# Will the Bug Ever Reach the End of the Stretching Rubber Band?

• Nathanael
In summary: B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7In summary, the conversation discusses a problem involving a 1 km long rubber band with one end attached to a wall and the other end in the hand. The bug crawls towards the hand at a rate of 1 cm/sec while the rubber band is being continuously stretched at a rate of 1 km/sec. The question is whether the bug will ever reach the hand and if so, in how much time. A suggested solution involves setting up an equation using the fraction of the rubber band that the bug has crawled behind him and using
Nathanael
Homework Helper

## Homework Statement

"You have a 1 km long rubber band with one end attached to the wall, and the other in your hand. The bug begins to crawl towards you on the rubber band, starting from the wall, at a rate of 1 cm/sec. As he crawls the first centimeter you extend the rubber band 1 km; when he crawls the second centimeter you extend the rubber band another 1 km, and so on, every second. The question is: Does the bug ever reach you, and if so, in how much time?"

## Homework Equations

$L=1km$
$v=1km/s$
$u=1 cm/s$
x = distance between the bug and the wall

## The Attempt at a Solution

I came up with the following equation:

$dx=(u+\frac{x}{L+vt})dt$

but I don't know how to solve it. I'm thinking of integrating it from $0$ to $T$ and then setting it equal to $L+vT$ and then solving for $T$ but I don't know how to do this (I don't have experience with differential equations)
Is it possible to get a value for $T$ with this equation, and if so, how would I do it?

Thanks.

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Nathanael said:
I came up with the following equation:

$dx=(u+\frac{x}{L+vt})dt$
Have you made a dimensional analysis to see if this equation makes sense?

Nathanael said:
"You have a 1 km long rubber band with one end attached to the wall, and the other in your hand. The bug begins to crawl towards you on the rubber band, starting from the wall, at a rate of 1 cm/sec. As he crawls the first centimeter you extend the rubber band 1 km; when he crawls the second centimeter you extend the rubber band another 1 km, and so on, every second. The question is: Does the bug ever reach you, and if so, in how much time?"
I'm not sure how to interpret the question. The blue highlighted part I interpret to mean that the rubber band is stretched continuously, while the red I interpret to mean that the rubber band is stretched instantaneously every second.

1 person
Nathanael said:
$dx=(u+\frac{x}{L+vt})dt$

Watch out here, you are adding a velocity and a dimensionless number.

I would suggest instead working with the fraction of the rubber band that the bug has behind him. When it reaches 1 the bug has arrived. The problem will then require only basic integration.

@DrClaude: In the classical formulation of this problem, the bug crawls 1 cm/s and the band is being extended at a rate of 1 km/s. The only difference is whether to consider the problem a difference equation or a differential equation.

1 person
Orodruin said:
Watch out here, you are adding a velocity and a dimensionless number.
DrClaude said:
Have you made a dimensional analysis to see if this equation makes sense?
Yes, it was just a typo. I meant to put $dx=(u+v\frac{x}{L+vt})dt$ (Sorry, I should type more carefully.)
That's the ratio of the distance behind him to the entire length, multiplied by the speed of stretching, so the dimensions are now correct.

DrClaude said:
I'm not sure how to interpret the question. The blue highlighted part I interpret to mean that the rubber band is stretched continuously, while the red I interpret to mean that the rubber band is stretched instantaneously every second.
It does seem a bit contradictory. It doesn't matter though, (it's not homework), so consider it to be continuously stretched. (That's the case that my equation applies to.)

Orodruin said:
I would suggest instead working with the fraction of the rubber band that the bug has behind him. When it reaches 1 the bug has arrived. The problem will then require only basic integration.

Let's say $r=\frac{x}{L+vt}$ then I get:

$dr=\frac{u}{L+vt}dt+\frac{vr}{L+vt}dt$

Then I would set the integral equal to 1.
I don't see how this helps. Maybe that's not what you meant, I'm tired. I think I'll go to bed.

