- #1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter koosh
- Start date

- #1

- #2

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Is the speed of the ant, u, relative to the rubber band (locally) or relative to the wall?

- #3

rubber band....

It seemed to me that it would be like a conveyor belt at an airport kind of scenario...

It seemed to me that it would be like a conveyor belt at an airport kind of scenario...

Last edited by a moderator:

- #4

would the ant be traveling at its velocity plus the velocity of the band?

- #5

Tide

Science Advisor

Homework Helper

- 3,089

- 0

... all you have to do is solve the ode:

[tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

where x is the distance from the starting position and which is equivalent to

[tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

so the ant will reach the end (x = L) when

[tex]t = \frac {L_0}{v} (e^{v/u} -1)[/tex]

- #6

Tide

Science Advisor

Homework Helper

- 3,089

- 0

- #7

Severian596

- 286

- 0

I tried to solve it a different way using strictly position equations, but I'm pretty sure it's wrong (compared to your fancy stuff!). Here's how I did it.

* Let L = the end of the rubber band

* Let v = velocity of L

* Let x = the ant

* Let u = velocity of x

* The wall is the left edge of this diagram and the motion of the band goes to the right, so after some time t, we have

|-------x---------L

Position equations for the rubber band and the ant, relative to us the observers:

s_T(t) = vt

s_x(t) = (u+v)t [

Set the two equations equal to each other to find out the time they merge:

vt = (u+v)t

vt = ut + vt

1 = ut

According to this "solution", if the band leaves the wall at 5 m/s and the ant travels at 2 m/s, the ant will reach the end of the band after

Thanks in advance, dude. I look forward to your feedback.

- #8

Severian596

- 286

- 0

vt = ut + vt

1 = ut

oops! This actually resolves to

1 = ut/vt + 1 or

0 = u/v

which resolves to total nonesense.

- #9

Tide

Science Advisor

Homework Helper

- 3,089

- 0

The speed of any point on the rubber band varies linearly with position - the point connected to the wall never moves while the other end moves at constant speed v. Let's call it w(x) so that [itex]w(x) = v \frac {x}{L}[/itex]

If the ant moves along at speed u with respect to the band then its speed is the sum of the two so that

[tex]\frac {dx}{dt} = u + w(x)[/tex]

which is just

[tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

Now, L depends only on time so I decided to turn it into the independent variable rather than using time so I wrote

[tex]\frac {dx}{dt} = \frac {dL}{dt} \frac {dx}{dL}[/tex]

but [itex]\frac {dL}{dt} = v[/itex] so

[tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

One way to solve this is to define a new variable y such that [itex]y = x / L[/itex] so that [itex]dx = L dy + y dL[/itex] from which

[tex]v L \frac {dy}{dL} + v y = u + v y[/tex]

or

[tex]v L \frac {dy}{dL} = u[/tex]

Integration gives [itex]\frac {v}{u} (y - y_0) = ln \frac {L}{L_0}[/itex]. Since the ant starts at the wall, [itex]y_0 = 0[/itex] and it reaches the end of the rubber band when [itex]y = 1[/itex]. Therefore,

[tex]\frac {L}{L_0} = e^{v/u}[/tex]

Finally, write L as [itex]L = L_0 + vt[/itex] and solve for t.

- #10

Severian596

- 286

- 0

[tex]w(x) = v \frac {x}{L}[/tex]

This was a key concept that I totally overlooked. Each point on the rubber band moves with a velocity proportionate to its position on the rubber band. The point at [itex]L[/itex] moves at [itex]v[/itex]. The point at [itex]\frac {L}{2}[/itex] moves at [itex]\frac {v}{2}[/itex]. Excellent! Furthermore, this is not a "position" equation because it's not a function of time. Instead it reveals the velocity of any given point as a function of that point's position on the band.

