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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?

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  • $\begingroup$ Are the ships always traveling in straight lines? If so then the $x$-coords are always $0$. $\endgroup$ – tilper Aug 10 '16 at 21:56
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    $\begingroup$ do you mean for the starting point of ship B to be at (k,0)? And, can we assume that the speeds of the boats equal? If they are not then the distance will either be 0, or infinite. $\endgroup$ – Doug M Aug 10 '16 at 21:59
  • $\begingroup$ Ship A always travels in a straight line towards $(0,\infty)$. Ship B always travels towards ship A, wherever ship A is located. $\endgroup$ – Zzcnick Aug 10 '16 at 21:59
  • $\begingroup$ We might as well let the speed be $1$. Ship $A$ is then at $(t,0)$ at time $t$. Ship $B$ is at $(x(t),y(t))$ and we are given $$\frac {dx}{dt}=\frac {t-x}{\sqrt{(t-x)^2+y^2}}\\\frac {dy}{dt}=\frac {-y}{\sqrt{(t-x)^2+y^2}}\\x(0)=0,y(0)=k$$ and we are looking for $$\lim_{t \to \infty} (t-x)$$ $\endgroup$ – Ross Millikan Aug 10 '16 at 22:00
  • $\begingroup$ @Rob Arthan, Doug M: Ah, apologies! Either ship B must be at $(k,0)$ or ship A traveling towards $(\infty,0)$. The question has been edited such that A travels towards $(\infty,0)$. $\endgroup$ – Zzcnick Aug 10 '16 at 22:12
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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.

$2d = k\\ d= \frac 12 k$

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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