Nathanael said:
Let's say $r=\frac{x}{L+vt}$ then I get:

$dr=\frac{u}{L+vt}dt+\frac{vr}{L+vt}dt$

[STRIKE]##dx/dt## is not equal to ##u## and the second term should have a minus sign since ##L+vt## was in the denominator. It also has a contribution from the stretch of the rubber band as you described in your first post (in fact you should be able to insert your expression for it here). An alternative is just making the change of variables ##r(L+vt) = x## in your (corrected) original equation.[/STRIKE]

Edit: I realize now that the above might not have been what you did. When you do the differentiation of the right-hand side of your substitution, remember that ##L+vt## is not a constant, but that ##d(L+vt) = v\, dt##.

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1 person
The problem is I don't know how to calculate the integral. It seems recursive. (I could write it as an infinite chain of fractions.)
x depends on the integral, which depends on x, (which depends on the integral, and so on and so forth indefinitely).

I've come across these infinite fraction chains before but I don't know what to do with them. I'm guessing it doesn't help to treat it as a continued fraction? And there's probably a simpler way to evaluate it? I just don't know how.

Your first derivative, $\frac{dx}{dt} = u + v \frac x{L+vt}$, is fairly easy to integrate. Clearing the denominator and rearranging yields $(L+vt)\frac {dx}{dt} - u(L+vt) = vx$. That's a first order ODE in normal linear form. Use an integrating factor and you're done.

Your second derivative in post #4 is even easier. You computed the derivative incorrectly. With $r = \frac x {L+vt}$, the derivative is $\frac {dr}{dt} = \frac{\dot x (L+vt) - vx}{(L+vt)^2}$. See if you can take it from there.

We've had this one several times before; can't find it because I don't know how to operate the PF advanced search (help!).

there's even a Wiki on it here

BvU said:
We've had this one several times before; can't find it because I don't know how to operate the PF advanced search (help!)

D H said:
Your first derivative, $\frac{dx}{dt} = u + v \frac x{L+vt}$, is fairly easy to integrate. Clearing the denominator and rearranging yields $(L+vt)\frac {dx}{dt} - u(L+vt) = vx$. That's a first order ODE in normal linear form. Use an integrating factor and you're done.

Thanks for your help but I still haven't made it to the answer. I'm sure it is simple for you guys but I'm still learning. I think I will save this problem for later while I learn more mathematics.

If you are in the US, you'll typically learn how to solve $\frac {dx} {dt} = u + v \frac {x(t)} {L+vt}$ in your second or third college calculus class.

Solving for $r(t)$ where $r(t) = \frac {x(t)} {L+vt}$ only requires high school AP calculus. You should be able to do that. Your problem is that you made an error in calculating the derivative of $r(t)$. You didn't use the quotient rule.

Nathanael said:
Yes, it was just a typo. I meant to put $dx=(u+v\frac{x}{L+vt})dt$ (Sorry, I should type more carefully.)

Could you please tell me how you got this equation?

Satvik Pandey said:
Could you please tell me how you got this equation?

I assumed that the rubber band is stretched uniformly. So the "speed of stretching" of a piece of the rubber band is propotional to the length of that piece, and to the speed that the entire rubber band is being stretched.
[[[EDIT:]]]
[[[I should've said it's proportional to the ratio of the piece's length with the entire length (not just to the piece's length)]]]
(I call the speed that the entire rubber band is being stretched "v" and the length of the entire band "L")

I guess the mathematical way to say this idea (of "uniform stretching") would be $dv=v\frac{dL}{L}$

For example, a piece of the rubber band of length $k$ would be stretched with a speed of $v\frac{k}{L}$Let's now call the position of the bug $x$ (as measured from the wall)

The length of rubber band as a function of time is $L+vt$ (where $L$ is the initial length)

The speed of the bug ($\frac{dx}{dt}$) would be the walking speed of the bug ($u$) plus the speed the rubber band behind the bug being stretched.

