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I'm having a hard time solving this problem:

Let there be a town $A$ in a shore of a river. Let $x=0$ be the shore. Let $(0,0)$ be the location of the town. Let $B$ be another town, in the oppossite shore, $x=b$, and let the town be in $(b,0)$. Suppose a person from town $A$ goes in a boat with velocity $v$ to $B$, always poiting at $B$ (see the picture), and let the river flow in the positive direction of $y$ with velocity $u$. Find the curve that gives the person's trayectory over time.

In an arbitrary point of the curve there will always be a system of three vectors, $v$, $u$ and $w=v+u$ in the following manner, where $w$, the tangent vector, is $u+v$. Their modulus is constant, thus since $v+u$ varies, the variables here are the angles and the modulus. Remember that $v$ always points at $B$. IN the image, the lengths of $v$ and $v_1$ and of $u$ and $u_1$ should be the same (since they're the same vector).

enter image description here

You can see that $v$ is "pushing" to get to $B$ but the vector $u$ will always modify $v$'s direction (and thus the man's), making the tangent actually be $w = u+v$. The dotted parallel lines are a reference to the tangent angle which is that between the dotted line and $w$. My approach is:

$$ \tan \theta = \frac{dy}{dx}$$

and the modulus of $v+u$ will satisfy, being a function of time.

$$|| v+u||(t) = \frac{ds}{dt}$$

I just need then to find a way of relating the modulus to the angle to find the solution.

Another thing that comes to mind is thinking as the solution in a parametric way, thus first finding $$\frac{dy}{dt} = f(t) $$ and $$\frac{dx}{dt} = g(t) $$ then taking the quotient and integrating.

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Just to clarify, does $v$ have constant magnitude in the problem statement, or does $u + v$ have constant magnitude? If $u + v$ has constant magnitude, then I would expect both the magnitude and direction of $v$ to vary with the position of the boat, because $u$ is a constant vector. –  Geoff Oxberry Feb 8 '12 at 5:01
    
$v$ and $u$ are constant. The magnitude of $v+u$ which I understand is the instat velocity of the boat varies depending on how we combine the vectors. –  Pedro Tamaroff Feb 8 '12 at 5:24
    
an answer that stays in the scope of the th[e]ory of (...) persecution problems... Never heard about such a theory. Sounds cool, though. –  Did Feb 13 '12 at 10:02
    
No, thanks... My point is to recommend some more restraint in the use of the word theory (which in this case is, I think, inappropriate). –  Did Feb 13 '12 at 12:06
    
@DidierPiau Well, in ODE courses persecution problems is a common topic to see when studying isogonal families of curves. I don't really appreaciate the ironical approach, you can be straightforward and I won't mind. –  Pedro Tamaroff Feb 13 '12 at 12:10

2 Answers 2

up vote 4 down vote accepted
+50

I assume that $|u|$ and $|v|$ are constant. Since the boat is always pointing towards $(b.0)$, we know that $$ \frac{v_y}{v_x}=\frac{-y}{b-x}\tag{1} $$ Since $|v|^2=v_x^2+v_y^2$, we can square $(1)$ and add $1$, or square the reciprocal of $(1)$ and add $1$ to get $$ \begin{array}{} v_x=|v|\frac{b-x}{\sqrt{(b-x)^2+y^2}}&\text{and}&v_y=|v|\frac{-y}{\sqrt{(b-x)^2+y^2}} \end{array}\tag{2} $$ Since the river flow is in the positive $y$ direction, $u_y=|u|$ and $u_x=0$. The slope of the trajectory is the ratio of the total $y$-velocity to the total $x$-velocity: $$ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &=\frac{v_y+u_y}{v_x+u_x}\\ &=\frac{v_y+|u|}{v_x}\\ &=\frac{|u|\sqrt{(b-x)^2+y^2}-|v|y}{|v|(b-x)}\tag{3} \end{align} $$ which becomes $$ |v|((b-x)\;\mathrm{d}y-y\;\mathrm{d}(b-x))=-|u|\sqrt{(b-x)^2+y^2}\;\mathrm{d}(b-x)\tag{4} $$ Dividing $(4)$ by $(b-x)^2$ and rearranging yields $$ 0=|v|\;\mathrm{d}\frac{y}{b-x}+|u|\sqrt{1+\left(\frac{y}{b-x}\right)^2}\;\mathrm{d}\log\left(b-x\right)\tag{5} $$ Dividing $(5)$ by $\sqrt{1+\left(\frac{y}{b-x}\right)^2}$ yields $$ |v|\;\mathrm{d}\operatorname{arcsinh}\left(\frac{y}{b-x}\right)+|u|\;\mathrm{d}\log(b-x)=0\tag{6} $$ which means $$ |v|\operatorname{arcsinh}\left(\frac{y}{b-x}\right)+|u|\log(b-x)=C|v|\tag{7} $$ which leads to $$ \begin{align} y &=(b-x)\sinh\left(C-\frac{|u|}{|v|}\log(b-x)\right)\\ &=(b-x)\sinh\left(\frac{|u|}{|v|}\log\left(\frac{b}{b-x}\right)\right)\\ &=\frac{b-x}{2}\left[\left(\frac{b}{b-x}\right)^{|u|/|v|}-\left(\frac{b-x}{b}\right)^{|u|/|v|}\right]\tag{8} \end{align} $$ Note that $$ \lim_{x\to b^-}y=\left\{\begin{array}{}0&\text{if }|u|<|v|\\b/2&\text{if }|u|=|v|\\\infty&\text{if }|u|>|v|\end{array}\right.\tag{9} $$


Path of Crossing:

All paths below, except $\dfrac{|u|}{|v|}=1$, end directly across the river from the starting point, but most are very steep at the end. For $\dfrac{|u|}{|v|}=1$, the boat never reaches the other side, but it limits to the point $\frac12$ the river width downstream.

