Session 15 - Optimizing Distance and Time | Tony's Intertwingled Memex

📅 Micro-Lecture

1. Beyond Static Geometry: The Physics of Motion, Clearance, and Time

Mathematical Foundation: In Session 14, we explored static geometric containers—enclosing fixed areas and packaging fixed volumes. In Session 15, we expand optimization into dynamic spatial coordinates and kinematic rates. Here, the decision variable often controls a spatial clearance, a trajectory transition point, or an operating velocity.
Engineering Context: Engineering systems are rarely static. Autonomous drones compute real-time collision clearances to surrounding obstacles; subsea civil engineers size pipeline routes across wetlands and rivers to balance environmental excavation costs; and logistics fleet managers program commercial cruise speeds to balance driver payroll against aerodynamic fuel consumption.

2. The Squared-Distance Technique & Geometric Orthogonality

Mathematical Foundation: The Euclidean distance $d(x) = \sqrt{(x - x_0)^2 + (y(x) - y_0)^2}$ contains an outer square root that complicates differentiation via the Chain Rule. Because the squaring function $g(u) = u^2$ is strictly monotonic for non-negative values $u \ge 0$ , minimizing the squared distance $D(x) = [d(x)]^2$ yields the exact same critical numbers as minimizing $d(x)$ .
The Normal Line Property: Calculus proves that the straight line connecting an external coordinate to the closest point on any smooth curve is strictly perpendicular to the curve's tangent line ( $m_{\text{normal}} \cdot m_{\text{tangent}} = -1$ ).

3. Fermat's Principle of Least Time & The Cost Equipartition Theorem

Mathematical Foundation: When an entity moves between two media with differing traversal speeds ( $v_1 \ne v_2$ ), the quickest path is not the straight Euclidean line, but a refracted trajectory. Differentiating the elapsed time function $T(x) = \frac{d_1(x)}{v_1} + \frac{d_2(x)}{v_2}$ proves Snell's Law of Refraction ( $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ ).
The Economic Trade-off: In transportation logistics, elapsed travel time scales as $1/v$ , while aerodynamic drag scales with velocity ( $v$ or $v^2$ ). At the mathematically optimal speed for linear power dissipation, total labor expenditure exactly equals total energy expenditure—a universal property known as Cost Equipartition.

Segment 1 of 8

Skill Block 1

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Minimum Distance to Functional Curves & The Squared-Distance Strategy

General Overview
Skill Block 1 focuses on finding the point on a curve closest to an external coordinate. Students learn to formulate Euclidean distance functions, apply the squared-distance optimization technique to eliminate radicals, determine interior critical points, and verify that the optimal displacement vector is orthogonal to the curve's tangent line.

Conceptual Framework

The Euclidean Distance Function:
- Let $(x_0, y_0)$ be a fixed external reference coordinate.
- Let $(x, f(x))$ be an arbitrary point along the curve $y = f(x)$ .
- Distance: $d(x) = \sqrt{(x - x_0)^2 + (f(x) - y_0)^2}$ .
The Squared-Distance Optimization Principle:
- Define $D(x) = [d(x)]^2 = (x - x_0)^2 + (f(x) - y_0)^2$ .
- By the Chain Rule, $D'(x) = 2 \cdot d(x) \cdot d'(x)$ .
- For any physical distance $d(x) > 0$ , $D'(x) = 0 \iff d'(x) = 0$ .
- Both functions share identical critical numbers and identical local/global extrema locations.
The Orthogonality (Normal Line) Principle:
- Setting $D'(x) = 0$ yields $(x - x_0) + (f(x) - y_0)f'(x) = 0$ .
- Rearranging reveals $\left[\frac{f(x) - y_0}{x - x_0}\right] \cdot f'(x) = -1$ .
- The line segment from the target point to the closest point on the curve is perpendicular to the tangent line at that point.

Key Equations & Structural Rules

Squared-Distance Objective Function: $D(x) = (x - x_0)^2 + (f(x) - y_0)^2$ (see Squared-Distance Optimization Technique)
Geometric Orthogonality Condition: $m_{\text{secant}} \cdot m_{\text{tangent}} = \left[\frac{f(x^*) - y_0}{x^* - x_0}\right] \cdot f'(x^*) = -1$ (see Squared-Distance Optimization Technique)

Worked Example A: Minimum Distance from a Point to a Parabola

Find the coordinates of the points on the parabola $y = x^2$ that are closest to the coordinate point $(0, 2)$ , and calculate the minimum distance.

💡 Reveal Worked Solution

Name the Governing Formulas: The Euclidean Distance Formula and the Squared-Distance Objective Function: (These optimization principles are defined in Squared-Distance Optimization Technique.)
Assign Variables and Define Objective Function: Let $(x, y)$ be an arbitrary point on the parabola $y = x^2$ .
The Euclidean distance from $(0, 2)$ to $(x, y)$ is: $d(x) = \sqrt{(x - 0)^2 + (y - 2)^2}$ Applying the Squared-Distance Optimization Technique, define the objective function $D(x) = [d(x)]^2$ : $D(x) = x^2 + (y - 2)^2$
Substitute the Curve Constraint: Substitute $y = x^2$ into the squared-distance function: $D(x) = x^2 + (x^2 - 2)^2 = x^2 + (x^4 - 4x^2 + 4) = x^4 - 3x^2 + 4$
Establish the Feasible Domain: Because the parabola extends infinitely across the real numbers, the algebraic domain is: $\text{Domain: } x \in (-\infty, \infty)$
Differentiate and Locate Critical Numbers: Take the first derivative using the Power Rule: $D'(x) = \frac{d}{dx}[x^4 - 3x^2 + 4] = 4x^3 - 6x$ Set $D'(x) = 0$ and factor: $2x(2x^2 - 3) = 0 \implies x = 0 \quad \text{or} \quad 2x^2 - 3 = 0 \implies x^2 = \frac{3}{2} \implies x = \pm\sqrt{\frac{3}{2}}$
Classify Extrema via the Second Derivative Test: Compute the second derivative:
$D''(x) = 12x^2 - 6$
- At $x = 0$ : $D''(0) = 12(0)^2 - 6 = -6 < 0 \implies$ Local Maximum distance among stationary points.
- At $x = \pm\sqrt{\frac{3}{2}}$ : $D''\left(\pm\sqrt{\frac{3}{2}}\right) = 12\left(\frac{3}{2}\right) - 6 = 18 - 6 = 12 > 0 \implies$ Local Minima.
Compute Optimal Coordinates and Minimum Distance: Substitute $x = \pm\sqrt{\frac{3}{2}}$ into the curve equation $y = x^2$ : $y = \left(\pm\sqrt{\frac{3}{2}}\right)^2 = \frac{3}{2}$ The two closest points on the parabola are: $\left(-\sqrt{\frac{3}{2}}, \frac{3}{2}\right) \quad \text{and} \quad \left(\sqrt{\frac{3}{2}}, \frac{3}{2}\right)$ Evaluate the squared distance at these critical points: $D\left(\pm\sqrt{\frac{3}{2}}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 4 = \frac{9}{4} - \frac{18}{4} + \frac{16}{4} = \frac{7}{4}$ Taking the square root gives the absolute minimum distance: $d_{\min} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \approx 1.323$ (Notice that at the local maximum $x = 0$ , $d = \sqrt{4} = 2 > 1.323$ , confirming that $(0, 0)$ is further from $(0, 2)$ than the two symmetrical troughs.)

