have come up with

x= v1*t+ integral[0;t]{v(x,t)dt}
v(x,t)= v2*(x/L(t)) = v2*x/(L0+v2*t)
with q=L0/v2
=> v(x,t)=x/(t+q)
integral[a;b]{x/(t+q)}dt = [x*ln(t+q)]_a^b*

=> x = v1*t + [x*ln(t+q)]₀^t
=> x = v1*t/(1+ln(1+(t/q)))

Unfortunately, this equation can't be brought to a form t= f(...) because one t is in the logarithm, the other isn't.
Thus with x=L0 one would have to solve an implicit equation to get the value of t, when the worm reaches the end of the rod.

*[f(x)]_a^b means f(b)-f(a)

[Swatopluk]

I just looked up the riddle on the net.
My thought process was obviously far too complicated.

The way is to start with a sum and then go to the integral.
First it is assumed that the rod does not expand continuously but in jumps.
The crawling distance for the worm can be calculated for each interval. first 1/x of the band, then 1/(2x), 1/(3x).....
This defines a sum.
If the time intervalls are shortened indefinitely the sum turns into an integral.
The solution for
v1= 1cm/sec
v2= 1m/sec
L=1 m
is
t = e¹⁰⁰⁰⁰⁰sec ~ 10⁴³⁴²² years
Old and grey he will be when reaching his goal.

[Sibling Chatty]

I say take the rod, the worm, some string...and go fishing!

Or at least sit around pretending to fish while you nap.

[Sibling Lambicus the Toluous]

The solution for
v1= 1cm/sec
v2= 1m/sec
L=1 m
is
t = e¹⁰⁰⁰⁰⁰sec ~ 10⁴³⁴²² years
Old and grey he will be when reaching his goal.

Is that with v1 and v2 defined as you did before?  If so, that seems too long.  Is there a unit prefix missing there?  1 m/s is a quick walking speed.

But look at it this way: if the worm were on the table next to the rod, rather than on the rod itself, it would take 1.01 seconds ( i.e. t = (v2-v1)/L  ) to reach the end of the rod.

Since in the original problem, the worm is going faster than v2, the time taken must be less than 1.01 s (assuming the units are okay).

[Swatopluk]

Sorry, confused v1 (rod) and v2 (worm)
And the solution quoted seems to imply a 1km rod, not a 1m one.

The solution looks confusing but the trick is to take L as a constant (tt: reduced length) and to change from an absolute speed of the worm to a relative one (i.e. in comparision to the reduced length L). Thus the term dependent on both x and t can be dropped and the worm term becomes essentially dimensionless. The integration is made from the time the rod reaches its initial length to t_end when the worm reaches it*. The integral is then set as 1 (= 1 length) and t isolated

* t=0 would be a point back in time when the rod had zero length.

I should have had that idea myself but I thought that a transformation of coordiantes would make the problem more not less difficult.

Now, where do we get such an elastic rod and a worm with sufficient life expectation?

[Griffin NoName]

I just washed my white towelling robe in deep blue detergent. Why is it still white?

I imagine you will tell me all about light. But can you phrase it in a way a three year old would understand?

[Swatopluk]

It is a dye that does not stick to the cloth. Thus it can be rinsed out completely.
If the little thingies that give the color would prefer to stay with the fibres instead of going for a swim, the situation could look different. Just don't ask me why the foam is white independent of the color of the stuff the foam came from.

[Sibling Chatty]

Silly...it's

MAGIC

[The Black Spot]

Okey Dokey then yer blaggards...

If you have to squeeze through a small gap you breath in.
Why should taking a large amount of air into your body make you smaller?

[Griffin NoName]

Hey Spot, it depends where you draw the air into 

[Swatopluk]

I think it's not breathing in but holding your breath.
To breathe in could be just a safety measure in case that you get stuck.
If you breathe out and then get stuck, you could be in trouble of suffocation.

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Let's say you have no idea about star constellations (or are on a planet with different ones from ours and no map), have no compass (neither magnetic nor gyroscopic) and no clock/watch, but want to find the exact direction towards the next geographic pole (no telephone pole ).
How could you do it with simple tools?

[Sibling Zono (anon1mat0)]

Erm... the sun? Or is it underground/always clowdy/in the forest/etc?

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I have a question: this happens to me from time to time, take a new book and start reading it slightly bending from the cover pages. If you put the book on its back the cover (and sometimes a few pages) is lifted. Now rest the book on its cover and -oh surprise- the back page begins to lift.

1. Does it happen to you (or is only me)?
2. well... why?

[Sibling Lambicus the Toluous]

1. Sometimes for paperbacks.  With hardcover books, I can't recall it happening.

2. Two possible reasons:

- you bent the back cover slightly while reading the book.

- as you read, the spine became bent by your turning the pages at the front.  This exerts force on the rest of the book.  Since the back cover and back page is light enough, it gets lifted when you put the book down.