Path of Crossing


Time to Cross:

As the relative speed of the stream gets closer to the speed of the boat, the time to cross the stream increases. Combining $(2)$ and $(8)$ yields $$ \begin{align} T &=\int_0^b\frac{\mathrm{d}x}{v_x}\\ &=\frac{1}{|v|}\int_0^b\frac{\sqrt{(b-x)^2+y^2}}{b-x}\;\mathrm{d}x\\ &=\frac{1}{|v|}\int_0^b\sqrt{1+\left(\frac{y}{b-x}\right)^2}\;\mathrm{d}x\\ &=\frac{1}{|v|}\int_0^b\sqrt{1+\frac14\left[\left(\frac{b}{b-x}\right)^{|u|/|v|}-\left(\frac{b-x}{b}\right)^{|u|/|v|}\right]^2}\;\mathrm{d}x\\ &=\frac{1}{2|v|}\int_0^b\left[\left(\frac{b}{b-x}\right)^{|u|/|v|}+\left(\frac{b-x}{b}\right)^{|u|/|v|}\right]\;\mathrm{d}x\\ &=\frac{1}{2|v|}\left[\frac{b}{1-|u|/|v|}+\frac{b}{1+|u|/|v|}\right]\\ &=\frac{b}{|v|}\frac{1}{1-(|u|/|v|)^2}\tag{10} \end{align} $$ Thus, with no cross-current ($|u|=0$), $T=\dfrac{b}{|v|}$, and as $|u|\to|v|$, $T\to\infty$; both as expected.

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everything seems ok except you're missing a factor of $(b-x)$ in your final answer. i.e, the solution should be $$\frac{{b - x}}{2}\left[ {{{\left( {\frac{b}{{b - x}}} \right)}^c} - {{\left( {\frac{{b - x}}{b}} \right)}^c}} \right]$$ where $c$ is the quotient. Could you expand a little on (2) and (3)? –  Pedro Tamaroff Feb 13 '12 at 15:54
    
A got it all except $$\frac{{dy}}{{dx}} = \frac{{{v_y} + \left| u \right|}}{{{v_x}}}$$ –  Pedro Tamaroff Feb 13 '12 at 16:18
    
are you around? –  Pedro Tamaroff Feb 13 '12 at 16:28
    
@Peter: I had already caught the $(b-x)$, but I missed the $\frac12$. Thanks for catching that. I will add a bit in the answer about $(2)$ and $(3)$. –  robjohn Feb 13 '12 at 16:33
    
I really appreciate your work. Thanks! –  Pedro Tamaroff Feb 13 '12 at 16:35

Edit

This gives a solution to what I understood was the problem before the OP edited it.


We have a river of width $w>0$. City $A$ is in the shore $y=0$ at the point $(0,0)$. City $B$ is in the opposite shore, which I assume to be $y=w$, at point $(b,w)$. A person rows from $A$ to $B$, pointing always in the direction of $B$ and with speed $v$ and has to fight against a current going from $b$ to $A$ of speed $u$. Let $(x(t),y(t))$ be the position of the boat at time $t$. The velocity that the rower gives to the boat at this time is in the direction of $(x,y)$ to $B$, that is, it is proportional to $(b-x,w-y)$. Since its modulus is equal to $v$, it must be $$ \frac{v}{r}\,(b-x,w-y),\quad r=\sqrt{(b-x)^2+(w-y)^2}. $$ Let $V=(V_x,V_y)$ be the real velocity of the boat. Then $$ V_x=\frac{v}{r}\,(b-x)-u,\quad V_y=\frac{v}{r}\,(w-y). $$ This leads to the following system of differential equations: $$\begin{align*} x'&=\frac{v}{r}\,(b-x)-u,\\ y'&=\frac{v}{r}\,(w-y), \end{align*}\tag1$$ together with the initial conditions $x(0)=y(0)=0$. The system is equivalent to the equation $$ \frac{dy}{dx}=\frac{w-y}{b-x-\dfrac{u}{v}\,\sqrt{(b-x)^2+(w-y)^2}},\quad y(0)=0.\tag2 $$ The change $z=w-y$, $t=b-x$ transforms it into an homogeneous equation, which can be solved explicitly. Observe that $$ v\,\frac{b}{\sqrt{b^2+w^2}}<u\implies \frac{dx}{dt}(t=0)<0, $$ and the boat will start going away from $B$. If $v$ is not too small, after a while the boat will turn and go towards $B$. In this case, system (1) and equation (2) are not equivalent.

This is how the trajectories of the boat look like

In the problem such as you stated it, the current seems to go perpendicular from one shore to the other. If this is what you really want, all you have to do is take the term $-u$ from the first equation in (1) to the second.

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you are placing $B$ in $(0,b)$ instead of $(b,0)$ –  Pedro Tamaroff Feb 13 '12 at 15:57
    
@Peter But if $B$ is in $(b,0)$ and $A$ is in $(0,0)$ and the shore is $y=0$, they are on the same shore. –  Julián Aguirre Feb 13 '12 at 16:03
    
The shore is the $y$ axis, not $y=0$. I'll correct it. –  Pedro Tamaroff Feb 13 '12 at 16:04

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