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain mathematically why minimizing the squared-distance function $D(x) = [d(x)]^2$ yields the exact same critical numbers and spatial coordinate locations as minimizing the radical distance function $d(x)$ .
(b) In Worked Example A, calculate the slope of the tangent line to $y = x^2$ at the point $\left(\sqrt{1.5}, 1.5\right)$ and the slope of the line segment connecting $(0, 2)$ to that point. Verify that their product equals $-1$ .
(c) In the linked engineering application Autonomous Drone LiDAR Obstacle Clearance (Computer Engineering), what physical safety clearance threshold must the quadcopter maintain from the pylon, and why does the embedded flight computer calculate $D(x)$ rather than $d(x)$ ?

Practice Space Solve this problem on paper or tablet.

Review the Squared-Distance Optimization Principle and the Normal Line Orthogonality Theorem in Squared-Distance Optimization Technique, and check the introduction of Autonomous Drone LiDAR Obstacle Clearance (Computer Engineering).

Answer: * (a) Because the squaring function $g(u) = u^2$ is strictly increasing for $u \ge 0$ , $D'(x) = 2d(x)d'(x) = 0 \iff d'(x) = 0$ for all $d(x) > 0$ .

(b) Tangent slope $m_{\text{tan}} = 2\sqrt{1.5}$ ; secant slope $m_{\text{normal}} = -\frac{1}{2\sqrt{1.5}}$ ; product is $(2\sqrt{1.5})\left(-\frac{1}{2\sqrt{1.5}}\right) = -1$ .
(c) Minimum threshold is $2.0\text{ dam}$ ( $20\text{ m}$ ); computing $D(x)$ avoids costly square-root calculations in real-time embedded firmware.

Step-by-Step Execution

(a) By the Chain Rule, $\frac{d}{dx}[d(x)^2] = 2d(x)d'(x)$ . Whenever physical distance is non-zero ( $d(x) > 0$ ), $2d(x)d'(x) = 0$ if and only if $d'(x) = 0$ . Because the squaring function is strictly monotonic for non-negative values, it preserves order: if $d(x_1) < d(x_2)$ , then $[d(x_1)]^2 < [d(x_2)]^2$ . Thus, every local and absolute extremum of $d(x)$ occurs at the identical location as that of $D(x)$ .
(b) For $y = x^2$ , the derivative is $y' = 2x$ . At $x = \sqrt{1.5}$ , the tangent slope is $m_{\text{tan}} = 2\sqrt{1.5}$ . The line segment connecting $(0, 2)$ to $(\sqrt{1.5}, 1.5)$ has slope $m_{\text{normal}} = \frac{1.5 - 2}{\sqrt{1.5} - 0} = \frac{-0.5}{\sqrt{1.5}} = -\frac{1}{2\sqrt{1.5}}$ . Multiplying the slopes: $m_{\text{tan}} \cdot m_{\text{normal}} = (2\sqrt{1.5})\left(-\frac{1}{2\sqrt{1.5}}\right) = -1$ . This confirms that the displacement vector is perpendicular to the tangent line.
(c) The autonomous quadcopter firmware requires a minimum safety clearance threshold of $d_{\min} \ge 2.0\text{ dam}$ ( $20\text{ meters}$ ) to prevent aerodynamic downwash destabilization and collision. Embedded flight microcontrollers calculate $D(x) = [d(x)]^2$ because polynomial evaluation requires fewer clock cycles and avoids floating-point square-root hardware latency during high-speed obstacle avoidance scans.

Practice Problem 2

Find the point on the curve $y = \sqrt{x}$ that is closest to the coordinate point $(4, 0)$ .

Practice Space Solve this problem on paper or tablet.

State the Squared-Distance Formula $D(x) = (x - x_0)^2 + (y - y_0)^2$ defined in Squared-Distance Optimization Technique. Substitute $y = \sqrt{x}$ and note the physical domain $x \ge 0$ .

