[Prof. Templeton]

An infinitely stretchable rubber-band has one end attached to the wall and the other end it attached to a scooter being driven Prof. Templeton. The good Prof. drives away from the wall at a constant speed of 1 meter per second. Starting from the wall and walking out onto the rubber-band is a bug. The bug travels at a constant speed of 1 cm per second. Both start moving at the same time in the same direction. The Prof. moves away from the wall perpendicularly and the bug moves along the stretching rubber-band away from the wall toward the Prof. Let’s assume the rubber-band is perpendicular to the wall with no angle of incidence (all along one plane) and the bug is a point on a line initially at distance 0 from the wall.

Will the bug ever reach Prof. Templeton, and if so can you tell how long it will take?

Edited August 19, 2008 by Prof. Templeton

[Guest]

I understand what you are saying, but I still don't think it's that complicated. Ultimately, we are looking at one of three possible movement scenarios...

The bug and the Professor both start at the same place and the bug will move with the professor all the time.

Given a 1m head start and assuming the bug is pulled along at 1m/s (same rate as the scooter) and also moves 1cm/s along the rubber band, he will reach the professor in less than 2 minutes

Given the head start as in Scenario 2, but the bug is always in the middle of the rubber band. Each second, he moves 50cm away from the wall, but also 49cm away from the professor (50 cm - 1 cm). Under this scenario, he can never move beyond 51% of the distance.

The way I read yours, and it may very well be an error on my part, you all were focusing too much on the displacement from the wall and not enough on the two objects in motion and their relative velocities to each other

[Guest]

Scenario 1 - The bug and the Professor both start at the same place and the bug will move with the professor all the time. boring!

Scenario 2 (Head-Start and Edge of Band) - Given a 1m head start and assuming the bug is pulled along at 1m/s (same rate as the scooter) and also moves 1cm/s along the rubber band, he will reach the professor in less than 2 minutes This is the band not attached to the wall and not stretching, but I agree with your solution

Scenario 3 - Given the head start as in Scenario 2, but the bug is always in the middle of the rubber band. Each second, he moves 50cm away from the wall, but also 49cm away from the professor (50 cm - 1 cm). Under this scenario, he can never move beyond 51% of the distance. I don't understand - he isn't in the middle any more, because he's 51% of the way along the band. Therefore, he moves 1cm/s under his own steam, plus 51% of the Prof's speed, and ends up at 51-point-something-% of the way along at the next second. He will catch eventually him up, won't he?

In my understanding, the OP was suggesting another scenario in which the bug started a one end of a rubber band that was attached to the wall. The other is moving away from the wall at 1m/s and the rest of the band is stretching in between. Therefore the middle of the band is moving at 0.5m/s and the bit at x cm from the wall is moving at x cm/s (after the start it is x/(t+1) m/s because the band is not longer 1m long).

at start,

bug(b) is moving at 1cm/s and Prof(p) is moving at 1m/s

b		  P

¦----------&


after one second, b is moving at 1cm/s + the bit of speed he gets from being 1/200th of the way along the band, ie 1/200th of 1m/s

 b				   P

¦-------------------&



after t seconds, b has moved t cm himself + the help he has had from the band stretching. Calling the total distance travelled x, he is moving at 1cm/s + x/t m/s

		b										 P

¦-------------------------------------------------&

You raise a good point, though - could it be easier to prove that the distance between the bug and the Prof tends -> 0, rather than looking at the distance travelled?

Edited August 21, 2008 by foolonthehill

[Guest]

You raise a good point, though - could it be easier to prove that the distance between the bug and the Prof tends -> 0, rather than looking at the distance travelled?

I think looking at distance from the wall is necessary, because the bug's velocity due to band stretching has to be related to how long the band is. You can demonstrate that they will meet using some basic logic, but I'm pretty confident there's no non-differential equation way to find a numerical answer.

[Guest]

I think looking at distance from the wall is necessary, because the bug's velocity due to band stretching has to be related to how long the band is. You can demonstrate that they will meet using some basic logic, but I'm pretty confident there's no non-differential equation way to find a numerical answer.

