% MATHS DIWALI ASSIGNMENT
% SOUMIK PAL
% 200901024
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\draftstring{SOUMIK PAL }
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\title{\textcolor{blue}{THE ANT ON A TAUT RUBBER ROPE PROBLEM MODEL}}
\begin{center}
\Huge
{\bf \textcolor{blue}{
{ DIWALI ASSIGNMENT ON CALCULUS } \\
}
}
\vspace{2 cm}
\huge\bf
{ Assigned by : Professor Manish K Gupta \\
COURSE : SC 105 \\
Calculus with complex variables\\
FALL 2009 \\
DA-IICT \\
GANDHINAGAR \\
}
\vspace{5 cm}
\LARGE {\it
{
Made by: SOUMIK PAL \\
ID : 200901024 \\
}
}
\end{center}
\author{
SOUMIK PAL\\
200901024,\\
DAIICT,\\
GANDHINAGAR,\\
INDIA\\
\texttt{200901024@daiict.ac.in}
} %\texttt formats the text to a typewriter style font
\date{\today}
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\maketitle
\begin{abstract}
\textcolor{red}
{
\large
{
In this article, I shall discuss the analytical way of solving the classical
\uwave {`ANT ON A TAUT RUBBER ROPE PROBLEM'}.
This document itself does not go into much depth, but is instead the output of a model
to demonstrate the structure of the universe and the principles of space-time.
}
}
\end{abstract}
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\newpage
\large
\section{\textcolor{blue}{INTRODUCTION}}
\textsf
`` Ant on a rubber rope '' is a mathematical puzzle with a solution that appears
counter-intuitive or paradoxical. It is sometimes given as a worm, or
inchworm, on a rubber or elastic band, but the principles of the puzzle
remain the same.
The details of the puzzle can vary,
but a typical form is as follows. \cite{1} \cite{3} \cite{2}
`` An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second
(relative to the rubber it is crawling on)created by. At the same time, the rope starts to stretch by
1 km per second (so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc).
Will the ant ever reach the end of the rope ? ''
At first consideration it seems that the ant will never reach the end of the rope,
but in fact it does (although in the form stated above the time taken is colossal).
In fact, whatever the length of the rope and the relative speeds of the ant and
the stretching, providing the ant's speed and the stretching remain steady
the ant will always be able to reach the end given sufficient time.
\section{\textcolor{blue}{ A FORMAL STATEMENT OF THE PROBLEM }}
\textsf
The problem as stated above requires some assumptions to be made. The
following fuller statement of the problem attempts to make most of those
assumptions explicit.
Consider a thin and infinitely stretchable rubber rope held taut
along an $x$-axis with a starting-point marked at $\textcolor{blue}{x = 0}$ and a
target-point marked at $\textcolor{blue}{x = c}$, $\textcolor{blue}{c > 0}$.
At time $\textcolor{blue}{t = 0}$ the rope starts to stretch uniformly and smoothly in
such a way that the starting-point remains stationary at $\textcolor{blue}{x = 0}$
while the target-point moves away from the starting-point with
constant speed $\textcolor{blue}{v > 0}$.
A small ant leaves the starting-point at time $\textcolor{blue}{t = 0}$ and walks
steadily and smoothly along the rope towards the target-point at a
constant speed $\textcolor{blue}{\alpha > 0}$ relative to the point on the rope where the ant
is at each moment.
Will the ant reach the target-point?
\newpage
\section{\textcolor{blue}{AN ANALYTICAL SOLUTION}}
\textsf
A key observation is that the speed of the ant at a given time $\textcolor{blue}{ t > 0 }$
is its speed relative to the rope, i.e. $\alpha$ , plus the speed of
the rope at the point where the ant is. The target-point moves with
speed $v$, so at timecreated by $t$, it is at $\textcolor{blue}{x = c + vt}$. Other points along
the rope move with proportional speed, so at time t the point on the
rope at $\textcolor{blue}{x = X}$ is moving with speed $\textcolor{blue}{\frac{vX}{c+vt}}$. So if we write
the position of the ant at time t as $y(t)$, and the speed of the
ant at time $t$ as $y'(t)$, we can write:
\[
\textcolor{blue}
{
\mathbf
{y'(t)=\alpha+\frac{v\,y(t)}{c+vt}}
}
\]
This is a \textcolor{blue}{first order linear differential equation}, and it can be
solved with standard methods. However, to do so requires some moderately
advanced calculus. A much simpler approach considers the ant's position
as a proportion of the distance from the starting-point to the
target-point.\cite{3}
Consider coordinates $\psi$ measured along the rope with the
starting-point at $\textcolor{blue}{\psi=0}$ and the target-point at $\textcolor{blue}{\psi = 1}$.