[tex]\frac {dx}{dt} = u + w(x)[/tex]

This is the first place I have trouble. Why is the ant's velocity [itex]\frac {dx}{dt}[/itex]? Thinking out loud, I think, "Okay, we know the velocity of any point x on the rubber band is w(x). And we know that the ant's speed on the rubber band at any given point x is the ant's velocity, u, plus that point's velocity, w(x). So therefore the ant's velocity is a function of x, so let's say [itex]a(x) = u + w(x)[/itex]." But where does taking the differential ([itex]\frac {dx}{dt}[/itex]) come into this line of thinking? You can determine the ant's velocity at any point on the band with a(x). Why differentiate with respect to t? Essentially I'm saying that I'd be stuck at this point. I can understand up to this point and I can understand creating a(x) to tell you the ant's velocity at a given point, but I wouldn't know where to go from there. It looks like you knew and you introduced [itex]\frac {dx}{dt}[/itex], but what prompted you to do that?

My guess is that [itex]\frac {dx}{dt} = u + w(x)[/itex] is another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance along the band that the ant walks equals u + w(x)." You could also write [itex]\frac {dx}{dt} = w(x)[/itex] without the term u. But this is only another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance in space that a point on the band travels equals w(x)." But we knew that. And trying to find the instantaneous velocity of a point on the band is pointless because we know that the velocity of any point on the band

Is this correct? If so then I've just learned something new! If not back to the drawing board.

[tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

Marching on and taking for granted that you're still reading this post (if so THANK YOU), and taking for granted that I haven't thought myself into a corner, this simply substitutes w(x). Thank god, I don't have trouble with this step.

Now, L depends only on time so I decided to turn it into the independent variable

This is a tough leap and it seems very intuitive. How did you know where to go from here, [itex]\frac {dx}{dt} = u + v \frac {x}{L}[/itex]? What do we gain from the transformation that happens after

[tex]\frac {dx}{dt} = \frac {dL}{dt} \frac {dx}{dL}[/tex]

? Frankly I'm almost totally lost, but I'll venture a guess at saying that this equation says, "The infinitismal distance that the ant walks up the band over an infinitismal time period equals the velocity of the band times the velocity of the point x on the band."

[tex]v = \frac {dL}{dt}[/tex]

This is another intuitive step, and I'm not sure where we are after we substitute it into get

[tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

Are we trying to find the point when the ant's velocity ([itex]u + v \frac {x}{L}[/itex]) equals the tip of the rubber band's velocity ([itex]v \frac {dx}{dL}[/itex])?

If you're still reading this I'm surprised but very thankful. I appreciate it, Tide.

- #11

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Severian,

You seem to have the right idea! The basic statement is that, relative to the wall, the ant's speed is equal to the speed of the rubber upon which it is standing (assuming it is not slipping!) plus the speed at which it is walking relative to the band. That's what

[tex]\frac {dx}{dt} = u + w(x)[/tex]

means. The equation

[tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

is a bit of a challenge because the right side depends on both x and t so it's not easy to integrate directly. Rather than getting bogged down in a lot of algebra trying to do that I noticed that L has a simple time dependence so I decided to make it the independent variable. This is where

[tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

came from. Notice that if you set u = 0 then it just represents the relationship between x and L of points on the rubber band even if it does look complicated.

I showed you one way of solving that equation in my previous post. Here's another that may be more revealing. First, rewrite the equation as

[tex]\frac {dx}{dL} - \frac {x}{L} = \frac {u}{v}[/tex]

You may notice the left side somewhat resembles the derivative of a product or quotient (notice the signs of the terms on the left are different!) In fact, dividing both sides by [itex]1/L[/itex] gives

[tex]\frac {1}{L} \frac {dx}{dL} - \frac {x}{L^2} = \frac {1}{L} \frac {u}{v}[/tex]

and now the left side is the derivative of [itex]x/L[/itex] so

[tex]\frac {d}{dL} \frac {x}{L} = \frac {1}{L} \frac {u}{v}[/tex]

so that

[tex]d \frac {x}{L} = \frac {u}{v} \frac {dL}{L}[/tex]

which can be integrated.

