What is the final distance between two ships if one heads towards ($\infty$,0) and the other ship always heads towards the first? I just finished taking BC Calculus this year, and I read an interesting question that prompted me to think up this one:
There are two ships, A and B, traveling at the same positive speed on the XY plane. Ship A is located at $(0,0)$ and heads towards $(\infty,0)$ while ship B is located at $(0,k)$ and always heads towards ship A. As ship A approaches $(\infty,0)$, what does the distance between ships A and B approach?
I tried the question with both ships traveling at 1 unit/second, but I couldn't parameterize the x and y coordinates of both ships with respect to time. From my understanding of calculus, if the path B takes can be represented by a function $f(x)$, the distance between the ships will end up being $$\int_{0}^{\infty}f(x) dx$$
I couldn't find an equation to model B's movement, but I created a computer program to solve for several values of k and I received the formula:

 $\frac{k}{2}$ = distance between A and B

How is the formula for the distance between ships A and B derived? 
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\half}{{1 \over 2}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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The velocities $\ds{\vec{v}_{i}\pars{t}\,,\ \pars{~i = A, B~}}$, satisfy
  $\ds{t}$-independent $\ds{v_{i} \equiv \verts{\vec{v}_{i}\pars{t}} \equiv v}$.
  $\ds{v}$ is the common rapidity.

\begin{align}
\vec{v}_{B}\pars{t} & =
{\vec{r}_{A}\pars{t} - \vec{r}_{B}\pars{t} \over
\verts{\vec{r}_{A}\pars{t} - \vec{r}_{B}\pars{t}}}\,v\,,\qquad
\vec{r}_{A}\pars{t} = vt\,\hat{x}
\end{align}

Lets $\ds{\vec{r}_{B}\pars{t} \equiv
\vec{r}_{A}\pars{t} + \rho\pars{t}\cos\pars{\theta\pars{t}}\,\hat{x} + \rho\pars{t}\sin\pars{\theta\pars{t}}\,\hat{y}}$:
$$
\imp\quad
\bracks{v + \cos\pars{\theta}\dot{\rho} - \rho\sin\pars{\theta}\dot{\theta}}\hat{x} 
+\bracks{\sin\pars{\theta}\dot{\rho} + \rho\cos\pars{\theta}\dot{\theta}}\hat{y} =
-v\cos\pars{\theta}\hat{x} - v\sin\pars{\theta}\hat{y}
$$

\begin{align}
&\left\lbrace\begin{array}{rcrcl}
\cos\pars{\theta}\dot{\rho} & - & \rho\sin\pars{\theta}\dot{\theta} & = &
-v\cos\pars{\theta} - v
\\[2mm]
\sin\pars{\theta}\dot{\rho} & + & \rho\cos\pars{\theta}\dot{\theta} & = &
-v\sin\pars{\theta}
\end{array}\right.
\\[5mm]
\imp\quad &
\left\lbrace\begin{array}{rcl}
\dot{\rho} & = &\bracks{-v\cos\pars{\theta} - v}\rho\cos\pars{\theta} -
\bracks{-v\sin\pars{\theta}}\bracks{-\rho\sin\pars{\theta}} =
\bracks{-v - v\cos\pars{\theta}}\rho
\\[2mm]
\dot{\theta} & = &\cos\pars{\theta}\bracks{-v\sin\pars{\theta}} -
\sin\pars{\theta}\bracks{-v\cos\pars{\theta} - v} = v\sin\pars{\theta}
\end{array}\right.
\end{align}

\begin{align}
{1 \over \rho}\,\totald{\rho}{\theta} & =
-\,{1 + \cos\pars{\theta} \over \sin\pars{\theta}}
\\[5mm] 
\ln\pars{\rho_{\infty} \over k} & =
\left.-2\ln\pars{\sin\pars{\theta \over 2}}\,\right\vert_{\pi/2}^{\pi} =
\ln\pars{\half}\quad\imp\quad
\color{#f00}{\rho_{\infty}} = \color{#f00}{k \over 2}
\end{align}
A: Assuming both ships travel at the same speed.
Ship $A$ is at $(t,0)$
Ship $B$ is at $(x(t),y(t))$
$d = \sqrt {(t-x)^2 + y^2}$
the velocity of $B = (x', y') =(\frac {(t-x)}d, -\frac yd)$
$u = t-x$
$d' = $$\frac {uu' + yy'}{d} \\ \frac {u - ux' + yy'}d\\ \frac uD - \frac {u^2 + y^2}{d^2} \\x' - 1$
$d'-x' = -1\\
d - x = -t + C\\
d(0) = k, x(0) = 0, C=k\\
d +t - x = k\\
d + u = k$
As $t\to\infty, u\to d$  That is $y$ is very small, and ship B is tracking directly behind ship A.
$2d  =  k\\
d= \frac 12 k$