I don't have time now, but I thought that we could look at it from the Prof's reference point:

To begin with the bug is 1m away and is moving away from him - after 1 sec, the bug is 1.99m away. But according to my solution, there is a point at which he begins to close the gap.

If we can show that there is a limit to the distance that the Prof can ever get from the bug, or that the acceleration of the bug wrt to the Prof is always towards him, then he is guaranteed to be caught.

[Guest]

Scooter: With the 1m headstart, the scooter will be 101 meters from the wall after 100 seconds (100 X 1 mps)

Bug: The bug will be 101 meters from the wall after 100 seconds. The rubber band will carry it the 100 meters in 100 seconds, while simultaneously, he will move 100 cms in 100 seconds.

This can all be seen by the formula below, where the bug is the left side and the scooter is the right:

1m(100s) + 1cm (100s) = 1m (100s) + 1m (0s)

I think looking at distance from the wall is necessary, because the bug's velocity due to band stretching has to be related to how long the band is. You can demonstrate that they will meet using some basic logic, but I'm pretty confident there's no non-differential equation way to find a numerical answer.

I've found two numerical answers without diffential equations...The first, proving how the professor can be reached, is given above. The second is when the professor cannot be reached and the bug, though perpetually in motion, will always remain 51% of the professor's distance from the wall.

For all of your computers and equations, it's a logic puzzle that can be reasonably thought out/argued without needing any of the fancy stuff, not that I don't respect your ability to think in that light, I just feel it's a bit excessive and unnecessary for what lies at hand

[k-man]

I've found two numerical answers without diffential equations...The first, proving how the professor can be reached, is given above. The second is when the professor cannot be reached and the bug, though perpetually in motion, will always remain 51% of the professor's distance from the wall.

For all of your computers and equations, it's a logic puzzle that can be reasonably thought out/argued without needing any of the fancy stuff, not that I don't respect your ability to think in that light, I just feel it's a bit excessive and unnecessary for what lies at hand

For every problem there is usually only one correct answer and an unlimited number of incorrect answers. The fact that you found 2 different mutually exclusive answers should tell you that at least one of them is wrong. I will try to explain why both of them are wrong using plain logic and no math.

The first answer is assuming that every second the bug is moving by 1.01 meters (1 meter with the band and .01 meter on its own). To help see the error in this assumption let's imagine that the rubber band is originally 1 meter long and has markings at each centimeter. The mark 0 is tied to the wall and the mark 100 is tied to the scooter. The bug is sitting on the mark 0. During the first second assuming the band is stretching evenly the marks on the rubber band are moving with different speeds - mark 0 is not moving at all (it's tied to the wall), mark 1 is moving at 1 cm/sec, mark 2 is moving at 2 cm/sec, and so on. After the first second the total length of the rubber band will be 2 meters and the distance between each mark will also increase from 1 cm to 2 cm. For the bug to be 1.01 meters away from the wall it should be between the marks 50 and 51. How is that possible given that the bug's speed is 1 cm/second? In reality, the bug will be somewhere between marks 0 and 1. It will be more than 1 cm away from the wall, but less than 2 cm away. Why will it not reach the mark 1 that is now 2 cm away from the wall? Because it's moving continuosly during the second and not in one leap. If we divide the second into smaller time periods you will see that each fraction of the second the bug moves slower than the mark 1. Therefore at the end of the first second the bug will not reach mark 1.

The second answer states that the bug will always be at 51% of the length of the rubber band. This would be true if the bug was initially placed in this spot (mark 51) and was not moving on its own. But it's moving, so once it gets to the mark 51 it will keep moving toward mark 52. At that time it will be many miles away, but the bug will eventually get there.

Unfortunately to mathematically prove that the bug will eventually catch the Prof. you need to use some math. And to calculate the exact time it will take you need to use differential equations. I came to the same equation that the foolonthehill posted, but was too lazy to actually solve it.