- #12

x(t+Dt) = x(t) + va*Dt + vc*x(t)/(1+t*vc)*Dt

then differentiated, messed up a few equations and solved some things... and ultimately got Lo*e^(u/v)-1 but did not divide by v...what you're showing is I'm now sure absolutely right.. so thanks again for the help!

- #13

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Great job! Incidentally, you should check your units for consistency. If you're looking for time then your units should reflect that and getting units of length is a clue you've messed up on the algebra somewhere along the line.

- #14

Severian596

- 286

- 0

Tide thank you very much for your willingness to explain these things to me. I know for sure now that I lack the ability to problem solve using advanced math, and I'm sure it's because I don't feel confident with some fundamentals. I hope I can learn the fundamentals that it takes to be as good as you are with equations some day...it doesn't help that I'm really out of practice (last college calc class was 5+ years ago). I should've committed myself to the concepts better back then.

For the moment, could you answer this? How does this equation represent the relationship between the velocity of the tip of the band and the velocity of any point on the band?

(1) [tex]v \frac {dx}{dL} = v \frac {x}{L}[/tex]

Assuming you divide by v you're left with

(2) [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

Maybe it's the notation I'm becomming confused with? They didn't stress Leibniz notation when I was in school...frankly I think they should've instead of simply saying, "The derivative is f(x) with a prime, f'(x), see?"

If I'm not mistaken does equation (2) mean the same thing as equation (3) below if you assign it some function lable?

(3) [tex]f(x) = \frac {x}{L}[/tex]

Or would f(x) actually be the antiderivative of [itex]\frac {x}{L}[/itex] with respect to L?

(4) [tex]f(x) = x\ ln(L) + C[/tex]

For the moment, could you answer this? How does this equation represent the relationship between the velocity of the tip of the band and the velocity of any point on the band?

(1) [tex]v \frac {dx}{dL} = v \frac {x}{L}[/tex]

Assuming you divide by v you're left with

(2) [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

Maybe it's the notation I'm becomming confused with? They didn't stress Leibniz notation when I was in school...frankly I think they should've instead of simply saying, "The derivative is f(x) with a prime, f'(x), see?"

If I'm not mistaken does equation (2) mean the same thing as equation (3) below if you assign it some function lable?

(3) [tex]f(x) = \frac {x}{L}[/tex]

Or would f(x) actually be the antiderivative of [itex]\frac {x}{L}[/itex] with respect to L?

(4) [tex]f(x) = x\ ln(L) + C[/tex]

Last edited:

- #15

Severian596

- 286

- 0

- #16

Tide

Science Advisor

Homework Helper

- 3,089

- 0

[tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

means

[tex]\frac {x}{L} = \frac {x_0}{L_0}[/tex]

which you can see by rewriting it as

[tex]\frac {dx}{x} = \frac {dL}{L}[/tex]

Integrate both sides to find

[tex]ln \frac {x}{x_0} = ln \frac {L}{L_0}[/tex]

etc.

- #17

Severian596

- 286

- 0

(1) [tex]\frac {dx}{dL} = \frac {x}{L}[/tex]

Is it true to say

[tex]\frac {df}{dL} = \frac {x}{L}[/tex]

where [itex]f[/itex] is the antiderivative of [itex]\frac {x}{L}[/itex] with respect to L? I'm getting the variable x and the function x messed up in the symmetrical equation (1) and it's making it hard for me to really understand what it means...

- #18

Tide

Science Advisor

Homework Helper

- 3,089

- 0

[tex]f{ '} = \frac {f}{L}[/tex]

where the prime means differentiation with respect to L. In other words,

the antiderivative of f/L is equal to f.

