MATH-181 (Calculus I) Curriculum

Session 15 - Optimizing Distance and Time

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Formulate Euclidean distance objective functions between external reference coordinates and functional curves, applying the Squared-Distance Optimization Technique to eliminate radical chain-rule differentiation while preserving critical number locations.
Verify geometric orthogonality between the displacement vector to the closest point on a curve and the curve's tangent line, proving that the target point lies on the normal line.
Model multi-medium transit scenarios using time-rate formulations ( $T = d/v$ ), identifying optimal boundary crossing points and discovering the calculus derivation of Snell's Law and Fermat's Principle of Least Time formalized in Fastest Path Across Media and Snell's Law.
Optimize municipal utility infrastructure routing (subsea/river pipelines and telecommunication cables) across disparate physical media, balancing per-meter excavation costs against path lengths using the Closed Interval Method.
Construct operational fleet transit cost functions balancing inverse-velocity time/labor overhead ( $A/v$ ) against nonlinear aerodynamic drag and fuel/energy expenses ( $Bv$ or $Bv^2$ ), validating the Cost Equipartition Theorem and the Single Critical Point Rule on open domains.

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.

Skill Block 1

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.

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 $d(x)$ ?

Solution Key:

(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.

Show your steps and calculations here.

Practice Problem 2

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

Solution Key:

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: $D(3.5) = (3.5)^2 - 7(3.5) + 16 = 12.25 - 24.5 + 16 = 3.75 = \frac{15}{4}$ $d_{\min} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \approx 1.936$

Show your steps and calculations here.

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.

Solution Key:

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.

Show your steps and calculations here.

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.

Skill Block 2

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.

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?

Solution Key:

(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 ( $\$835,224$ ).

Show your steps and calculations here.

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.

Solution Key:

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.

Show your steps and calculations here.

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?

Solution Key:

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 $86.67\text{ seconds}$ .

Show your steps and calculations here.

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 ( $x = 0$ ).

Skill Block 3

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.

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?

Solution Key:

(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.

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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.

Solution Key:

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 $\$30,000.00$ .

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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.

Solution Key:

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.)

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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.

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.