Edited August 21, 2008 by k-man

[Prof. Templeton]

In the problem we are given that A = 1m and B = 0.01m and I will make an assumption that the prof is given a 1m headstart:

t = 1 + e^100

t = 2.7 x 10^43 seconds!

the Prof would have covered an equal distance in metres, which considering the Milky Way is about 10^21 m across (I think), is a little further than he is likely to have got before running out of petrol!

For interest, another couple of scenarios where the bug is going a little bit faster:

for B = 0.1m, 22027s (about 6 hours)

for B = 1m, 3.72s.

There is every chance that my maths is wrong here somewhere so please feel freee to correct, but it feels as though its about right....

s' = B + s / u

s' + (-1/u)s = B [rearrange]

(1/u)s' + (-1/u²) s = (1/u)B [multiply through by an integrating factor of 1/u]

[(1/u)s]' = (1/u)B

s/u = B . ∫ 1/u .dt [integrating both sides]

s/u = B . (ln u + C)

s = Bu (ln u + C)

phew

we get

s = Bu(ln u + C), with C as a constant of integration.

Knowing that at u=d/A, s=0

s = Bk(ln d/A + C) = 0

C = -ln d/A

and leaving us with:

s = Bu (ln u - ln d/A)

s = Bu ln (Au/d)

to find the point where they meet, we solve

s_bug = s_prof

Bu ln (Au/d) = Au

ln (Au/d) = A/B

Au/d = e^(A/B)

u = d/A . e^(A/B)

so,

t = u + d/A

t = d/A [1 + e^(A/B)]

You guys have proven that the bug will indeed reach the scooter regardless of an initial head-start. If the scooter and bug both start at the wall as stated in the OP, than the time it takes is 0. If we give the scooter a head-start of 1 meter, than it will take

2.7 x 10^43 seconds, as foolonthehill has correctly figured, or 8.5 x 10^35 years

I apologize for the complicated math, but I thought the question itself would lead to some interesting discussion. I'm glad to see the correct answer posted so quickly.

[Prime]

This isn't going to be pretty, but I think I have a solution with one small modification to the problem:

The Prof needs to start a distance d from the wall initially, ie. the prof needs a head start. Otherwise, the bug has already caught up and my equations won't work as they are undefined for d=0 at t=0.

The speed of the bug is his starting speed, plus a proportion of the Prof's speed:

(I define s as the distance from the wall, and s' as the speed)

s'_bug = 0.01 + s_bug/(t+d)

so, the acceleration (s''_bug) is

s''_bug = d/dt [ s_bug/(t+d) ]

s''_bug = s'_bug/(t+d) - s_bug/(t+d)²

s''_bug = 0.01/(t+d), substituing the equation for s'_bug above

This is positive for all t>0, as long as d>0. Therefore, the bug is always accelerating (though at a decreasing rate!). As the Prof has a fixed speed, the bug must catch up the Prof at some t.

But the question says 'how long'....

For the general solution, we define the problem as a function of the distance from the wall, s, wrt time, t:

s_prof = At+d, where A is the speed of the scooter.

s_bug = ?

s'_prof = A (' denote derivates, ie speed)

s'_bug = B + s'_prof x (s_bug / s_prof), where B is the speed of the bug in his frame of reference.

s''_prof = 0, cleary!

s''_bug = B/t

the s'_bug is simply his personal speed, B, plus a proportion of the prof's speed as a faction of his distance along the band.

the s'_bug is our differential equation which needs solving, so I have dropped the 'bug' bit for clarity (ie s = s_bug)

s' = B + s / (t + d/A)

s' = B + s / u, where u = t +d/A

using u will simply make everything else easier and has no effect since u' = t'.

With a bit of Maths I have largely forgotton (which is why this took me so long!)

Spoiler for complicated maths:

s' = B + s / u

s' + (-1/u)s = B [rearrange]

(1/u)s' + (-1/u²) s = (1/u)B [multiply through by an integrating factor of 1/u]

[(1/u)s]' = (1/u)B

s/u = B . ∫ 1/u .dt [integrating both sides]

s/u = B . (ln u + C)

s = Bu (ln u + C)

phew

we get

s = Bu(ln u + C), with C as a constant of integration.

