Baywatch

There are two different strategies to minimize the total running time $T$ for the lifeguard:

• Minimize the overall path length
• Minimize line segment inside the water

While the straight path in (C) optimizes the overall path length, the route in (E) has the shortest swim distance. The route in (D) is a compromise between the opposing strategies and represents the true optimum.

To be more precise we consider a path between the points $Q = (0,0)$ (lifeguard) and $P = (X,Y)$ (drowning man) with an intermediate point $R = (x,y)$ at the store. The distance of both line segments on the beach and in the water are $s_1 = \sqrt{x^2 + y^2} \qquad \text{and} \qquad s_2 = \sqrt{(X-x)^2 + (Y-y)^2}$ The total time $T$ , that the lifeguard needs to reach the man, reads $T = t_1 + t_2 = \frac{s_1}{v_1} + \frac{s_2}{v_2} = \frac{\sqrt{x^2 + y^2}}{v_1} + \frac{\sqrt{(X-x)^2 + (Y-y)^2}}{v_1}$ with the velocities $v_1$ and $v_2$ for running and swimming (with $v_1 = 2 v_2$ ). The only free parameter is the coordinate $x$ , where the lifeguard reaches the shore. To minimize the recue time $T$ , we zero the derivative $\frac{dT}{dx} = \frac{x}{v_1 \sqrt{x^2 + y^2}} - \frac{X - x}{v_1 \sqrt{(X-x)^2 + (Y-y)^2}} \stackrel{!}{=} 0 \qquad \text{(*)}$ Rewriting $\sin \alpha = \frac{x}{v_1 \sqrt{x^2 + y^2}}, \quad \sin \beta = \frac{X - y}{\sqrt{(X-x)^2 + (Y-y)^2}}$ with the angle $\alpha$ and $\beta$ of incidence and refraction, the equation (*) reads $\frac{\sin \alpha}{v_1} = \frac{\sin \beta}{v_2} \quad \text{or} \quad \frac{\sin \alpha}{\sin \beta} = \frac{v_1}{v_2}$ This is identical to Snell's law, which decribes the light refraction at boundary between two different isotropic media. In our case, the optimal path is gives by the condition $\sin \alpha = 2 \sin \beta$ , so that the relation $\alpha > \beta > 0$ must be fulfilled. This geometric constraint is only true for the path (D).