The problem of finding the minimum meeting time is significantly simplified by analyzing the motion in the reference frame of the river's water. In this frame, the velocity of the river flow, $u$, has no effect on the relative motion of the two boats, as both are subject to the same drift.

Let's set up a coordinate system with the origin at the initial position of the first boat (Boat 1), the y-axis pointing directly across the river, and the x-axis along the river bank in the direction of the flow. The initial position of Boat 1 is $\vec{r}_{1,0} = (0, 0)$. The initial position of Boat 2 on the opposite bank is $\vec{r}_{2,0} = (x_0, d)$. The line connecting them makes an angle $\alpha$ with the bank, so $\tan \alpha = d/x_0$, which gives $x_0 = d \cot \alpha$. Thus, $\vec{r}_{2,0} = (d \cot \alpha, d)$.

The initial separation vector is $\vec{L} = \vec{r}_{2,0} - \vec{r}_{1,0} = (d \cot \alpha, d)$. The magnitude of the initial separation (the distance between the boats) is:

$$L = |\vec{L}| = \sqrt{(d \cot \alpha)^2 + d^2} = d\sqrt{\cot^2\alpha + 1} = d\sqrt{\csc^2\alpha} = \frac{d}{\sin\alpha}$$

In the water's reference frame, the boats' velocities are their velocities relative to the water, $\vec{v}_{1,w}$ and $\vec{v}_{2,w}$, with fixed magnitudes $|\vec{v}_{1,w}| = v_1$ and $|\vec{v}_{2,w}| = v_2$. For the boats to meet at time $t$, their positions in this frame must be equal:

$$\vec{r}_{1,0} + \vec{v}_{1,w} t = \vec{r}_{2,0} + \vec{v}_{2,w} t$$

Rearranging the terms, we get:

$$\vec{r}_{2,0} - \vec{r}_{1,0} = (\vec{v}_{1,w} - \vec{v}_{2,w}) t$$ $$\vec{L} = \vec{v}_{rel} t$$

where $\vec{v}_{rel} = \vec{v}_{1,w} - \vec{v}_{2,w}$ is the relative velocity of Boat 1 with respect to Boat 2.

The time to meet is found by taking the magnitude:

$$t = \frac{L}{|\vec{v}_{rel}|} = \frac{L}{|\vec{v}_{1,w} - \vec{v}_{2,w}|}$$

To minimize the time $t$, the magnitude of the relative velocity, $|\vec{v}_{1,w} - \vec{v}_{2,w}|$, must be maximized. The magnitudes $v_1$ and $v_2$ are constant, but the boats can choose their directions. The magnitude of the difference of two vectors is maximized when the vectors are anti-parallel (point in opposite directions).

$$|\vec{v}_{1,w} - \vec{v}_{2,w}|_{max} = v_1 + v_2$$

This occurs when $\vec{v}_{1,w}$ is directed opposite to $\vec{v}_{2,w}$.

Furthermore, the equation $\vec{L} = \vec{v}_{rel} t$ requires the relative velocity vector $\vec{v}_{rel}$ to be in the same direction as the initial separation vector $\vec{L}$. To satisfy both conditions, Boat 1 must row directly towards the initial position of Boat 2, and Boat 2 must row directly towards the initial position of Boat 1. This sets $\vec{v}_{1,w}$ parallel to $\vec{L}$ and $\vec{v}_{2,w}$ anti-parallel to $\vec{L}$, fulfilling the anti-parallel condition for maximum relative speed.

The minimum time is therefore:

$$t_{min} = \frac{L}{|\vec{v}_{rel}|_{max}} = \frac{d/\sin\alpha}{v_1 + v_2}$$ $$t_{min} = \frac{d}{(v_1 + v_2)\sin\alpha}$$

The angles at which they should row are determined by the direction of the initial separation line. The vector $\vec{L}$ from Boat 1 to Boat 2 makes an angle $\alpha$ with the river bank (the x-axis).

• Boat 1 must row towards Boat 2, so its velocity vector relative to the water, $\vec{v}_{1,w}$, must make an angle $\theta_1 = \alpha$ with the river bank.
• Boat 2 must row towards Boat 1, so its velocity vector relative to the water, $\vec{v}_{2,w}$, must be in the opposite direction, making an angle $\theta_2 = \alpha + \pi$ with the river bank.