Knowing that at u=d/A, s=0

s = Bk(ln d/A + C) = 0

C = -ln d/A

and leaving us with:

s = Bu (ln u - ln d/A)

s = Bu ln (Au/d)

to find the point where they meet, we solve

s_bug = s_prof

Bu ln (Au/d) = Au

ln (Au/d) = A/B

Au/d = e^(A/B)

u = d/A . e^(A/B)

so,

t = u + d/A

t = d/A [1 + e^(A/B)]

In the problem we are given that A = 1m and B = 0.01m and I will make an assumption that the prof is given a 1m headstart:

t = 1 + e^100

t = 2.7 x 10^43 seconds!

the Prof would have covered an equal distance in metres, which considering the Milky Way is about 10^21 m across (I think), is a little further than he is likely to have got before running out of petrol!

For interest, another couple of scenarios where the bug is going a little bit faster:

for B = 0.1m, 22027s (about 6 hours)

for B = 1m, 3.72s.

There is every chance that my maths is wrong here somewhere so please feel freee to correct, but it feels as though its about right....

That's a good solid effort supplying Bug with necessary differential equations helping to estimate the time it would take to catch up with the scooter. It does seem that the time required is to the tune of e¹⁰⁰. However, I have some problems with the solution.

1. I disagree with the conjecture that "The Prof needs to start a distance d from the wall initially..." just because "Otherwise, the bug has already caught up and my equations won't work as they are undefined for d=0 at t=0."

It is a very viable everyday occurrence that Prof starts at 0 metres from the wall instantly attaining speed of 1m/sec and the magic rubber band and zero-dimention Bug spring into existence. If equations don't work for that situation, we must find different equations.

2. You found a function describing distance traveled by Bug: s = Bu*ln(Au/d), or expanding variable "u":

s = B*(t+d/A)*ln(tA/d+1)

Where B is Bug's speed; A is the speed of scooter; d is initial distance of scooter from the wall; and t is elapsed time.

In our problem we set d=1 metre arbitrarily. But what if d was much smaller? Something like d=10^-44? Then after 1 second (t=1), Bug would have traveled more than 1 metre -- further than the scooter! That disagrees with our notion of how long it takes Bug to catch up!

It seems that function describing the distance traveled by Bug: s = B(t+1)*ln(t+e) fits the bill. But if we take the derivative s' = B*ln(t+e) + B(t+1)/(t+e) -- it does not agree with our concensus what instanteneous speed for Bug should be. (Even though I suggested instanteneous speed myself post#15).

P.S. I have a feeling, there is a way to solve that problem without use of differential equations.

[Prime]

Correction:

I meant B*t*ln(t+e) as the function for the distance traveled by Bug (starting at d=0 from the wall). And the instanteneous speed: B*ln(t+e) + B*t/(t+e).

The puzzle here, to find justification for the instanteneous speed formula.

[Guest]

Correction:

I meant B*t*ln(t+e) as the function for the distance traveled by Bug (starting at d=0 from the wall). And the instanteneous speed: B*ln(t+e) + B*t/(t+e).

The puzzle here, to find justification for the instanteneous speed formula.

hate math..

infinite time - it will catch up!

I'm assuming that the physical barriers of reality are removed - ie failure equations of material, natural life cycle etc - purely on the math

[Yoruichi-san]

I apologize for the complicated math, but I thought the question itself would lead to some interesting discussion. I'm glad to see the correct answer posted so quickly.

Actually, I wouldn't mind seeing more problems that can be solved with differential equations...I think they are the way to make math actually useful, i.e. by being able to translate what we observe into math...and the ones that are actually solvable are fun to solve ;P

[Guest]

As the rubber band expands, the bug moves along with it, but its position as a percentage of distance covered between the wall and the professor never changes. For example, a bug that starts at 0.42m from the wall on a 1m rubber band can be said to be at 42%, and will always stay at 42% if it stands still. So we can throw the bug's movement due to the rubber band's expansion out the window.

