Qwen3.6-Plus：迈向真实世界 Agent

Course Notes

Optimization Problem: Minimum Distance Between Two Boats

Qwen3.6 April 1, 2026

Video Author/Channel : integralCALC

Video Duration : 2:30 Contents

1 Problem Statement 2

1.1 Scenario Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Mathematical Modeling 2

2.1 Coordinate System Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Defining the Time Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Position Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Optimization Process 4

3.1 Distance Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.2 Expanding the Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.3 Finding the Critical Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4 Solution and Interpretation 5

4.1 Calculating the Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54.2 Calculating the Minimum Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

5 Summary and Extension 6

5.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65.2 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611 Problem Statement

The video presents a classic calculus optimization problem involving related rates and distance minimiza-tion. The scenario involves two boats moving in perpendicular directions relative to a central dock.

Figure 1: Problem statement displayed on screen. 1

1.1 Scenario Details

• Boat 1: Leaves the dock at 2:00 p.m. and travels South at a constant speed of 20 km/h. • Boat 2: Has been heading East at a constant speed of 15 km/h and is scheduled to reach the dock at 3:00 p.m.

1.2 Objective

The goal is to determine the exact time at which the two boats are closest to each other.

2 Mathematical Modeling

To solve this problem, we must translate the physical movement of the boats into mathematical functions. This requires setting up a coordinate system and defining the position of each boat as a function of time.

2.1 Coordinate System Setup

We place the dock at the origin (0 , 0) of a Cartesian coordinate system. • The positive y-axis points North, and the negative y-axis points South. • The positive x-axis points East, and the negative x-axis points West.

1Video frame time interval: 00:13–00:27.

2Figure 2: Diagram showing the coordinate system, boat positions, and velocity vectors. 2

2.2 Defining the Time Variable

Let t represent the time in hours elapsed since 2:00 p.m. • At 2:00 p.m., t = 0 .• At 3:00 p.m., t = 1 .

Time Definition Caution

Ensure that t = 0 corresponds to the start time of the first boat (2:00 p.m.). The second boat reaches the dock at 3:00 p.m., which corresponds to t = 1 . This offset is crucial for correctly modeling the second boat’s position.

2.3 Position Functions

Boat 1 (Southbound):

• Starts at (0 , 0) at t = 0 .• Moves in the negative y direction at 20 km/h. • Position at time t: P1(t) = (0 , −20 t).

Boat 2 (Eastbound):

• Moves in the positive x direction at 15 km/h. • Reaches the origin (0 , 0) at t = 1 .• Since it is moving East towards the origin from the West, its x-coordinate is negative for t < 1.• The position function is x(t) = 15( t − 1) .

– Check: At t = 1 , x = 0 (at dock).

– Check: At t = 0 , x = −15 (15 km West of dock). • Position at time t: P2(t) = (15( t − 1) , 0) .

2Video frame time interval: 00:47–01:12.

33 Optimization Process

We need to minimize the distance between P1(t) and P2(t).

3.1 Distance Function

The distance d between the two boats at time t is given by the distance formula:

d = √(x2 − x1)2 + ( y2 − y1)2

To simplify the calculation, we minimize the square of the distance, d2. The value of t that minimizes

d2 will also minimize d, since distance is non-negative.

d2 = [15( t − 1) − 0] 2 + [0 − (−20 t)] 2

d2 = (15 t − 15) 2 + (20 t)2

3.2 Expanding the Function

Expanding the terms:

d2 = (225 t2 − 450 t + 225) + 400 t2

d2 = 625 t2 − 450 t + 225

Figure 3: Derivation of the distance squared function. 3

3.3 Finding the Critical Point

To find the minimum distance, we take the derivative of d2 with respect to t and set it to zero.

ddt [d2] = ddt (625 t2 − 450 t + 225)

ddt [d2] = 1250 t − 450

Setting the derivative to zero:

1250 t − 450 = 0 1250 t = 450

3Video frame time interval: 01:50–01:54.

4t = 450 1250 = 45 125 = 925 hours

Figure 4: Solving for t using the derivative. 4

4 Solution and Interpretation

4.1 Calculating the Time

The boats are closest when t = 925 hours. We convert this to minutes to find the clock time. Minutes = 925 × 60 = 540 25 = 21 .6 minutes Since 0.6 minutes is 0.6 × 60 = 36 seconds, the time elapsed is 21 minutes and 36 seconds. Adding this to the start time of 2:00 p.m.: Time of closest approach = 2:21:36 p.m.

4.2 Calculating the Minimum Distance

To find the actual distance, we substitute t = 925 back into the distance squared equation.

d2 = 625

d2 = 625

− 18(9) + 225

d2 = 81 − 162 + 225

d2 = 144

d = √144 = 12 km

4Video frame time interval: 01:55–02:01.

5Figure 5: Final calculation of the minimum distance. 5

Final Result

The boats are closest together at 2:21:36 p.m. , at which point the distance between them is 12 km .

5 Summary and Extension

5.1 Summary

This problem demonstrates a standard approach to optimization in calculus: 1. Model the situation: Define a coordinate system and express the positions of moving objects as functions of time. 2. Formulate the objective function: Create an equation for the quantity to be optimized (in this case, distance squared). 3. Differentiate and solve: Find the critical points by setting the derivative to zero. 4. Interpret the result: Convert the mathematical solution back into the context of the problem (time and distance).

5.2 Extension

This type of problem is a specific case of ”related rates” or ”optimization of distance”. Similar problems might involve: • Objects moving at angles other than 90 degrees (requiring the Law of Cosines). • Objects accelerating (requiring quadratic or higher-order position functions). • Minimizing time or cost instead of distance. For further practice with calculus problems, you can visit integralCALC.com .

5Video frame time interval: 02:11–02:15.