So you get:

$\frac{dx}{dt}=u+v\frac{x}{L+vt}$I hope this was clear (I am sometimes bad at explaining)

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1 person
Nathanael said:
I assumed that the rubber band is stretched uniformly. So the "speed of stretching" of a piece of the rubber band is propotional to the length of that piece, and to the speed that the entire rubber band is being stretched.
[EDIT:]
[I should've said it's proportional to the ratio of the piece's length with the entire length (not just to the piece's length)]
(I call the speed that the entire rubber band is being stretched "v" and the length of the entire band "L")

I guess the mathematical way to say this idea (of "uniform stretching") would be $dv=v\frac{dL}{L}$

For example, a piece of the rubber band of length $k$ would be stretched with a speed of $v\frac{k}{L}$

Let's now call the position of the bug $x$ (as measured from the wall)

The length of rubber band as a function of time is $L+vt$ (where $L$ is the initial length)

The speed of the bug ($\frac{dx}{dt}$) would be the walking speed of the bug ($u$) plus the speed the rubber band behind the bug being stretched.

So you get:

$\frac{dx}{dt}=u+v\frac{x}{L+vt}$

I hope this was clear (I am sometimes bad at explaining)

Thank you Nathanael.I got that.
[

I haven't officially taken calculus, yet, (my high school didn't have it) and I've only read the first 5 chapters of a calculus text (strang's). I've been intending for a little while to finish reading Calculus, but I've made little progress because physics is far more fun. My understanding of calculus (however unthrorough) comes from physics problems.

D H said:
Solving for $r(t)$ where $r(t) = \frac {x(t)} {L+vt}$ only requires high school AP calculus.
What didn't make sense to me in solving this, is that, I only know x(t) as "the solution to that other equation" so how am I supposed to integrate an ambiguous function?
[[[EDIT:]]]
[[[It makes more sense to me how it's possible, because although x(t) is ambiguous, the integral of x(t) (from 0 to the "T" which satisfies the problem) is known to be L+vT; but I still don't know how to use this knowledge]]]
D H said:
If you are in the US, you'll typically learn how to solve $\frac {dx} {dt} = u + v \frac {x(t)} {L+vt}$ in your second or third college calculus class.
I've written it as the limit of a series, but there are two problems for figuring it out. The series is:

$lim_{\Delta t\rightarrow 0} [^{(k/\Delta t)}_{(n=0)}\Sigma(x_n)]=vk\Delta t+L$
Where $x_n=nu\Delta t+\frac{x_{n-1}}{L+nv\Delta t}$

The first problem is the way I've defined k (the upper limit). It is not explicitly stated (I couldn't think of a way to write it explicitly, so I just wrote the relationship). This makes it quite confusing to deal with.

The second problem is, (as I touched on in an earlier post,) the sequence is recursive. The definition of $x_n$ depends on $x_{n-1}$. This isn't theoretically a problem (because we know $x_0=0$) but it makes it too complicated for me. The only "solvable" (non recursive) way I can think of to write it is, as long chain of fractions (which becomes infinite in the limit).

That is beyond my capabilities, so I will just save this problem for a future day.
EDIT:
I probably should've written " k " as " T " because that's what it represents in the equation (it is the solution of the problem)

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Satvik Pandey said:
Thank you Nathanael.

You're welcome

P.S. (to D H)

D H said:
If you are in the US, you'll typically learn how to solve $\frac {dx} {dt} = u + v \frac {x(t)} {L+vt}$ in your second or third college calculus class.

If you don't mind, can you show me how you would solve this?
Even if I don't entirely understand, it could possibly give me something to think about.
(I like learning math from physical ideas like this, where you can actually picture what's happening)

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Diffrential of form
$\frac { dy }{ dx }$ +Py=Q
(where P is constant and Q can be constant as well as function of y)
can be solved as

$y{ e }^{ \int { Pdx } }$=$\int { \{ Q.{ e }^{ Pdx } } \} dx\quad$ +C

$\frac{dx}{dt} -v\frac{x}{L+vt}$ =u

Can it be solved like this?