The following assumes that the professor starts at 1m from the wall.

Let's take a look at what % the bug covers by walking in the one second. The % part can be calculated as (distance bug walks)/(length of rubber band). We know that the bug walks 0.01m in one second, but the length of the rubber band isn't constant. However, we know that the professor is traveling at 1m/s and so the rubber band is expanding at a rate of 1m/s. So the length of the rubber band is always t+1. So the rate at which the bug walks is 0.01/(t+1) in % per second.

Now that we have a speed function in %/s, let's integrate it to find the total % covered in some amount of time. We integrate 0.01/(t+1) and get 0.01*log(t+1)+C, where C is an integration constant (0 in this case, since we know that the bug starts at 0m and so 0% from the wall). That is a distance function for calculating the % covered by the bug at any time t. So let's plug in 100%, which is 1. This will tell us how long it takes for the bug to travel 100% of the rubber band's length and reach the professor.

1 = 0.01*log(t+1)

100 = log(t+1) [multiplying both sides by 100]

10^100 = t+1 [changing to exponents]

10^100-1 = t

So the bug reaches the professor in 10^100-1 seconds, or just under a googol seconds.

Someone please check my math and reasoning.

[Prime]

Someone please check my math and reasoning.

As the rubber band expands, the bug moves along with it, but its position as a percentage of distance covered between the wall and the professor never changes. For example, a bug that starts at 0.42m from the wall on a 1m rubber band can be said to be at 42%, and will always stay at 42% if it stands still. So we can throw the bug's movement due to the rubber band's expansion out the window.

The following assumes that the professor starts at 1m from the wall.

Let's take a look at what % the bug covers by walking in the one second. The % part can be calculated as (distance bug walks)/(length of rubber band). We know that the bug walks 0.01m in one second, but the length of the rubber band isn't constant. However, we know that the professor is traveling at 1m/s and so the rubber band is expanding at a rate of 1m/s. So the length of the rubber band is always t+1. So the rate at which the bug walks is 0.01/(t+1) in % per second.

Now that we have a speed function in %/s, let's integrate it to find the total % covered in some amount of time. We integrate 0.01/(t+1) and get 0.01*log(t+1)+C, where C is an integration constant (0 in this case, since we know that the bug starts at 0m and so 0% from the wall). That is a distance function for calculating the % covered by the bug at any time t. So let's plug in 100%, which is 1. This will tell us how long it takes for the bug to travel 100% of the rubber band's length and reach the professor.

1 = 0.01*log(t+1)

100 = log(t+1) [multiplying both sides by 100]

10^100 = t+1 [changing to exponents]

10^100-1 = t

So the bug reaches the professor in 10^100-1 seconds, or just under a googol seconds.

I did not understand the reasoning with respect to "So we can throw the bug's movement due to the rubber band's expansion out the window". If it were true, the bug wouldn't catch up to the scooter, and we already proved that it does. (See my post 15).

The math appears incorrect. I don't see how the integral of 0.01/(t+1) is 0.01*log(t+1) + C. To verify the result of your integration, take a derivative of the resulting function: (0.01*log(t+1) +C)' = 0.01/((t+1)*ln10) -- not the same as your original function.

However, the idea of MEASURING BUG'S MOVEMENT IN TERMS OF THE PERCENTAGE OF THE STRING TRAVERSED is very interesting. That may be the way to solve the problem without use of differential equations. Although, you still end up deriving a series approximating natural logarithm.

[Guest]

I did not understand the reasoning with respect to "So we can throw the bug's movement due to the rubber band's expansion out the window". If it were true, the bug wouldn't catch up to the scooter, and we already proved that it does. (See my post 15).

The math appears incorrect. I don't see how the integral of 0.01/(t+1) is 0.01*log(t+1) + C. To verify the result of your integration, take a derivative of the resulting function: (0.01*log(t+1) +C)' = 0.01/((t+1)*ln10) -- not the same as your original function.

However, the idea of MEASURING BUG'S MOVEMENT IN TERMS OF THE PERCENTAGE OF THE STRING TRAVERSED is very interesting. That may be the way to solve the problem without use of differential equations. Although, you still end up deriving a series approximating natural logarithm.