Answer:

\text{Point: } \left(\frac{7}{2}, \sqrt{\frac{7}{2}}\right) = (3.5, \approx 1.871), \quad \text{Minimum Distance: } d_{\min} = \frac{\sqrt{15}}{2} \approx 1.936

Step-by-Step Execution

Name the Governing Formula: The Squared-Distance Objective Function: (Defined in Squared-Distance Optimization Technique.) $D(x) = (x - x_0)^2 + (y - y_0)^2$
Substitute Curve and Coordinate: Given $(x_0, y_0) = (4, 0)$ and $y = \sqrt{x}$ : $D(x) = (x - 4)^2 + (\sqrt{x} - 0)^2 = (x^2 - 8x + 16) + x = x^2 - 7x + 16$
Establish Physical Domain: Because $y = \sqrt{x}$ requires non-negative radicands: $\text{Domain: } x \in [0, \infty)$
Differentiate and Find Critical Numbers: $D'(x) = \frac{d}{dx}[x^2 - 7x + 16] = 2x - 7$ Set $D'(x) = 0$ : $2x - 7 = 0 \implies x = \frac{7}{2} = 3.5$ $x = 3.5$ lies strictly within the domain $[0, \infty)$ .
Verify Global Minimum: $D''(x) = 2 > 0 \quad \text{for all } x$ Because $D''(x) > 0$ everywhere and $x = 3.5$ is the unique critical point on $[0, \infty)$ , it is the absolute minimum by the Single Critical Point Rule.
Compute Coordinates and Minimum Distance: $y = \sqrt{3.5} = \sqrt{\frac{7}{2}}$ The closest point is $\left(\frac{7}{2}, \sqrt{\frac{7}{2}}\right) \approx (3.5, 1.871)$ .
Minimum distance: $D(3.5) = (3.5)^2 - 7(3.5) + 16 = 12.25 - 24.5 + 16 = 3.75 = \frac{15}{4}$ $d_{\min} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \approx 1.936$

Practice Problem 3

Find the point on the line $y = 2x + 1$ that is closest to the point $(4, 2)$ , and verify that the line connecting these points is perpendicular to the boundary line.

Practice Space Solve this problem on paper or tablet.

Use the Squared-Distance Formula $D(x) = (x - x_0)^2 + (y - y_0)^2$ from Squared-Distance Optimization Technique. Substitute $y = 2x + 1$ .

Answer:

\text{Point: } \left(\frac{6}{5}, \frac{17}{5}\right) = (1.2, 3.4), \quad \text{Minimum Distance: } d_{\min} = \sqrt{\frac{49}{5}} = \frac{7}{\sqrt{5}} \approx 3.130

Step-by-Step Execution

Name the Governing Formula: The Squared-Distance Objective Function: (Defined in Squared-Distance Optimization Technique.) $D(x) = (x - x_0)^2 + (y - y_0)^2$
Substitute Curve and Point: Given $(x_0, y_0) = (4, 2)$ and $y = 2x + 1$ : $D(x) = (x - 4)^2 + ((2x + 1) - 2)^2 = (x - 4)^2 + (2x - 1)^2$ Expand both terms: $D(x) = (x^2 - 8x + 16) + (4x^2 - 4x + 1) = 5x^2 - 12x + 17$ Domain: $x \in (-\infty, \infty)$ .
Differentiate and Locate Critical Numbers: $D'(x) = 10x - 12$ Set $D'(x) = 0$ : $10x - 12 = 0 \implies x = \frac{12}{10} = \frac{6}{5} = 1.2$
Verify Global Minimum: $D''(x) = 10 > 0$ The function is strictly concave up, confirming an absolute minimum at $x = 1.2$ .
Calculate Point and Verify Orthogonality: Substitute $x = 1.2$ into the line equation: $y = 2(1.2) + 1 = 2.4 + 1 = 3.4 = \frac{17}{5}$ The closest point is $(1.2, 3.4)$ .
Minimum distance: $d_{\min} = \sqrt{(1.2 - 4)^2 + (3.4 - 2)^2} = \sqrt{(-2.8)^2 + (1.4)^2} = \sqrt{7.84 + 1.96} = \sqrt{9.8} = \frac{7}{\sqrt{5}} \approx 3.130$ Check orthogonality: The boundary line $y = 2x + 1$ has slope $m_1 = 2$ .
The segment from $(4, 2)$ to $(1.2, 3.4)$ has slope: $m_2 = \frac{3.4 - 2}{1.2 - 4} = \frac{1.4}{-2.8} = -0.5 = -\frac{1}{2}$ Since $m_1 \cdot m_2 = 2 \cdot \left(-\frac{1}{2}\right) = -1$ , the segment is perpendicular to the line.

Segment 2 of 8

📅 Regroup 1

The Power of Monotonicity: Minimizing $D(x) = [d(x)]^2$ rather than $d(x)$ is an essential engineering simplification. It eliminates square roots and chain-rule fractions without altering critical numbers.
The Normal Line Property: If an exam problem asks for the point on a curve closest to an external point, you can immediately check your work: the line passing through both points must be perpendicular to the curve's tangent line.
Algebraic Artifacts: When solving $D'(x) = 0$ on higher-degree polynomials (such as the cubic in Worked Example A), always use the Second Derivative Test to distinguish between local minima and local maxima.

Segment 3 of 8

Skill Block 2

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Fastest Paths Across Media & Refraction (Time & Infrastructure Routing)

General Overview
Skill Block 2 addresses path optimization across heterogeneous media with contrasting traversal velocities or construction costs. Students learn to model travel time functions ( $T = d/v$ ), differentiate radical expressions, solve for interior transition coordinates, and derive Snell's Law and Fermat's Principle of Least Time.

Conceptual Framework

Travel Time Across Dual Media:
- Let $v_1$ be the propagation speed in Medium 1, and $v_2$ be the speed in Medium 2 ( $v_1 \ne v_2$ ).
- Total travel time is the sum of transit times across both legs: $T(x) = \frac{d_1(x)}{v_1} + \frac{d_2(x)}{v_2}$
The Refraction Mechanism:
- If $v_1 > v_2$ , the entity moves faster in Medium 1.
- To minimize total elapsed time, the optimal path covers greater distance in the high-speed medium, refracting at the boundary rather than traveling along a straight Euclidean line.
The Calculus Derivation of Snell's Law:
- Differentiating $T(x) = \frac{\sqrt{h_1^2 + x^2}}{v_1} + \frac{\sqrt{h_2^2 + (L - x)^2}}{v_2}$ and setting $T'(x) = 0$ yields: $\frac{x}{v_1 \sqrt{h_1^2 + x^2}} = \frac{L - x}{v_2 \sqrt{h_2^2 + (L - x)^2}}$
- Recognizing $\sin \theta_1 = \frac{x}{\sqrt{h_1^2 + x^2}}$ and $\sin \theta_2 = \frac{L - x}{\sqrt{h_2^2 + (L - x)^2}}$ gives: $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$
Infrastructure Cost Equivalence:
- When laying utility pipelines or cables across water and land, replacing velocities $1/v$ with unit installation costs $C$ ( $\$ / \text{m}$ ) produces identical mathematical equations.