$y{ e }^{ \int { \frac { v dt }{ L+vt } } }$=$\int { u.{ e }^{ \int { \frac { v dt }{ L+vt } } } } dt\quad$ +C

1 person
Nathanael said:
$\frac{dx}{dt}=u+v\frac{x}{L+vt}$

Nathanael, I understand that you do not want to apply Maths methods you do not know and understand yet, but you might be curious about the solution, are you not? You can cheat a bit then and look at wolframalpha.com, but first bring the equation to a simpler form, by introducing dimensionless quantities where possible.

So choose y=x/L and denote a=u/L and b=v/L. You get

$\frac{dy}{dt}=a+\frac{by}{1+bt}$ with a= u/L and b=v/L and the initial condition y=0 at t=0.

I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0. That is a separable equation, it can be written as

$\frac{df}{dt}=-P(t)f$ and the remaining part is $fg'+Q(t)=0$

I rearrange it (it does not seem legal, but it will be)

$$\frac{df}{f}=-P(t) dt$$

and integrate

$$\int {\frac{df}{f}}=\int{-P(t) dt}$$

Do not bother with the integration constant. Substitute f into the remaining part of the equation:
$fg'+Q(t)=0$: $g'=-Q(t)/f=0$
Integrate, but add the integration constant now. Substitute back into y=fg.

The method is equivalent with the integrating factor method in principle, but I never can remember that

Try, just for fun. ehild

1 person
Was my method correct,ehild?

Yes, it can be solved by the integrating factor method, but what is∫(vdt/(L+vt))?

ehild

ehild said:
Yes, it can be solved by the integrating factor method, but what is∫(vdt/(L+vt))?

ehild

Substituting (L+vt)=a.So da=vdt.

=$\int { \frac { da }{ a } }$ =$\log { a }$ =$\log { (L+vt) }$

ehild said:

Let R=$\log { (L+vt) }$
At t=0 R= log(L)
Is it right?

Well, go on...

1 person
Take care with the minus: P=-v/(L+vt)

ehild said:
Well, go on...

$y{ e }^{ -log { (L+vt) } }$=$u\int { { e }^{ -log(L+vt) } } dt\quad +C$

How to solve $u\int { { e }^{- log(L+vt) } } dt\quad +C$?

Should I substitute log(L+vt)?

I am not very comfortable in integrating logarithmic and exponential function.

Why we have not considered constant of ∫(vdt/(L+vt))?

What is ##e^{-\log(x)}##

1 person
ehild said:
but first bring the equation to a simpler form, by introducing dimensionless quantities where possible.

So choose y=x/L and denote a=u/L and b=v/L. You get

$\frac{dy}{dt}=a+\frac{by}{1+bt}$ with a= u/L and b=v/L and the initial condition y=0 at t=0.
I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)

ehild said:
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0. That is a separable equation, it can be written as

$\frac{df}{dt}=-P(t)f$ and the remaining part is $fg'+Q(t)=0$
Ok that is pretty cool, I just don't quite understand how you did f'g+P(t)fg=0? That is just for convenience? Why are you at liberty to set up that equation arbitrarily?

ehild said:
I rearrange it (it does not seem legal, but it will be)

$$\frac{df}{f}=-P(t) dt$$

and integrate

$$\int {\frac{df}{f}}=\int{-P(t) dt}$$

Do not bother with the integration constant.

Why do we not bother with the integration constant? Is it because it will "mix" with the integration constant in the next part?

Integrating both sides I get $ln(f)=ln(bt+1)$ and thus (taking e^ to both sides) $f=bt+1$? (This doesn't seem quite right)

Thank you very much for your reply ehild! It was very interesting to see how it's done I have no idea how you looked at the equation and this is what came to mind!

