1. Problem Geometry and Conditions

The boat needs to travel from A to C. Point B is directly opposite A. The path AC makes an angle $\alpha$ with the river bank (horizontal).

From the diagram, we have: $\tan \alpha = \frac{b}{a}$ and $\sin \alpha = \frac{b}{\sqrt{a^2 + b^2}}$.

To minimize the boat’s speed $v_b$, its velocity vector $\vec{v}_b$ must be perpendicular to the resultant velocity $\vec{v}_{net}$, which lies along the path AC.

2. Minimum Speed and Steering Direction

We construct a velocity triangle with $\vec{v}_w$ (river velocity), $\vec{v}_b$ (boat velocity), and $\vec{v}_{net}$ (resultant velocity). For minimum $v_b$, the triangle is right-angled, with $\vec{v}_w$ as the hypotenuse and $\vec{v}_b$ perpendicular to $\vec{v}_{net}$.

Minimum Speed ($v_b$):

From the right triangle, $v_b = v_w \sin \alpha$. Substituting $\sin \alpha$:

$$ v_b = \frac{v_w b}{\sqrt{a^2 + b^2}} $$

Steering Direction ($\phi$):

The angle $\phi$ is measured upstream from the line AB (the vertical). From the geometry of the velocity triangle, this angle is equal to $\alpha$.

$$ \phi = \alpha = \tan^{-1}\left(\frac{b}{a}\right) $$