Key Equations & Structural Rules

Travel Time Objective Function: $T(x) = \frac{d_1(x)}{v_1} + \frac{d_2(x)}{v_2} = \frac{x}{v_1} + \frac{\sqrt{h^2 + (L - x)^2}}{v_2}$ (see Fastest Path Across Media and Snell's Law)
Snell's Law of Refraction: $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ (see Fastest Path Across Media and Snell's Law)

Worked Example A: The Coastal Transit / Lifeguard Path

A lifeguard at point $A$ on a straight shoreline needs to reach an offshore swimmer at point $B$ . Point $B$ is located $30\text{ meters}$ offshore and $80\text{ meters}$ downshore from point $A$ . 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.

💡 Reveal Worked Solution

Name the Governing Formula: The Multi-Medium Travel Time Function: (Defined in Fastest Path Across Media and Snell's Law.) $T = \frac{d_1}{v_1} + \frac{d_2}{v_2}$
Assign Variables and Define Objective Function: Let $x$ be the distance run along the shoreline (in meters).
The remaining downshore distance is $80 - x$ .
The diagonal swimming distance to the offshore platform is: $d_{\text{swim}}(x) = \sqrt{30^2 + (80 - x)^2}$ Total transit time $T(x)$ is: $T(x) = \frac{x}{5} + \frac{\sqrt{30^2 + (80 - x)^2}}{3} = \frac{1}{5}x + \frac{1}{3}\sqrt{900 + (80 - x)^2}$
Establish Context-Driven Domain: The lifeguard must enter the water between the starting position ( $x = 0$ ) and the point directly opposite the swimmer ( $x = 80$ ): $\text{Domain: } x \in [0, 80]\text{ meters}$
Differentiate and Locate Critical Numbers: Differentiate $T(x)$ using the Chain Rule: $T'(x) = \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{2\sqrt{900 + (80 - x)^2}} \cdot 2(80 - x)(-1) = \frac{1}{5} - \frac{80 - x}{3\sqrt{900 + (80 - x)^2}}$ 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 $u = 80 - x$ : $\frac{u}{\sqrt{900 + u^2}} = \frac{3}{5}$ Square both sides: $\frac{u^2}{900 + u^2} = \frac{9}{25} \implies 25u^2 = 9(900 + u^2) = 8100 + 9u^2$ $16u^2 = 8100 \implies u^2 = \frac{8100}{16} \implies u = \frac{90}{4} = 22.5\text{ meters}$ Solve for $x$ : $x = 80 - 22.5 = 57.5\text{ meters}$ $x = 57.5\text{ m}$ lies strictly inside the physical domain $[0, 80]$ .
Evaluate Extrema via the Closed Interval Method: Evaluate $T(x)$ at the domain endpoints and the interior critical point:
- At $x = 0$ (swimming immediately along direct line): $T(0) = \frac{0}{5} + \frac{\sqrt{30^2 + 80^2}}{3} = \frac{\sqrt{900 + 6400}}{3} = \frac{\sqrt{7300}}{3} \approx 28.48\text{ seconds}$
- At $x = 80$ (running to perpendicular point then swimming straight): $T(80) = \frac{80}{5} + \frac{\sqrt{30^2 + 0^2}}{3} = 16 + \frac{30}{3} = 16 + 10 = 26.00\text{ seconds}$
- At $x = 57.5$ (optimal refracted path): $T(57.5) = \frac{57.5}{5} + \frac{\sqrt{30^2 + (22.5)^2}}{3} = 11.5 + \frac{\sqrt{900 + 506.25}}{3} = 11.5 + \frac{\sqrt{1406.25}}{3} = 11.5 + \frac{37.5}{3} = 11.5 + 12.5 = 24.00\text{ seconds}$ The absolute minimum transit time is $24.0\text{ seconds}$ , achieved by running $57.5\text{ meters}$ along the shore and swimming $37.5\text{ meters}$ diagonally.

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) In time-minimization problems across two media with different speeds ( $v_1 > v_2$ ), why does the optimal path bend at the interface boundary rather than following a single Euclidean straight line?
(b) In Worked Example A, let $\theta$ represent the angle between the diagonal swimming path and the shoreline. Explain how setting $T'(x) = 0$ directly reveals $\cos \theta = \frac{v_{\text{swim}}}{v_{\text{run}}}$ .
(c) In the linked engineering application Regional Wastewater Effluent Pipeline Routing (Civil & Environmental Engineering), what construction factor makes wetland pipeline installation more expensive than dry bank trenching, and how much money is saved by optimizing the crossing point?

Practice Space Solve this problem on paper or tablet.

Review the derivation of Snell's Law in Fastest Path Across Media and Snell's Law and inspect the cost analysis in Regional Wastewater Effluent Pipeline Routing (Civil & Environmental Engineering).

Answer: * (a) Because speed is faster in Medium 1, spending more distance in Medium 1 reduces total transit time even though overall path length increases.

(b) The geometric ratio $\frac{80 - x}{\sqrt{30^2 + (80 - x)^2}}$ is $\cos \theta$ ; setting $T'(x) = 0$ equates this ratio to $\frac{v_2}{v_1} = \frac{3}{5}$ .
(c) Wetland directional drilling costs $\$400/\text{m}$ vs $\$240/\text{m}$ for dry trenching; optimizing saves $\$48,000$ compared to perpendicular crossing.

