Ant on a rubber string, mathematical series

[prehisto]

Homework Statement

Ideal rubber stirng with length 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 ?

Homework 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.
Please,help here ?

The Attempt at a Solution

 

[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.

Spoiler

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.

 

[dauto]

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.

Spoiler

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.

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

 

[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.

 

[Chestermiller]

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 r_A 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

 

[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 ?

 

[prehisto]

I red the first post and that was enough for me at the time. And forgot about this thread :(
But thanks guys.
Now I am looking into latest posts and they are interesting!