# 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?

• Are the ships always traveling in straight lines? If so then the $x$-coords are always $0$. – tilper Aug 10 '16 at 21:56
• 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. – Doug M Aug 10 '16 at 21:59
• Ship A always travels in a straight line towards $(0,\infty)$. Ship B always travels towards ship A, wherever ship A is located. – Zzcnick Aug 10 '16 at 21:59
• 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 $$\lim_{t \to \infty} (t-x)$$ – Ross Millikan Aug 10 '16 at 22:00
• @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 $(\infty,0)$. – Zzcnick Aug 10 '16 at 22:12

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 $B = (x', y') =(\frac {(t-x)}d, -\frac yd)$

$u = t-x$

\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}