The bug is still moved by the rubber band, but that part of its movement plays no part in my calculations, since it doesn't affect the %. I'm not saying it doesn't move, but simply that it doesn't matter that it does move. In fact, my first point in that post was to say that the bug does get moved by the rubber band's expansion.

As for the function, you're right. I got my log and ln mixed up. It's supposed to say 0.01*ln(t+1), which yields t = e^100-1.

[Guest]

The bug is still moved by the rubber band, but that part of its movement plays no part in my calculations, since it doesn't affect the %. I'm not saying it doesn't move, but simply that it doesn't matter that it does move. In fact, my first point in that post was to say that the bug does get moved by the rubber band's expansion.

As for the function, you're right. I got my log and ln mixed up. It's supposed to say 0.01*ln(t+1), which yields t = e^100-1.

It's very common in math for people to use "log" to mean the natural logarithm (ie log base e). For people who don't deal with measurements (like pH, dB, etc), a base 10 logarithm has very little utility.

Interesting side note: Back in 2002, around the whole "Freedom Fries" fiasco, we used to call natural logs "freedom logs," since "ln" comes from the way French puts nouns before adjectives.

[Guest]

That's a good solid effort supplying Bug with necessary differential equations helping to estimate the time it would take to catch up with the scooter. It does seem that the time required is to the tune of e¹⁰⁰. However, I have some problems with the solution.

1. I disagree with the conjecture that "The Prof needs to start a distance d from the wall initially..." just because "Otherwise, the bug has already caught up and my equations won't work as they are undefined for d=0 at t=0."

It is a very viable everyday occurrence that Prof starts at 0 metres from the wall instantly attaining speed of 1m/sec and the magic rubber band and zero-dimention Bug spring into existence. If equations don't work for that situation, we must find different equations. You are right that we should choose different equations and we can: we just define s=0 at t=0! We can never find a nice smooth equation to cover this and t>0 because it is, by definition, discontinuous at t=0: the acceleration is infinite because a magic rubber band has just appeared!

2. You found a function describing distance traveled by Bug: s = Bu*ln(Au/d), or expanding variable "u":

s = B*(t+d/A)*ln(tA/d+1)

Where B is Bug's speed; A is the speed of scooter; d is initial distance of scooter from the wall; and t is elapsed time.

In our problem we set d=1 metre arbitrarily. But what if d was much smaller? Something like d=10^-44? Then after 1 second (t=1), Bug would have traveled more than 1 metre -- further than the scooter! That disagrees with our notion of how long it takes Bug to catch up! This is something I can deal with easier - if d=10^-44 then the equation is correct that s>1, the mistake is ours in assuming that the equation is still valid. My equations are only valid when the bug is on the band and hasn#t already caught the Prof, ie, 0<s<=(At+d). If you calculate the time at which the bug and Prof meet, you find t=d/A . (e^A/B -1) = 0.27s [see below for my correction]. Therefore, if they start so close, the bug catches the Prof after just over quarter of a second...

It seems that function describing the distance traveled by Bug: s = B(t+1)*ln(t+e) fits the bill. But if we take the derivative s' = B*ln(t+e) + B(t+1)/(t+e) -- it does not agree with our concensus what instanteneous speed for Bug should be. (Even though I suggested instanteneous speed myself post#15).

I think there is an error with your equation for A=d=1, which gives u=t+1:

s = Bu ln (u)

s = B(t + 1) ln(t + 1)

s'= Bln(u) + Bu/u

s' = Bln(u) + B

s' = Bs/u + B

which is the instantaneous speed I originally defined.

P.S. I have a feeling, there is a way to solve that problem without use of differential equations.

[Guest]

Someone please check my math and reasoning.

As the rubber band expands, the bug moves along with it, but its position as a percentage of distance covered between the wall and the professor never changes. For example, a bug that starts at 0.42m from the wall on a 1m rubber band can be said to be at 42%, and will always stay at 42% if it stands still. So we can throw the bug's movement due to the rubber band's expansion out the window.

