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A friend of mine posed this question to me:

A boat with a maximum speed of $5$ meters per second crosses a river that is flowing at $3$ meters per second, and is fifteen meters wide. The boat always faces its target $(0,15)$. Does the boat reach its target? If so, how long does it take? What is the path of the boat?

If the boat is at a point $(x,y)$, then the angle $\theta$ to the target is $\theta(x,y) = \arctan\left(\frac{15 - y}{x}\right)$ and the velocity is $v(\theta) = (-3 + 5 \cos \theta) \hat{\imath} + (5 \sin \theta) \hat{\jmath}$.

I couldn't find parametric equations, so I wrote a small program to solve it for me. The program steps through small intervals of time, moves the boat by $v\,\mathrm dt$, and recalculates the velocity. The program halts when the boat has arrived at the target.

The graph of the path of the boat generated by this program should approximate the actual graph. The program found that the time taken is about $4.7$ seconds.

How can I find parametric equations for the path of the boat?

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On this site, you enclose $\LaTeX$ in dollar signs. – Ross Millikan Oct 22 '11 at 19:32
If the river is flowing to the left, as indicated by the given $v$, then $x$ will be negative through the whole process. Therefore it seems that your formula for $\theta$ has the wrong sign. – Christian Blatter Oct 22 '11 at 20:35
I believe this originally appeared in a set of problems given by Feynman - I'm almost certain. – mixedmath Oct 23 '11 at 6:40
Feynman it is - see – Gerry Myerson Oct 23 '11 at 11:22
The solutions given there are worth a look. The one by Diego Scarabelli is particularly nice, as it gives a simple and elegant calculation for the duration without first determining the trajectory. The one by Riccardo Borghi is how I approached this. By the way, their result coincides with Heike's. Note that this method of crossing the river is slower than the optimal method by the same factor as the optimal method is slower than crossing without a current. – joriki Oct 23 '11 at 16:21
up vote 11 down vote accepted

From $v(\theta)$ it follows that $$ \frac{\mathrm{d}x}{\mathrm{d}y}=-\frac{3}{5}\operatorname{cosec}\theta+\cot\theta = -\frac{3}{5}\sqrt{1+\cot^2\theta}+\cot\theta $$ As Christian Blatter remarked correctly, $\theta$ should be equal to $\theta=-\arctan((15-y)/x)$ since $x\leq 0$ all along the path. Then $\cot\theta=-x/(15-y)$ and $$ \frac{\mathrm{d}\cot\theta}{\mathrm{d}y} = -\frac{1}{15-y}\frac{\mathrm{d}x}{\mathrm{d}y} -\frac{x}{(15-y)^2}=\frac{3}{5(15-y)}\sqrt{1+\cot^2\theta} $$ We can integrate this equation by using separation of variables. After rearranging the result we find $$ \cot\theta = \frac{1}{2}\left(\left(\frac{15}{15-y}\right)^{3/5}- \left(\frac{15-y}{15}\right)^{3/5}\right), $$ where we used the condition $\cot\theta=0$ at $y=0$. By plugging this expression back into the first equation and integrating with respect to $y$ we find with $x=0$ at $y=0$ $$ x(y)=-\frac{(15-y)^{2/5}}{2\cdot 15^{3/5}} \left(1-\left(\frac{15-y}{15}\right)^{6/5}\right) $$ The graph of this function looks like

path of the boat

Finally with $$\dot y=5\sin\theta = \frac{5}{\sqrt{1+\cot^2\theta}} = \frac{10}{\left(\frac{15}{15-y}\right)^{3/5}+ \left(\frac{15-y}{15}\right)^{3/5}} $$ we find that the time it takes for the boat to cross the stream is equal to $$ T = \int_0^{15}\frac{\mathrm{dy}}{\dot y} = \frac{1}{10}\int_0^{15} \left(\frac{15}{15-y}\right)^{3/5}+ \left(\frac{15-y}{15}\right)^{3/5} \mathrm{d}y = \frac{75}{16} $$

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EDIT: If Heike's solution is correct (and it looks good to me) then what I've done below is based on a misunderstanding of the problem and is best ignored.

If there were no current, the boat could get to $(3,4)$ in a second, since that point is 5 meters away from the origin (which, I assume, is where the boat starts). With the current, which pushes the boat 3 units to the left during that second, it winds up at $(0,4)$ after one second (by attempting to reach $(3,4)$). Now multiply everything by 15/4. The boat still tries to go in the direction $(3,4)$, and reaches its goal, $(0,15)$, after $15/4=3.75$ seconds.

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Why is the boat heading towards (3,4)? The question says the boat always faces its target (0,15)... – user7530 Oct 23 '11 at 6:26
Yes, that bothered me, too. If the boat heads toward its target it will always be pushed downstream and never reach its target, so I decided that the boat can face toward $(0,15)$ as long as it heads toward $(3,4)$. Maybe real boats don't work that way, but that's the only way I can make sense of the problem. You got a better idea? – Gerry Myerson Oct 23 '11 at 6:42
Are you sure it would never reach? If it starts directly horizontal of the target it will make it (since it is faster than the current) so I would expect it to be able to "spiral in" if it starts elsewhere on the river. You can set up the obvious system of nonlinear ODEs but I can't see how to solve them... – user7530 Oct 23 '11 at 7:37
@user7530, I believe you are right, and I have warned readers off my answer. – Gerry Myerson Oct 23 '11 at 11:19
It should not be ignored: Even if you didn't answer the original question your answer describes the optimal solution to the practical problem of crossing this river: The solution is optimal when the boat traverses a straight segment with respect to the water masses, and this is the case in your setup. – Christian Blatter Oct 23 '11 at 12:38

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