Fastest Path Across Media and Snell's Law

Core Definition

The Fastest Path Across Media problem optimizes the trajectory of an entity traveling between two points situated in differing physical media where propagation speeds contrast ( $v_1 \ne v_2$ ). Grounded in Fermat's Principle of Least Time, calculus demonstrates that the quickest path is not a single straight Euclidean line, but a piecewise-linear path refracted at the interface boundary satisfying Snell's Law:

\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}

Concept Explanation

1. Travel Time Formulation ( $T = d / v$ )

When traversing a homogeneous medium at a constant velocity $v$ , elapsed time $T$ is given by distance divided by rate:

T = \frac{d}{v}

When an object travels from point $A(0, y_1)$ in Medium 1 (speed $v_1$ ) to point $B(L, -y_2)$ in Medium 2 (speed $v_2$ ), crossing a straight boundary along the $x$ -axis at coordinate $(x, 0)$ :

           Medium 1 (Speed v1)
   A(0, y1)
      \
       \   d1 = sqrt(x^2 + y1^2)
        \
  -------*------------------------ (Interface Boundary, y = 0)
       (x, 0)
           \
            \   d2 = sqrt((L - x)^2 + y2^2)
             \
          B(L, -y2)
           Medium 2 (Speed v2)

The total elapsed travel time function is:

T(x) = T_1(x) + T_2(x) = \frac{\sqrt{x^2 + y_1^2}}{v_1} + \frac{\sqrt{(L - x)^2 + y_2^2}}{v_2}

The physical domain is restricted to the interval between the two horizontal projections: $x \in [0, L]$ .

2. Calculus Derivation of Snell's Law

To minimize transit time, differentiate $T(x)$ with respect to the crossing coordinate $x$ :

\frac{dT}{dx} = \frac{1}{v_1} \cdot \frac{2x}{2\sqrt{x^2 + y_1^2}} + \frac{1}{v_2} \cdot \frac{2(L - x)(-1)}{2\sqrt{(L - x)^2 + y_2^2}}

\frac{dT}{dx} = \frac{x}{v_1 \sqrt{x^2 + y_1^2}} - \frac{L - x}{v_2 \sqrt{(L - x)^2 + y_2^2}}

Setting $\frac{dT}{dx} = 0$ :

\frac{x}{v_1 \sqrt{x^2 + y_1^2}} = \frac{L - x}{v_2 \sqrt{(L - x)^2 + y_2^2}}

Now examine the trigonometry of the two travel legs:

Let $\theta_1$ be the angle between the travel path in Medium 1 and the interface normal (vertical axis). Then $\sin \theta_1 = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + y_1^2}}$ .
Let $\theta_2$ be the angle between the travel path in Medium 2 and the interface normal. Then $\sin \theta_2 = \frac{L - x}{\sqrt{(L - x)^2 + y_2^2}}$ .

Substituting these trigonometric ratios directly into the derivative equation yields:

\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}

This is Snell's Law of Refraction, proving that light travels along paths that minimize total transit time.

3. Coastal / Highway Boundary Variant

When an entity travels along a defined boundary (such as a runner along a beach road or a utility pipe along a riverbank) before cutting diagonally across an off-road terrain:

T(x) = \frac{x}{v_{\text{road}}} + \frac{\sqrt{d_{\text{off}}^2 + (L - x)^2}}{v_{\text{field}}}

Differentiating and setting to zero yields:

\frac{1}{v_{\text{road}}} - \frac{L - x}{v_{\text{field}}\sqrt{d_{\text{off}}^2 + (L - x)^2}} = 0 \implies \cos \alpha = \frac{L - x}{\sqrt{d_{\text{off}}^2 + (L - x)^2}} = \frac{v_{\text{field}}}{v_{\text{road}}}

Because $\cos \alpha \le 1$ , a distinct interior turn-off point exists whenever $v_{\text{field}} < v_{\text{road}}$ . If the off-road speed equals or exceeds the road speed, the optimal strategy degenerates to a direct Euclidean straight line ( $x = 0$ ).

Engineering Applications & Context

In UNLV engineering practice, the mathematical framework of the fastest path applies directly to infrastructure routing and resource transport:

Civil & Environmental Engineering: Sizing subsea and terrestrial wastewater pipelines, oil pipelines, and municipal water mains across wetlands, rivers, and dry easements where underwater trenching costs $\$ / \text{m}$ differ drastically from overland excavation costs.
Electrical & Computer Engineering: Deploying submarine high-voltage direct current (HVDC) power cables and transatlantic fiber-optic telecommunications lines.
Robotics & Autonomous Ground Vehicles: Motion planners on unstructured off-road terrains navigate variable surface traction (e.g., pavement vs. sand vs. mud) by balancing path length against allowable speed.
Acoustics & Geophysics: Seismic wave refraction modeling through stratified rock strata for subsurface geothermal and structural soil exploration.

Example Problems

Example Problem 1: Coastal Transit Optimization (The Lifeguard Problem)

A lifeguard at point $A$ on a straight sandy beach must reach an offshore swimmer at point $B$ . Point $B$ is located $30\text{ meters}$ offshore and $80\text{ meters}$ downshore from $A$ . The lifeguard can run along the shoreline at $5\text{ m/s}$ and swim through the water at $3\text{ m/s}$ . Find the distance $x$ along the shoreline the lifeguard should run before entering the water to minimize total rescue time.