In these coordinates, all points on the rope remain at a fixed position (in
terms of $\psi$ ) as the rope stretches. At time $\textcolor{blue}{t\ge 0}$ , a point at
$\textcolor{blue}{ x = X }$ is at $\textcolor{blue}{\mathbf{\psi= \frac{X}{c+vt}}}$,
and a speed A, relative to the rope in terms of x, is equivalent to a speed
$\textcolor{blue}{\frac{A}{c+vt}}$, in terms of $\psi$. So if we write the position of
the ant in terms of $\psi$ at time t as $\phi(t)$, and the speed
of the ant in terms of $\psi$ at time t as $\phi'(t)$, we can write:
\[
\textcolor{blue}
{
\mathbf{\phi'(t)=\frac{\alpha}{c+vt}}
}
\]
\[
\textcolor{blue}
{
\mathbf{\therefore \phi(t)=\int{\frac{\alpha}{c+vt}\,dt}=\frac{\alpha}{v}\log(c+vt)+\kappa}
}
\]
where $\kappa$ is a constant of integration.
Now, $\textcolor{blue}{\phi(0)=0}$ which gives $\textcolor{blue} {\kappa=-\frac{\alpha}{v}\log{c}}$, so
\[
\textcolor{blue}
{
\mathbf
{\phi(t)=\frac{\alpha}{v}\log{\left(\frac{c+vt}{c}\right)}}.
}
\]
\newpage
If the ant reaches the target-point (which is at $\textcolor{blue}{\psi = 1}$) at time
$\textcolor{blue}{t = T}$, we must have $\textcolor{blue}{\phi(T) = 1}$ which gives us:
\[
\textcolor{blue}
{
\mathbf
{ \frac{\alpha}{v}\log{\left(\frac{c+vT}{c}\right)}=1 }
}
\]
\[
\textcolor{blue}
{ \mathbf
{\therefore T=\frac{c}{v}\left(e^{v/\alpha}-1\right)}
}
\]
As this gives a finite value $T$ for all finite
$\textcolor{blue}{c , v ,\alpha (v>0, \alpha>0 )}$, this means that,
given sufficient time, the ant will complete the journey to the target-point.
This formula can be used to find out how much time is required.
\\
For the problem as originally stated,
$\textcolor{blue}{c=1\,\mathrm{km}}$,
$\textcolor{blue}{v=1\,\mathrm{km}/\mathrm{s}}$ and
$\textcolor{blue}{\alpha=1\,\mathrm{cm}/\mathrm{s}}$, which gives
\[
\textcolor{blue}
{
\mathbf
{T=(e^{100,000}-1)\,\mathrm{s} \approx2.8\times10^{43,429}\,\mathrm{s}}
}
\]
This is a truly vast timespan, vast even in comparison to the estimated
age of the created byuniverse, and the length of the rope after such a time is
similarly huge, so it is only in a mathematical sense that the ant can
ever reach the end of this particular rope.
$\therefore$ THE PROBLEM IS SOLVED.
\section{\textcolor{blue}{APPLICATIONS OF THE PROBLEM}}
\textsf
This problem has a bearing on the question of whether light from distant
galaxies can ever reach us if the universe is expanding.
If the universe is expanding uniformly, this means
that galaxies that are far enough away from us will have an apparent
relative motion greater than the speed of light.\cite{2} This does not violate
the relativistic constraint of not travelling faster than the speed of
light, because the galaxy is not "travelling" as such -- it is the space
between us and the galaxy which is expanding and making new distance.
The question is whether light leaving such a distant galaxy can ever
reach us, given that the galaxy appears to be receding at a speed
greater than the speed of light.
By thinking of light photons as ants crawling along the rubber rope of
space between the galaxy and us, it can be seen that just as the ant
will eventually reach the end of the rope, given sufficient time, so the
lightted by from the distant galaxy will eventually reach earth, given
sufficient time.
\section{\textcolor{blue}{CONCLUSION}}
\textbf
To summarise, the answer to the original question as stated at the beginning of the project is:
yes, the ant will always reach the end of the string provided that the string is finite and
is expanding at a finite rate and the ant has finite speed.
Similarly, light from the distant galaxies will always eventually reach earth ,
given sufficient time.
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