Step-by-Step Execution

(a) A straight Euclidean line minimizes spatial distance, but not travel time. Because the entity travels faster in Medium 1 ( $v_1 > v_2$ ), the time saved by traveling farther through the higher-speed medium outweighs the small increase in total geometric distance, up to the critical refraction angle where marginal time saved equals marginal time added.
(b) In the right triangle formed by the swimming leg, the side adjacent to the shoreline has length $80 - x$ and the hypotenuse is $\sqrt{30^2 + (80 - x)^2}$ . Therefore, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{80 - x}{\sqrt{30^2 + (80 - x)^2}}$ . Differentiating $T(x) = \frac{x}{v_1} + \frac{d_2}{v_2}$ and setting to zero gives $\frac{1}{v_1} - \frac{\cos \theta}{v_2} = 0 \implies \cos \theta = \frac{v_2}{v_1} = \frac{3}{5}$ .
(c) Wetland pipeline installation requires specialized horizontal directional drilling (HDD) rigs, biological environmental monitoring, and secondary leak containment liners costing $\$400/\text{m}$ , whereas dry easement trenching costs $\$240/\text{m}$ . Optimizing the crossing point to $x = 450\text{ m}$ produces a total cost of $\$672,000$ , saving $\$48,000$ ( $6.7\%$ ) compared to a perpendicular crossing ( $\$720,000$ ) and saving $\$163,224$ ( $19.5\%$ ) compared to a direct diagonal line ( $\$835,224$ ).

Practice Problem 5

A municipal water pipeline connects a pumping station $P$ on the bank of a straight $400\text{-meter}$ wide river to a treatment plant $F$ located $1000\text{ meters}$ downstream on the opposite bank. Laying pipe underwater costs $\$100/\text{meter}$ , while laying pipe along the dry riverbank costs $\$60/\text{meter}$ . The pipe runs underwater in a straight diagonal line from $P$ to a point $Q$ on the opposite bank, and then runs along the bank to $F$ . Determine the distance $x$ downstream where the pipe should reach the opposite bank to minimize total construction cost.

Practice Space Solve this problem on paper or tablet.

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

Answer:

\text{Optimal Crossing Distance: } x = 300\text{ meters}, \quad \text{Minimum Total Cost: } C(300) = \$92,000

Step-by-Step Execution

Name the Governing Formula: The Dual-Medium Cost Objective Function: (Defined in Fastest Path Across Media and Snell's Law.) $C(x) = C_1 \sqrt{w^2 + x^2} + C_2 (L - x)$
Assign Parameters and Formulate Objective: River width $w = 400\text{ m}$ , total downstream distance $L = 1000\text{ m}$ , water cost $C_1 = \$100/\text{m}$ , land cost $C_2 = \$60/\text{m}$ . $C(x) = 100\sqrt{400^2 + x^2} + 60(1000 - x) \quad \text{for } x \in [0, 1000]$
Differentiate and Locate Critical Numbers: $C'(x) = 100 \cdot \frac{2x}{2\sqrt{160,000 + x^2}} - 60 = \frac{100x}{\sqrt{160,000 + x^2}} - 60$ Set $C'(x) = 0$ : $\frac{100x}{\sqrt{160,000 + x^2}} = 60 \implies \frac{x}{\sqrt{160,000 + x^2}} = \frac{60}{100} = \frac{3}{5}$ Square both sides: $\frac{x^2}{160,000 + x^2} = \frac{9}{25} \implies 25x^2 = 9(160,000 + x^2) = 1,440,000 + 9x^2$ $16x^2 = 1,440,000 \implies x^2 = 90,000 \implies x = 300\text{ meters}$ $x = 300\text{ m}$ is a valid critical number in $[0, 1000]$ .
Evaluate via the Closed Interval Method:
- At $x = 0$ (perpendicular crossing): $C(0) = 100(400) + 60(1000) = 40,000 + 60,000 = \$100,000$
- At $x = 1000$ (direct straight line underwater): $C(1000) = 100\sqrt{400^2 + 1000^2} = 100\sqrt{160,000 + 1,000,000} = 100\sqrt{1,160,000} \approx 100(1077.03) = \$107,703$
- At $x = 300$ (optimal landing point): $C(300) = 100\sqrt{400^2 + 300^2} + 60(1000 - 300) = 100(500) + 60(700) = 50,000 + 42,000 = \$92,000$ The absolute minimum construction cost is $\$92,000$ , achieved by landing the underwater pipe $300\text{ meters}$ downstream.

Practice Problem 6

A planetary rover at point $A$ on a compacted dirt road needs to reach an exploration target $B$ located $600\text{ meters}$ east and $200\text{ meters}$ north across a boulder plain. The road runs directly east-west. Rover speed is $10\text{ m/s}$ along the road and $6\text{ m/s}$ across the boulder plain. How far $x$ along the road should the rover travel before turning diagonally into the boulder plain to reach target $B$ in minimum time?

Practice Space Solve this problem on paper or tablet.

State the Transit Time Formula $T(x) = \frac{x}{v_{\text{road}}} + \frac{\sqrt{d_{\text{north}}^2 + (L - x)^2}}{v_{\text{plain}}}$ from Fastest Path Across Media and Snell's Law.

Answer:

\text{Distance along road: } x = 450\text{ meters}, \quad \text{Minimum Travel Time: } T = 86.67\text{ seconds}