- #19

Severian596

- 286

- 0

Interesting...spoken in plain english this equation says, "The derivative of f with respect to L isTide said:[tex]f{ '} = \frac {f}{L}[/tex]

[tex]f(x) = x^2+3[/tex]

I don't think I've ever seen an equation like the one you posted there. It has f on both sides, for god's sake!! In fact...now that I look back I realize that this is the keystone that's been bothering me all along. The equation you initially wrote for the ant on the rubber band was:

[tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

and essentially you're defining the derivative of x in terms that use x itself...I don't get this. Or rather, I kept mixing up what 'x' meant. Did it mean a point on the rubber band that's distance x from the wall? Or a change in the position of the ant who's walking on the rubber band? Or both?

- #20

Severian596

- 286

- 0

You know what, after changing the way the equation looks from (1) to (2)

(1) [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

(2) [tex]dx = (u + v \frac {x}{L})dt[/tex]

It's rather easy to see that you have an equation to represent the ant's progress. Because the progress of the ant down the rubber band equals**rate** times **time**. The rate of the ant is its speed relative to the rubber band (u) plus the speed of the point where it stands on the rubber band at the moment ([itex]v\frac{x}{L}[/itex]), and the time is the infinitismal quantity dt.

(1) [tex]\frac {dx}{dt} = u + v \frac {x}{L}[/tex]

(2) [tex]dx = (u + v \frac {x}{L})dt[/tex]

It's rather easy to see that you have an equation to represent the ant's progress. Because the progress of the ant down the rubber band equals

Last edited:

- #21

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Well, I'm glad that's settled! :)

- #22

Severian596

- 286

- 0

Holy crap! woot! I feel like a student again, thanks so much Tide...your patience have had an impact on this student today, thanks!Tide said:Well, I'm glad that's settled! :)

Just in case v or u is really fast...

[tex]dx = (\frac{u + v \frac {x}{L}}{1+\frac{uv\frac{x}{L}}{c^2}})dt[/tex]

You see how I make things difficult for myself? I learn the difference between Newtonian velocity addition and Relativistic velocity addition, but I still don't understand Leibniz notation...

- #23

- #24

Actually, where I am having trouble with your method is that little y = x/L subsitution and how dx= Ldy + ydL came from... It seems really cool, but if you could explain what exactly happened there I'd be appreciative.Tide said:[tex]v \frac {dx}{dL} = u + v \frac {x}{L}[/tex]

One way to solve this is to define a new variable y such that [itex]y = x / L[/itex] so that [itex]dx = L dy + y dL[/itex] from which

[tex]v L \frac {dy}{dL} + v y = u + v y[/tex]

or

[tex]v L \frac {dy}{dL} = u[/tex]

- #25

Doc Al

Mentor

- 45,447

- 1,907

Try writing it as x = yL, then use the product rule for derivatives.koosh said:Actually, where I am having trouble with your method is that little y = x/L subsitution and how dx= Ldy + ydL came from... It seems really cool, but if you could explain what exactly happened there I'd be appreciative.

Alternatively, use differentials: x + dx = (y + dy)(L + dL)

Multiply it out and discard higher orders of infinitesimals.

- #26

Tide

Science Advisor

Homework Helper

- 3,089

- 0

koosh said:

Yes, that is correct. I see Doc Al filled you in on one of the details (thanks, Al!)

I think the problem you posed is pretty neat! Where did you find it?

- #27

I was actually suprised that I solved it with that (delta) everything equation I posted earlier, because I don't consider myself very good at physics, but rather calculus.

I definitely appreciate your help in showing me this better method.

- #28

Severian596

- 286

- 0

Your skill at math shows, koosh. Keep up the good work, trust me! At this point in my life I wish I'd given math more attention when I was your age.koosh said:I was actually suprised that I solved it with that (delta) everything equation I posted earlier, because I don't consider myself very good at physics, but rather calculus.

Also, in my opinion this problem has almost zero bearing on physics. It's math all the way. Even though the problem takes place in some physical world, it's a world where you can stretch rubber bands indefinitely and ants are determined to walk the length of a rubber band at a constant velocity...

