Vector physics boat problem

Question

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?

Answer 1

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

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 $$ T = \int_0^{15}\frac{\mathrm{dy}}{\dot y} = \frac{1}{10}\int_0^{15} \left(\frac{15}{15-y}\right)^{3/5}+ \left(\frac{15-y}{15}\right)^{3/5} \mathrm{d}y = \frac{75}{16} $$

Answer 2

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.