Step-by-Step Execution

Name the Governing Formula: The Multi-Medium Travel Time Function: (Defined in Fastest Path Across Media and Snell's Law.) $T(x) = \frac{x}{v_1} + \frac{\sqrt{h^2 + (L - x)^2}}{v_2}$
Assign Parameters: Road speed $v_1 = 10\text{ m/s}$ , off-road speed $v_2 = 6\text{ m/s}$ , lateral offset $h = 200\text{ m}$ , east distance $L = 600\text{ m}$ . $T(x) = \frac{x}{10} + \frac{\sqrt{200^2 + (600 - x)^2}}{6} \quad \text{for } x \in [0, 600]$
Differentiate and Find Critical Numbers: $T'(x) = \frac{1}{10} - \frac{600 - x}{6\sqrt{40,000 + (600 - x)^2}}$ Set $T'(x) = 0$ : $\frac{600 - x}{6\sqrt{40,000 + (600 - x)^2}} = \frac{1}{10} \implies \frac{600 - x}{\sqrt{40,000 + (600 - x)^2}} = \frac{6}{10} = \frac{3}{5}$ Let $u = 600 - x$ : $\frac{u^2}{40,000 + u^2} = \frac{9}{25} \implies 25u^2 = 360,000 + 9u^2 \implies 16u^2 = 360,000$ $u^2 = 22,500 \implies u = 150\text{ meters}$ Solve for $x$ : $x = 600 - 150 = 450\text{ meters}$
Evaluate via the Closed Interval Method:
- At $x = 0$ (direct diagonal across boulders): $T(0) = \frac{\sqrt{200^2 + 600^2}}{6} = \frac{\sqrt{400,000}}{6} \approx \frac{632.46}{6} \approx 105.41\text{ s}$
- At $x = 600$ (road to perpendicular point, then due north): $T(600) = \frac{600}{10} + \frac{200}{6} = 60 + 33.33 = 93.33\text{ s}$
- At $x = 450$ (optimal turn-off point): $T(450) = \frac{450}{10} + \frac{\sqrt{200^2 + 150^2}}{6} = 45 + \frac{250}{6} = 45 + 41.67 = 86.67\text{ s}$ The rover should travel $450\text{ meters}$ along the road before turning toward target $B$ , minimizing total transit time to $86.67\text{ seconds}$ .

Segment 4 of 8

📅 Regroup 2

The Derivative Pattern: Notice that in every dual-medium problem, differentiating $\frac{\sqrt{h^2 + u^2}}{v}$ always generates the exact trigonometric projection $\frac{u}{v \sqrt{h^2 + u^2}} = \frac{\sin \theta}{v}$ .
Algebraic Substitution Trick: When solving equations like $\frac{L - x}{\sqrt{h^2 + (L - x)^2}} = k$ , always substitute $u = L - x$ before squaring. This prevents expanding unwieldy quadratic expressions $(L - x)^2 = L^2 - 2Lx + x^2$ until after $u$ is solved.
Physical Boundary Sanity Check: What happens if $v_1 \le v_2$ ? If the off-road speed is faster than or equal to the road speed ( $v_2 \ge v_1$ ), the ratio $\frac{v_2}{v_1} \ge 1$ . Since $\sin \theta \le 1$ , no interior critical point exists, meaning the optimal path is a direct straight line ( $x = 0$ ).

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Skill Block 3

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Velocity, Fuel Economy & Operational Transit Cost Optimization

General Overview
Skill Block 3 extends optimization to kinematic operating costs and energy consumption. Students learn to model trip transit time as $t = D/v$ , balance inverse-velocity hourly labor expenses ( $wD/v$ ) against direct-velocity aerodynamic drag expenses ( $k D v$ or $k D v^2$ ), verify absolute extrema on open intervals, and apply the Cost Equipartition Theorem.

Conceptual Framework

The Transit Trade-off Model:
- Let $D$ be total trip distance. Elapsed transit time is $t(v) = \frac{D}{v}$ .
- Time-dependent expenses (crew wages, vehicle depreciation) scale as $C_{\text{time}} = w \cdot \frac{D}{v} \propto \frac{1}{v}$ .
- Energy/fuel consumption required to overcome aerodynamic or hydrodynamic drag scales nonlinearly with speed ( $P \propto v^3 \implies \text{Energy/time} \propto v^2 \text{ or } v^3$ ).
- Energy expense over distance $D$ scales as $C_{\text{fuel}} = k D v^{n-1}$ (typically linear or quadratic in $v$ ).
The Cost Equipartition Theorem:
- When hourly fuel cost is quadratic ( $R_{\text{fuel}} = k v^2$ ), total trip cost is $C(v) = \frac{A}{v} + Bv$ .
- Differentiating and setting to zero yields $-\frac{A}{v^2} + B = 0 \implies \frac{A}{v} = Bv$ .
- At the optimal velocity, total labor/time expenditure exactly equals total fuel expenditure.
Verification on Open/Unbounded Domains:
- In vehicle speed problems, $v > 0$ . The Second Derivative Test confirms: $C''(v) = \frac{2A}{v^3} > 0 \quad \text{for all } v > 0$
- By the Single Critical Point Rule (formalized in First Derivative Test), an interior critical point with $C''(v^*) > 0$ is guaranteed to be the unique absolute minimum on $(0, \infty)$ .

Key Equations & Structural Rules

Fleet Transit Cost Objective Function: $C(v) = \frac{w D}{v} + k D v^{n-1}$ (see Cruising Velocity and Operational Transit Cost Optimization)
The Cost Equipartition Condition: $C_{\text{time}}(v^*) = C_{\text{fuel}}(v^*) \iff \frac{w D}{v^*} = k D (v^*)^{n-1}$ (see Cruising Velocity and Operational Transit Cost Optimization)

Worked Example A: Commercial Freight Truck Fleet Optimization

A long-haul freight truck operates on a $300\text{-mile}$ highway route. The commercial driver receives a flat wage of $\$36/\text{hour}$ . When cruising at a steady speed $v$ (in $\text{mph}$ ), diesel fuel costs $\frac{v^2}{100}$ dollars per hour. Fleet governors restrict cruising velocity to the closed interval $v \in [40, 75]\text{ mph}$ . Find the cruising speed that minimizes the total cost of the trip, and calculate the cost at that speed.

