Solution to Question 10
1. Problem Formulation
We need to minimize the time taken for the angler to travel from his house $H$ to an unspecified spot on the bank $OB$. The path involves walking on ground (speed $v$) to a point $P$ on bank $OA$, and then swimming in the bay (speed $u$) to a point $Q$ on bank $OB$.
According to Fermat’s Principle (or the principle of least time), the optimal path must satisfy two conditions:
- Refraction at $OA$: The path must obey Snell’s Law at the interface between the ground and the water.
- Arrival at $OB$: Since the destination can be anywhere on the line $OB$, the time is minimized when the path in the water is perpendicular to the boundary $OB$.
Figure 2: Optimal path geometry. The angler walks from H to P, then swims from P to Q. To minimize time in the water reaching the line OB, the path PQ must be perpendicular to OB.
2. Mathematical Derivation
Step A: Geometry of Refraction
Let the normal to the bank $OA$ be vertical. The path $PQ$ in the water is perpendicular to bank $OB$.
From geometry, if the angle between banks $OA$ and $OB$ is $\theta$, and $PQ \perp OB$, then the angle the path $PQ$ makes with the normal to $OA$ is exactly $\theta$.
Therefore, the angle of refraction is $r = \theta$.
Using Snell’s Law at the interface (ground/water): $$ \frac{\sin i}{v} = \frac{\sin r}{u} \implies \frac{\sin i}{v} = \frac{\sin \theta}{u} $$ $$ \sin i = \frac{v}{u} \sin \theta $$ Consequently, we can express $\cos i$: $$ \cos i = \sqrt{1 – \sin^2 i} = \sqrt{1 – \frac{v^2}{u^2}\sin^2 \theta} = \frac{1}{u}\sqrt{u^2 – v^2 \sin^2 \theta} $$
Step B: Calculating Total Time
Let $H$ be at coordinates $(x_H, d)$ relative to $O(0,0)$, where $x_H = \sqrt{l^2 – d^2}$.
Let $P$ be the point where the angler enters the water. The horizontal distance from $H$ to $P$ is determined by the angle of incidence $i$:
$$ x_{HP} = d \tan i $$
The walking distance $HP$:
$$ d_{walk} = \frac{d}{\cos i} $$
The remaining horizontal distance from $P$ to $O$:
$$ x_P = x_H – d \tan i = \sqrt{l^2 – d^2} – d \tan i $$
The swimming distance $PQ$ is the perpendicular distance from $P$ to line $OB$:
$$ d_{swim} = x_P \sin \theta = (\sqrt{l^2 – d^2} – d \tan i) \sin \theta $$
Step C: Assembling the Time Equation
Total time $T = T_{walk} + T_{swim}$:
$$ T = \frac{d}{v \cos i} + \frac{(\sqrt{l^2 – d^2} – d \tan i) \sin \theta}{u} $$
Rearranging terms:
$$ T = \frac{d}{v \cos i} – \frac{d \sin i}{u \cos i} \sin \theta + \frac{\sqrt{l^2 – d^2} \sin \theta}{u} $$
Factor out $\frac{d}{\cos i}$:
$$ T = \frac{d}{\cos i} \left( \frac{1}{v} – \frac{\sin i \sin \theta}{u} \right) + \frac{\sqrt{l^2 – d^2} \sin \theta}{u} $$
Substitute $\sin i = \frac{v}{u} \sin \theta$ into the parenthesis:
$$ \left( \frac{1}{v} – \frac{\frac{v}{u}\sin\theta \cdot \sin\theta}{u} \right) = \frac{1}{v} – \frac{v \sin^2 \theta}{u^2} = \frac{u^2 – v^2 \sin^2 \theta}{v u^2} $$
Now substitute this back into the expression for $T$, along with the expression for $\cos i$:
$$ T = \frac{d}{\frac{1}{u}\sqrt{u^2 – v^2 \sin^2 \theta}} \cdot \frac{u^2 – v^2 \sin^2 \theta}{v u^2} + \frac{\sqrt{l^2 – d^2} \sin \theta}{u} $$
Simplify the first term:
$$ T = \frac{d u}{\sqrt{u^2 – v^2 \sin^2 \theta}} \cdot \frac{u^2 – v^2 \sin^2 \theta}{v u^2} $$
$$ T = \frac{d (u^2 – v^2 \sin^2 \theta)}{u v \sqrt{u^2 – v^2 \sin^2 \theta}} $$