:tongue:

- #29

Severian596

- 286

- 0

(sad thing is koosh is still in high school and already knows how...bummer)

- #30

NateTG

Science Advisor

Homework Helper

- 2,452

- 7

Let's say we keep track of distance in terms of the length of the rubber band, rather than using a fixed scale.

Now, we know that the the length of the rubber band is:

[tex]l_{\rm{initial}}+v(t)[/tex]

So, the ant's relative speed is going to be:

[tex]u \frac{l_{\rm{initial}}}{l_{\rm{initial}}+vt}[/tex]

Now, if you like calculus, you can look at:

[tex]\int_{0}^{\infty} u\frac{l_{\rm{initial}}}{l_{\rm{initial}}+vt}dt[/tex]

which goes to positive infinity for [itex]u>0[/itex] so the ant makes it.

Without calculus it's not a whole lot harder. Clearly, the ant's speed is always increasing, so we can find a lower bound on how far the ant has traveled at time t:

[tex]\sum_{i=0}^{t}u \frac{l_{\rm{initial}}}{l_{\rm{initial}}+vt}[/tex]

Which is also unbounded, since it's very easy to bound below by the harmonic series.

I will concur that proving the unboundedness is trickier than the physics, but, at the same time, physicists generally spend more time dealing with differential equations than mathematicians.

- #31

Tide

Science Advisor

Homework Helper

- 3,089

- 0

That ant's speed relative to the rubber band is u, a constant. Also, the speed of the point on the band upon which the ant is standing at any instant depends on where that point started. Since the ant is moving relative to the band, the speed of the point it is standing on varies so the numerator in your expression cannot be a constant. (I.e., the instantaneous point upon which the ant stands changes from one step to the next and each of those points started off at a different location on the initial band.) Finally, you are assuming the (incorrect) answer by setting the upper limit of your integral to infinity. The ant reaches the end of the band in a finite time.

- #32

NateTG

Science Advisor

Homework Helper

- 2,452

- 7

Tide said:

That ant's speed relative to the rubber band is u, a constant. Also, the speed of the point on the band upon which the ant is standing at any instant depends on where that point started. Since the ant is moving relative to the band, the speed of the point it is standing on varies so the numerator in your expression cannot be a constant. (I.e., the instantaneous point upon which the ant stands changes from one step to the next and each of those points started off at a different location on the initial band.) Finally, you are assuming the (incorrect) answer by setting the upper limit of your integral to infinity. The ant reaches the end of the band in a finite time.

It's probably my fault, but it's clear that you didn't understand what I wrote.

I was taking the (unorthodox) approach of changing from an absolute unit of length to using the rubber band's length as a unit of length. This has the disadvantage that some quantities - like the ant's speed relative to the rubber band - which were constant, are now variable, but has the advantage that other quantities - the length of the rubber band - become constants.

Regarding the use of the improper (only in the sense of limits of integration) integral:

The question whether the ant reaches the end of the rubber band is quite similar to asking whether, for:

[tex]f(x)=\int_{0}^{x}\frac{1}{y}dy[/tex]

it is ever true that

[tex]f(x)>a[/itex]

for some arbitrary [itex]a[/itex]

This is equivalent to asking wether:

[tex]\int_{0}^{\infty}\frac{1}{y}dy[/tex]

tends to positive infinity.

- #33

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Nate,

Yes, I did miss that the first time around. I like the approach - nice!

Yes, I did miss that the first time around. I like the approach - nice!

Share:

- Replies
- 2

- Views
- 316

- Replies
- 2

- Views
- 565

- Last Post

- Replies
- 4

- Views
- 296

- Last Post

- Replies
- 3

- Views
- 345

- Replies
- 14

- Views
- 190

- Last Post

- Replies
- 8

- Views
- 566

- Last Post

- Replies
- 16

- Views
- 691

- Replies
- 51

- Views
- 1K

- Last Post

- Replies
- 25

- Views
- 677

- Replies
- 30

- Views
- 578