💡 Reveal Worked Solution

Name the Governing Formula: The Fleet Transit Cost Objective Function: (Defined in Cruising Velocity and Operational Transit Cost Optimization.) $C(v) = t(v) \cdot \left[ R_{\text{labor}} + R_{\text{fuel}}(v) \right] = \frac{D}{v} \left( w + \frac{v^2}{k} \right) = \frac{w D}{v} + \frac{D}{k} v$
Assign Parameters and Formulate Objective: Route distance $D = 300\text{ miles}$ , wage rate $w = \$36/\text{hr}$ , fuel coefficient $k = 100$ . $C(v) = \frac{300}{v}\left(36 + \frac{v^2}{100}\right) = \frac{10,800}{v} + 3v \quad \text{for } v \in [40, 75]$ Rewrite in power form for differentiation: $C(v) = 10,800 v^{-1} + 3v$
Differentiate and Find Critical Numbers: $C'(v) = \frac{d}{dv}[10,800 v^{-1} + 3v] = -10,800 v^{-2} + 3 = -\frac{10,800}{v^2} + 3$ Set $C'(v) = 0$ : $\frac{10,800}{v^2} = 3 \implies 3v^2 = 10,800 \implies v^2 = 3600 \implies v = 60\text{ mph}$ $v = 60\text{ mph}$ lies strictly within the allowable speed interval $[40, 75]$ .
Evaluate Extrema via the Closed Interval Method: Evaluate $C(v) = \frac{10,800}{v} + 3v$ at the boundaries and the critical point:
- At minimum governor speed ( $v = 40\text{ mph}$ ): $C(40) = \frac{10,800}{40} + 3(40) = 270 + 120 = \$390.00$
- At maximum speed limit ( $v = 75\text{ mph}$ ): $C(75) = \frac{10,800}{75} + 3(75) = 144 + 225 = \$369.00$
- At critical speed ( $v = 60\text{ mph}$ ): $C(60) = \frac{10,800}{60} + 3(60) = 180 + 180 = \$360.00$ The absolute minimum trip cost is $\$360.00$ , achieved at a cruising velocity of $60\text{ mph}$ . (Notice that total driver wage $\$180$ exactly equals total fuel cost $\$180$ , validating the Cost Equipartition Theorem.)

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why the labor component of transit cost decreases as $1/v$ while the aerodynamic fuel component increases with $v$ .
(b) In Worked Example A, verify that at $v = 60\text{ mph}$ , the marginal rate of labor cost savings ( $-\frac{dC_{\text{labor}}}{dv}$ ) exactly equals the marginal rate of fuel cost increase ( $\frac{dC_{\text{fuel}}}{dv}$ ).
(c) In the linked engineering application Electric Delivery Van Highway Cruising Optimization (Mechanical Engineering), what cruising speed minimizes the delivery shift cost, and how does wage inflation alter that optimal speed?

Practice Space Solve this problem on paper or tablet.

Review the mathematical model in Cruising Velocity and Operational Transit Cost Optimization and inspect the conclusion of Electric Delivery Van Highway Cruising Optimization (Mechanical Engineering).

Answer: * (a) Elapsed trip time $t = D/v$ is inversely proportional to speed, but aerodynamic drag forces scale with $v^2$ , making fuel power scale with $v^3$ and cost per mile scale with $v$ .

(b) $-\frac{d}{dv}\left[\frac{10,800}{v}\right] = \frac{10,800}{3600} = 3\text{ \$/(mph)}$ ; $\frac{d}{dv}[3v] = 3\text{ \$/(mph)}$ .
(c) Optimal speed is $60\text{ mph}$ ; wage inflation increases time-based overhead, shifting the optimal speed higher to reduce paid hours.

Step-by-Step Execution

(a) Because transit distance $D$ is constant, driving faster decreases the total duration of the trip according to $t = D/v$ , causing hourly labor expenses to decline as $wD/v$ . Conversely, overcoming aerodynamic drag requires mechanical power proportional to the cube of velocity ( $P \propto v^3$ ). Dividing power by velocity reveals that fuel or battery energy consumed per unit distance increases with speed ( $E/D \propto v^2$ or $v$ ), causing energy costs to climb.
(b) Total trip cost is $C(v) = C_{\text{labor}}(v) + C_{\text{fuel}}(v) = \frac{10,800}{v} + 3v$ . Taking derivatives of the individual components:
- Marginal labor savings rate: $-\frac{dC_{\text{labor}}}{dv} = \frac{10,800}{v^2}$ . At $v = 60$ , this equals $\frac{10,800}{3600} = 3\text{ dollars per mph}$ .
- Marginal fuel expense rate: $\frac{dC_{\text{fuel}}}{dv} = 3\text{ dollars per mph}$ . At $v = 60\text{ mph}$ , the marginal savings from driving faster exactly balances the marginal cost of additional fuel, which is why the net derivative $C'(v) = -3 + 3 = 0$ .
(c) For the electric delivery van fleet operating on the $240\text{-mile}$ Interstate-15 corridor, the optimal cruising speed is $60\text{ mph}$ , resulting in an operating cost of $\$360$ per shift. If commercial driver wages rise (e.g., due to overtime premiums from $\$45/\text{hr}$ to $\$60/\text{hr}$ ), the numerator of the time-cost function increases ( $wD = 60 \times 240 = 14,400$ ), shifting the optimal speed higher to $v^* = \sqrt{14,400 / 3} = \sqrt{4,800} \approx 69.3\text{ mph}$ to minimize paid operating hours.

Practice Problem 8

A commercial container cargo vessel navigates a $400\text{-nautical-mile}$ voyage. Ship crew and onboard hotel electrical systems cost a flat $\$500/\text{hour}$ . Heavy bunker fuel costs $0.25 v^3$ dollars per hour when the vessel cruises at velocity $v$ (in knots). Navigational safety limits cruising velocity to $v \in [6, 18]\text{ knots}$ . Find the cruising speed that minimizes the total voyage cost, and calculate the cost at that speed.

Practice Space Solve this problem on paper or tablet.

Formulate $C(v) = \frac{D}{v}(w + k v^3) = \frac{w D}{v} + k D v^2$ using Cruising Velocity and Operational Transit Cost Optimization. Differentiate and set to zero.