The following assumes that the professor starts at 1m from the wall.

Let's take a look at what % the bug covers by walking in the one second. The % part can be calculated as (distance bug walks)/(length of rubber band). We know that the bug walks 0.01m in one second, but the length of the rubber band isn't constant. However, we know that the professor is traveling at 1m/s and so the rubber band is expanding at a rate of 1m/s. So the length of the rubber band is always t+1. So the rate at which the bug walks is 0.01/(t+1) in % per second.

Now that we have a speed function in %/s, let's integrate it to find the total % covered in some amount of time. We integrate 0.01/(t+1) and get 0.01*log(t+1)+C, where C is an integration constant (0 in this case, since we know that the bug starts at 0m and so 0% from the wall). That is a distance function for calculating the % covered by the bug at any time t. So let's plug in 100%, which is 1. This will tell us how long it takes for the bug to travel 100% of the rubber band's length and reach the professor.

1 = 0.01*log(t+1)

100 = log(t+1) [multiplying both sides by 100]

10^100 = t+1 [changing to exponents]

10^100-1 = t

So the bug reaches the professor in 10^100-1 seconds, or just under a googol seconds.

Damn. You got a far better/simpler answer than mine! The "rubber band" frame of reference was what I was trying to propose in my later posts and, as you show, it works much better because you don't have the extra term in your equation for s.

You also identified an error in my solution: I mistakely put t = u + d/A. It actually t = u - d/A, and so t = d/A(e^A/B - 1). Therefore, I get the same t=e¹⁰⁰ - 1.

[Prime]

The bug is still moved by the rubber band, but that part of its movement plays no part in my calculations, since it doesn't affect the %. I'm not saying it doesn't move, but simply that it doesn't matter that it does move. In fact, my first point in that post was to say that the bug does get moved by the rubber band's expansion.

As for the function, you're right. I got my log and ln mixed up. It's supposed to say 0.01*ln(t+1), which yields t = e^100-1.

OK, I understand, we view Bug's advancement as a percentage of the total length of the string it traversed.

Let's agree on the nomenclature. We'll use some of it as Foolonthehill suggested, except I want to change “d” (initial distance) to H (head start), so that I don't confuse it with “delta”:

A = speed of the scooter.

B = speed of the bug.

H = initial distance of the scooter from the wall

t = time elapsed.

Your theory is that instantaneous speed of the bug in terms of ratio (percentage) of the string is

B/(At+H), or slightly modified B/A(1/(t+H/A).

(That is ratio of the bug's speed to the total length of the rubber band at time t).

An integral of an instantaneous speed is the function for the distance traveled. Integrating, we obtain: B/A * ln(t+H/A) + C

Now, plugging in the initial condition for the bug to be at zero point at t=0, we have: B/A * ln(H/A) + C = 0 and then C = B/A * ln(A/H)

In that way our formula for the bug's distance in terms of the ratio (percentage) of the rubber band traversed becomes:

B/A * ln(t+H/A) + B/A * ln(A/H). Simplifying: B/A * ln(t*A/H + 1)

As we look for the total ratio of the string traversed equal to 1 (the whole string) then we solve the equation: B/A * ln(t*A/H + 1) = 1

Yielding: t = H/A * (e^(A/B) - 1)

That is identical to what FonH had found, but the differential equation is simpler.

Interesting point here:

I thought I was joking when suggested that Bug could simply hang on to the rear fender if we didn't give Prof. T some headstart. But it turns out to be the case. If there was no head start, then it is impossible to decide which point of the rubber band the bug is initially on.

So we should not give Prof. T such a huge handicap in the beginning. Give him something like 10^(-33) meters, and the bug will catch up in a matter of just few years.

[Guest]

So we should not give Prof. T such a huge handicap in the beginning. Give him something like 10^(-33) meters, and the bug will catch up in a matter of just few years.

Yeah, the diameter of a proton is about 10^-15m. Maybe a better idea would be to increase the bug's speed, or decrease the scooter's speed.