Ant on a rubber rope problem confusion

[Kumar8434]

I was reading this wikipedia article: https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

In the analytic solution section, it says that, at time $t$, a point at $x=X$, i.e. initially at a distance $X$ from the starting fixed point on a uniformly expanding rope moves with a speed $\frac{vX}{c+vt}$, where $c$ is the length of the rope and $v$ is the speed with which the end point moves.

I don't think that is correct. Suppose the distance of the point A initially distance $X$ from the starting point O at a time $t$ is given by the function $X(t)$. The end point E is initially at $x=c$ and since it is moving at a speed $v$ so at time $t$ the end point is at $c(t)=c+vt$. Also, since the rope is uniformly expanding so the ratio $\frac{OA}{OE}$ must be maintained. This ratio is initially $\frac{X}{c}$.

So, $\frac{X(t)}{c(t)}=\frac{X}{c}$
$\frac{X(t)}{c+vt}=\frac{X}{c}$
So, $X(t)=(c+vt)\frac{X}{c}$
which gives $X'(t)=\frac{Xv}{c} \neq \frac{vX}{c+vt}$ as mentioned in the article. Did I do something wrong?
Also, I don't think the speeds of the points on a uniformly expanding rope can depend on time because it would mean the points are accelerating.
In the article, it is mentioned in the solution to the problem that fraction of the rope that the ant moves in a time $t$ is preserved regardless of the expansion of the rope. So, $\frac{OA}{OE}$ must be the same for all points O's even if the rope expands with time.

 

[jbriggs444]

In the analytic solution section, it says that, at time $t$, a point at $x=X$, i.e. initially at a distance $X$ from the starting fixed point on a uniformly expanding rope moves with a speed $\frac{vX}{c+vt}$, where $c$ is the length of the rope and $v$ is the speed with which the end point moves.

The formula does not apply to a point initially at X. It applies to a point currently at X.

 

[Kumar8434]

The formula does not apply to a point initially at X. It applies to a point currently at X.

Oh, thanks a lot. Terrible mistake.