Answer:

\text{Optimal Speed: } v = 10\text{ knots}, \quad \text{Minimum Total Cost: } C(10) = \$30,000

Step-by-Step Execution

Name the Governing Formula: The Cubic-Drag Transit Cost Objective Function: (Defined in Cruising Velocity and Operational Transit Cost Optimization.) $C(v) = \frac{D}{v}(w + k v^3) = \frac{w D}{v} + k D v^2$
Assign Parameters: Voyage distance $D = 400\text{ nmi}$ , hourly crew overhead $w = \$500/\text{hr}$ , fuel coefficient $k = 0.25$ . $C(v) = \frac{400}{v}(500 + 0.25 v^3) = \frac{200,000}{v} + 100 v^2 \quad \text{for } v \in [6, 18]$ Rewrite in power form: $C(v) = 200,000 v^{-1} + 100 v^2$
Differentiate and Find Critical Numbers: $C'(v) = \frac{d}{dv}[200,000 v^{-1} + 100 v^2] = -\frac{200,000}{v^2} + 200 v$ Set $C'(v) = 0$ : $200 v = \frac{200,000}{v^2} \implies 200 v^3 = 200,000 \implies v^3 = 1000 \implies v = 10\text{ knots}$ $v = 10\text{ knots}$ lies strictly inside $[6, 18]$ .
Evaluate via the Closed Interval Method:
- At minimum safe speed ( $v = 6\text{ knots}$ ): $C(6) = \frac{200,000}{6} + 100(6)^2 \approx 33,333.33 + 3,600 = \$36,933.33$
- At maximum safe speed ( $v = 18\text{ knots}$ ): $C(18) = \frac{200,000}{18} + 100(18)^2 \approx 11,111.11 + 32,400 = \$43,511.11$
- At critical speed ( $v = 10\text{ knots}$ ): $C(10) = \frac{200,000}{10} + 100(10)^2 = 20,000 + 10,000 = \$30,000.00$ The optimal cruising speed is $10\text{ knots}$ , minimizing total voyage expenses to $\$30,000.00$ .

Practice Problem 9

An autonomous delivery quadcopter flies a courier route on the open speed domain $v \in (0, 25]\text{ m/s}$ . The net electrical power required to sustain flight is $P(v) = 200 + 2v^2$ watts (where $200\text{ W}$ accounts for hover and avionics base load and $2v^2$ accounts for aerodynamic drag). Find the forward air speed that minimizes the total energy consumed per meter of flight ( $e(v) = \frac{P(v)}{v}$ ), and prove global optimality using the Second Derivative Test.

Practice Space Solve this problem on paper or tablet.

State the Energy-per-Distance Formula $e(v) = \frac{P(v)}{v} = \frac{200}{v} + 2v$ from Cruising Velocity and Operational Transit Cost Optimization. Differentiate and evaluate $e''(v)$ .

Answer:

\text{Optimal Air Speed: } v = 10\text{ m/s} \; (36\text{ km/h}), \quad \text{Minimum Energy: } e(10) = 40\text{ J/meter}

Step-by-Step Execution

Name the Governing Formula: The Specific Energy Objective Function: (Defined in Cruising Velocity and Operational Transit Cost Optimization.) $e(v) = \frac{P(v)}{v}$
Formulate Objective Function: Given power $P(v) = 200 + 2v^2$ (in watts, where $1\text{ W} = 1\text{ J/s}$ ): $e(v) = \frac{200 + 2v^2}{v} = \frac{200}{v} + 2v = 200 v^{-1} + 2v \quad \text{for } v \in (0, 25]$
Differentiate and Locate Critical Numbers: $e'(v) = \frac{d}{dv}[200 v^{-1} + 2v] = -\frac{200}{v^2} + 2$ Set $e'(v) = 0$ : $\frac{200}{v^2} = 2 \implies 2v^2 = 200 \implies v^2 = 100 \implies v = 10\text{ m/s}$ $v = 10\text{ m/s}$ ( $36\text{ km/h}$ ) is the unique stationary point in $(0, 25]$ .
Prove Global Minimum via Second Derivative Test: Compute the second derivative: $e''(v) = \frac{d}{dv}[-200 v^{-2} + 2] = \frac{400}{v^3}$ Evaluate at $v = 10$ : $e''(10) = \frac{400}{10^3} = \frac{400}{1000} = 0.4 > 0$ Because $e''(10) > 0$ and $v = 10$ is the unique critical point on the open-bounded interval $(0, 25]$ , it is the absolute minimum by the Single Critical Point Rule.
Evaluate Minimum Energy Expenditure: $e(10) = \frac{200}{10} + 2(10) = 20 + 20 = 40\text{ Joules per meter}$ (Notice that base hover energy $20\text{ J/m}$ equals aerodynamic drag energy $20\text{ J/m}$ , again confirming cost equipartition.)

Segment 6 of 8

📅 Regroup 3

The Drag Penalty: In vehicle systems, notice why higher speeds become rapidly unaffordable: drag power scales as $v^3$ . Driving $20\%$ faster does not cost $20\%$ more in fuel—it increases fuel burn rate by nearly $73\%$ ( $1.2^3 = 1.728$ ).
The Cost Equipartition Shortcut: Whenever a total cost function has the form $C(v) = \frac{A}{v} + Bv$ , the minimum always occurs where the two terms are equal ( $\frac{A}{v} = Bv \implies v = \sqrt{A/B}$ ). Use this shortcut on homework and exams to check your derivative algebra instantly.
Open Domain Verification: When problems do not specify closed bracket endpoints (e.g., $v \in (0, \infty)$ ), always evaluate the second derivative $C''(v)$ to formally guarantee that your stationary point is an absolute minimum rather than an inflection point.

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📅 Synthesis Wrap-up

Core Takeaways from Session 15:
1. Squared-Distance Simplification: Minimizing $D = d^2$ eliminates radical expressions and simplifies polynomial differentiation while preserving exact critical points.
2. Refracted Fastest Paths: Traversal between media with differing speeds demands bent trajectories; setting $dT/dx = 0$ derives Snell's Law $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ .
3. Velocity & Operating Cost Balance: Optimal speed balances inverse-speed time overhead ( $1/v$ ) against direct-speed aerodynamic drag ( $v$ or $v^2$ ), achieving cost equipartition.
Knewton Alta Bridge: This session completes the theoretical and computational foundation for Knewton Alta homework: Optimization Problems in Other Contexts.
Course Trajectory: Having completed constrained optimization of static geometry (Session 14) and dynamic motion/cost (Session 15), we transition in Session 16 to Curve Sketching, where first and second derivatives combine to map full system responses.

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