# Ant on a rubber string, mathematical series

1. Apr 15, 2014

### prehisto

1. The problem statement, all variables and given/known data
Ideal rubber stirng with lenght L=1km.Ant is takng a walk on the string with speed v=1cm/s
After every minute(Δt=60s) ,string is getting longer by ΔL=1km.
1)Will ant get to the end of string?
2)If yes,then how long it will take ?

2. Relevant equations
1) So i used series expansion
$\frac{s}{l}$=$\frac{vΔt}{L}$+$\frac{vΔt}{L+ΔL}$+..=
vΔt∑$\frac{1}{L+iΔL}$

The ant will get to the end when s/l=1
So i got that.

2)But I have problem figuring out how to get the time needed to get there.
I have tip to use integral test,but i do not understand how to use it,beacuse as far as i know,tests are used to test if series converges.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 15, 2014

### BvU

Your series expansion can be simplified to
$$\frac{s}{l}=\frac{vΔt}{L}\Bigl ( {1\over 1} + {1\over 2} + {1\over 3}+ {1\over 4} ... \Bigr )$$
If you (mentally) replace the summation by an integration you get a hunch that this sum is very close to a logarithm.
The expression in brackets is a harmonic number link
Mathematicians will claim the ant gets there.
Physicists are much more realistic: they take the lifetime of the universe into consideration and claim it doesn't get there.

3. Apr 15, 2014

### dauto

Are you sure about that? That series does not converge, I don't think.

4. Apr 16, 2014

### BvU

There is no question of convergence: the summation doesn't have to be continued ad infinitum. In fact, how far it has to be continued is OP part 2) of the exercise.

5. Apr 16, 2014

### Staff: Mentor

I would like to offer a different analysis of this problem.

Let L represent the total length of the string at time t, and let r represent the rate at which the length is increasing (1 km/min). Then,
$$L=L_0+rt$$
where L0 is the initial length (1 km). The velocity of the far end of the string is dL/dt = r. The velocity at location y along the string is
$$v_s(y)=r\frac{y}{L}=r\frac{y}{L_0+rt}$$
When the ant is at location y, its velocity is
$$\frac{dy}{dt}=r_A+v_s(y)=r_A+\frac{ry}{L_0+rt}$$
where rA is the velocity of the ant relative to the string.

The solution to this differential equation subject to the initial conditions is:
$$\frac{y}{L_0+rt}=\frac{r_A}{r}\ln \left(\frac{L_0+rt}{L_0}\right)$$
The ant reaches the end of the string when $y=L_0+rt$. Therefore, the time required is obtained from:

$$t=\frac{L_0}{r}\left(\exp(\frac{r}{r_A})-1\right)$$

Chet

6. Apr 17, 2014

### BvU

Interesting factor $\gamma$ between the "("faster!") discrete case and the continuous case !

With $\gamma = 0.57721566490153286060651209008240243104215933593992$ the discrete ant gets there 3*10723 minutes before the continuous ant !

But I'm afraid we've lost our prehisto friend here. Still listening in ?

7. Apr 21, 2014