# Row, Row, Row your boat

A boat can travel at a speed of $3\text{ ms}^{-1}$ on still water. A boatman wants to cross a river whilst covering the shortest possible distance. In what direction should he row with respect to the bank if the speed of the water is

1. $2\text{ ms}^{-1}$
2. $4\text{ ms}^{-1}$

Assume that the speed of the water is the same everywhere.

Problem Source : 200 Puzzling Physics Problem

We know that the direction for the shortest possible distance must be perpendicular to the river bank. So for the first sub-problem, the horizontal partial speed must be the same as the speed of the water.

\begin{aligned} 3\cos \theta &= 2 \\ \theta &= 48^{\circ}11' \end{aligned}

For the second sub-problem, no matter in which direction the boatman rows, he cannot reach the river bank in a perpendicular direction. This can be seen that there is no such $\theta$ fulfill $3\cos \theta = 4$.

Therefore, we solve it in a slightly different way. First, we let the river be still. Relatively, our horizontal partial speed will be $4 - 3 \cos \theta$ and vertical partial speed $3\sin \theta$.

Let the direction of the boat be $\alpha$, so that

$\displaystyle \tan \alpha = \frac{3\sin \theta}{4 - \cos \theta}$

Note that the direction of $\alpha$ is the angle we move when we row at angle $\theta$. Our goal is to find the value of $\theta$ such that $\alpha$ is maximized.

\begin{aligned} \frac{\mathrm d}{\mathrm d \theta} \left( \frac{3\sin \theta}{4 - \cos \theta} \right) &= 0 \\ \frac{3\cos \theta(4 - 3 \cos \theta)-3\sin \theta(3\sin \theta)}{(4 - \cos \theta)^2} &= 0 \\ 12\cos \theta - 9 &= 0 \\ \cos \theta &= \frac{3}{4} \\ \theta &= 41^{\circ}25' \end{aligned}