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}

Row, Row, Row your boat

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