(I went through the whole solution but I made a mistake somewhere, because my answer was wrong.)
(I am too tired to do it again tonight, (it's 1:30am) but I will try again, more carefully, tomorrow.)

Nathanael said:
I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)Ok that is pretty cool, I just don't quite understand how you did f'g+P(t)fg=0? That is just for convenience? Why are you at liberty to set up that equation arbitrarily?

Yes y=x/L is the same Orodruin suggested.
fg=y, but you can multiply one of them with something and divide the other with the same something. You choose f so that f'g+P(t)fg=0.
Nathanael said:
Why do we not bother with the integration constant? Is it because it will "mix" with the integration constant in the next part?
Yes, one integration constant is enough, and you include it in the next part. But you may add an integration constant in the first part. It might be useful if you get some logarithmic function.
Nathanael said:
Integrating both sides I get $ln(f)=ln(bt+1)$ and thus (taking e^ to both sides) $f=bt+1$? (This doesn't seem quite right)
Yes, f=1+bt. Why do you think it is not right?
Nathanael said:
(I went through the whole solution but I made a mistake somewhere, because my answer was wrong.)
(I am too tired to do it again tonight, (it's 1:30am) but I will try again, more carefully, tomorrow.)
OK, have a good night. ehild

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1 person
ehild said:
What is ##e^{-\log(x)}##

##{ e }^{ -log(x) }=\frac { 1 }{ x } ##

Using this

##\int { { e }^{ -log(L+vt) } } dt=\int { \frac { dt }{ L+vt } } ##

Let ##L+vt=a##.
So ## dt=\frac { da }{ v } ##

##\frac { 1 }{ v } \int { \frac { da }{ a } } =\frac { \log { (L+vt) } }{ v } ##

##\frac { x }{ (L+vt) } =\frac { u\log { (L+vt) } }{ v } +C##

##x=(L+vt)\frac { u\log { (L+vt) } }{ v } +C##

At ##t=0 x=0##

##0=\frac { Lu log(L) }{ v } +C##

As ##L=1##

So ##log(1)=0##

So ##C=0##

Is it right?

Satvik Pandey said:
##x=(L+vt)\frac { u\log { (L+vt) } }{ v } +C##

don't you miss parentheses?

ehild

Nathanael said:
I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

But it is not bad that you have understood how to solve a linear first-order differential equation.

ehild

ehild said:
Substitute f into the remaining part of the equation:
$fg'+Q(t)=0$: $g'=-Q(t)/f=0$
Integrate, but add the integration constant now. Substitute back into y=fg.

So:

$g=\frac{a}{b}ln(bt+1)+C$

And:

$y=(bt+1)(\frac{a}{b}ln(bt+1)+C)$

From the initial condition we see that C=0

Replacing a, b, and y, with their respective definitions (and multiplying by L) I get:

$x=(vt+L)(\frac{u}{v}ln(\frac{vt}{L}+1))=L+vt$

Which yields:

$t=e^{100000}-1$

Which is the correct answer[Edit: much easier when I'm not tired hehe ]

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1 person
ehild said:
Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.
It's amazing how you guys have such an intuitive grasp of differential equations. It is very impressive to me, I don't understand it at all.

ehild said:
But it is not bad that you have understood how to solve a linear first-order differential equation.
I would hardly say I've understood it! I would have never done that in a million years! I only "understood" because you were nice enough to walk me through it like a baby

ehild said:
I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0.

Does this always work? (For first order linear DE's)
If not then how often does it work?

Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?
(Or do you have to be clever about it?)

Nathanael said:

Does this always work? (For first order linear DE's)
If not then how often does it work?

This method always works for linear first order differential equations, just like the integrating factor method.

Nathanael said:
Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?

That belongs to the method. As you wrote, '"it is the trick" of the method.

Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

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