Solution

Step 1: State governing formulas

Transit Time Function: $T(x) = \frac{d_{\text{run}}}{v_{\text{run}}} + \frac{d_{\text{swim}}}{v_{\text{swim}}} = \frac{x}{v_1} + \frac{\sqrt{h^2 + (L - x)^2}}{v_2}$ (Governing formulas defined in Fastest Path Across Media and Snell's Law.)

Step 2: Assign parameters and formulate objective
Given $v_1 = 5\text{ m/s}$ , $v_2 = 3\text{ m/s}$ , offshore distance $h = 30\text{ m}$ , total downshore distance $L = 80\text{ m}$ .

T(x) = \frac{x}{5} + \frac{\sqrt{30^2 + (80 - x)^2}}{3}

Physical domain:

x \in [0, 80]\text{ meters}

Step 3: Differentiate and find critical numbers

T'(x) = \frac{1}{5} + \frac{1}{3} \cdot \frac{2(80 - x)(-1)}{2\sqrt{900 + (80 - x)^2}} = \frac{1}{5} - \frac{80 - x}{3\sqrt{900 + (80 - x)^2}}

Set

T'(x) = 0

\frac{80 - x}{3\sqrt{900 + (80 - x)^2}} = \frac{1}{5} \implies \frac{80 - x}{\sqrt{900 + (80 - x)^2}} = \frac{3}{5}

Let

u = 80 - x

\frac{u}{\sqrt{900 + u^2}} = \frac{3}{5} \implies \frac{u^2}{900 + u^2} = \frac{9}{25}

25u^2 = 9(900 + u^2) = 8100 + 9u^2

16u^2 = 8100 \implies u^2 = \frac{8100}{16} \implies u = \frac{90}{4} = 22.5\text{ meters}

Solving for

x

x = 80 - 22.5 = 57.5\text{ meters}

Step 4: Evaluate endpoints and critical point (Closed Interval Method)

Run directly into water ( $x = 0$ ): $T(0) = \frac{0}{5} + \frac{\sqrt{30^2 + 80^2}}{3} = \frac{\sqrt{900 + 6400}}{3} = \frac{\sqrt{7300}}{3} \approx 28.48\text{ s}$
Run to perpendicular point then swim ( $x = 80$ ): $T(80) = \frac{80}{5} + \frac{\sqrt{30^2 + 0^2}}{3} = 16 + 10 = 26.00\text{ s}$
Optimal split ( $x = 57.5$ ): $T(57.5) = \frac{57.5}{5} + \frac{\sqrt{30^2 + (22.5)^2}}{3} = 11.5 + \frac{\sqrt{900 + 506.25}}{3} = 11.5 + \frac{\sqrt{1406.25}}{3} = 11.5 + \frac{37.5}{3} = 11.5 + 12.5 = 24.00\text{ s}$

Absolute minimum transit time is $24.0\text{ seconds}$ , achieved by running $57.5\text{ meters}$ on shore before swimming $37.5\text{ meters}$ through water.

Example Problem 2: Infrastructure Cost Routing Across a Waterway

A utility pipeline connects a source facility $S$ on the shore of a $500\text{-meter}$ wide canal to a distribution plant $D$ located $1200\text{ meters}$ downstream on the opposite shore. Subsea pipe laying costs $\$250/\text{meter}$ , while overland pipe laying along the bank costs $\$150/\text{meter}$ . Determine the downstream landing point $x$ that minimizes total pipeline construction costs.

Solution

Step 1: State governing formulas

Cost Objective Function: $C(x) = C_{\text{water}} \sqrt{w^2 + x^2} + C_{\text{land}} (L - x)$ (Defined in Fastest Path Across Media and Snell's Law.)

Step 2: Assign parameters
Given width $w = 500\text{ m}$ , downstream distance $L = 1200\text{ m}$ , $C_{\text{water}} = \$250/\text{m}$ , $C_{\text{land}} = \$150/\text{m}$ .

C(x) = 250\sqrt{500^2 + x^2} + 150(1200 - x) \quad \text{for } x \in [0, 1200]

Step 3: Differentiate and find critical numbers

C'(x) = 250 \cdot \frac{x}{\sqrt{500^2 + x^2}} - 150

Set

C'(x) = 0

\frac{250x}{\sqrt{250,000 + x^2}} = 150 \implies \frac{x}{\sqrt{250,000 + x^2}} = \frac{150}{250} = \frac{3}{5}

Squaring both sides:

\frac{x^2}{250,000 + x^2} = \frac{9}{25} \implies 25x^2 = 9x^2 + 2,250,000

16x^2 = 2,250,000 \implies x^2 = 140,625 \implies x = 375\text{ meters}

Step 4: Execute Closed Interval Method

At $x = 0$ (perpendicular water crossing): $C(0) = 250(500) + 150(1200) = 125,000 + 180,000 = \$305,000$
At $x = 1200$ (direct diagonal underwater line): $C(1200) = 250\sqrt{500^2 + 1200^2} = 250(1300) = \$325,000$
At $x = 375$ (optimal angled crossing): $C(375) = 250\sqrt{500^2 + 375^2} + 150(1200 - 375) = 250(625) + 150(825) = 156,250 + 123,750 = \$280,000$

The minimum cost is $\$280,000$ , achieved by landing the subsea pipe $375\text{ meters}$ downstream.

Connections & References

Parent Concepts:
Related Optimization Techniques:
- Squared-Distance Optimization Technique
- Cruising Velocity and Operational Transit Cost Optimization
Course Links:
- Calc-1 Session 14 - Optimizing Areas and Volumes
- Calc-1 Session 15 - Optimizing Distance and Time