The motion of the boat is described by the vector addition of velocities: the velocity of the boat relative to the ground ($\vec{v}_{b/g}$) is the sum of the river's velocity ($\vec{u}$) and the boat's velocity relative to the water ($\vec{v}_{b/w}$). In the chosen coordinate system (y-axis North, x-axis East), the velocity components are:

$\vec{v}_{b/g} = (v_x, v_y) = (u - v\sin\theta, v\cos\theta)$.

The time to cross the river of width $D$ is $t = \frac{D}{v_y} = \frac{D}{v\cos\theta}$. The downstream drift distance is $x = v_x t = (u - v\sin\theta) \frac{D}{v\cos\theta} = D\left(\frac{u}{v\cos\theta} - \tan\theta\right)$.

[Q1] To find the angle $\theta$ that minimizes the downstream drift $x$, we analyze the drift equation. Let $k = x/D$ and $s = \sin\theta$. The equation can be written as $k = \frac{u/v - s}{\sqrt{1-s^2}}$. Squaring and rearranging gives a quadratic equation for $s$:

$$(1+k^2)s^2 - 2\left(\frac{u}{v}\right)s + \left(\left(\frac{u}{v}\right)^2 - k^2\right) = 0$$

For this equation to have a real solution for $s$, the discriminant must be non-negative ($\Delta \ge 0$), which simplifies to:

$$k^2 \ge \left(\frac{u}{v}\right)^2 - 1$$

This inequality determines the minimum possible drift.

Case 1: $v > u$ (boat faster than current) Since $u/v < 1$, $(u/v)^2 - 1 < 0$. The minimum non-negative value for the drift squared, $k^2$, is $k^2 = 0$. This corresponds to a minimum drift of $x_{min} = 0$. The angle $\theta$ for this case is found by setting $k=0$ in the quadratic equation, which gives $s = \sin\theta = u/v$. Thus, $\theta = \arcsin(u/v)$.

Case 2: $v < u$ (boat slower than current) Since $u/v > 1$, $(u/v)^2 - 1 > 0$. The drift $x$ (and $k$) is always positive. The minimum value for $k^2$ is $k_{min}^2 = (u/v)^2 - 1$. This gives a minimum drift of $x_{min} = D \sqrt{(u/v)^2 - 1} = \frac{D\sqrt{u^2-v^2}}{v}$. The optimal angle occurs when the discriminant is zero, giving a unique solution for $s = \sin\theta = \frac{u/v}{1+k_{min}^2} = v/u$. Thus, $\theta = \arcsin(v/u)$.

[Q2] The optimal angle $\theta_{opt}$ for minimum drift depends on $v$ as follows: If $0 \le v \le u$: $\theta_{opt} = \arcsin(v/u)$. If $v > u$: $\theta_{opt} = \arcsin(u/v)$.

Analyzing the behavior:

• As $v \to 0$, $\theta_{opt} \to \arcsin(0) = 0$.
• As $v$ increases to $u$, $\theta_{opt}$ increases to $\arcsin(1) = \pi/2$.
• As $v$ increases from $u$ to $\infty$, $\theta_{opt}$ decreases from $\pi/2$ back to $\arcsin(0) = 0$. So, the angle starts at 0, increases to a maximum of $90^\circ$ (straight upstream) when $v=u$, and then decreases back to 0 for very high speeds.

[Q3] The boat's path makes an angle $\alpha$ with the downstream bank, so the direction of its ground velocity $\vec{v}_{b/g}$ is fixed. The ratio of velocity components is constant:

$$\frac{v_y}{v_x} = \frac{v\cos\theta}{u-v\sin\theta} = \tan\alpha$$

Solving for the boat speed $v$:

$$v\cos\theta = (u-v\sin\theta)\tan\alpha \implies v(\cos\theta + \sin\theta\tan\alpha) = u\tan\alpha$$

Using trigonometric identities, this simplifies to $v\cos(\theta-\alpha) = u\sin\alpha$, so:

$$v = \frac{u\sin\alpha}{\cos(\theta-\alpha)}$$

To minimize $v$, the denominator must be maximized. The maximum value of $\cos(\theta-\alpha)$ is 1. This gives the minimum speed:

$$v_{min} = u\sin\alpha$$

[Q4] For $v < u$, the boat always drifts downstream ($x>0$). The locus of reachable points $(x, D)$ is determined by the range of $x(\theta)$ for $\theta \in (-\pi/2, \pi/2)$. From [Q1], the minimum drift is $x_{min} = \frac{D\sqrt{u^2-v^2}}{v}$. As the steering angle $\theta$ approaches $\pm\pi/2$, the crossing time $t = D/(v\cos\theta) \to \infty$. The x-component of velocity $v_x = u-v\sin\theta$ remains positive (since $v $$y=D, \quad x \ge \frac{D\sqrt{u^2-v^2}}{v}$$

[Q5] The time to cross the river is $t(\theta) = \frac{D}{v\cos\theta}$. To minimize this time, the denominator $v\cos\theta$ must be maximized. Since $v$ is constant, we must maximize $\cos\theta$. The maximum value is $\cos\theta = 1$, which occurs at $\theta=0$. This corresponds to pointing the boat straight across the river (due North). The minimum crossing time is:

$$t_{min\_time} = \frac{D}{v}$$

This result is independent of the condition $v < u$.