Calculus Lecture: Application of Derivatives

Topic: Optimization Problem with a River Crossing

Problem Statement

• Objective: Determine the optimal landing point on the opposite bank of a river to minimize the time taken to reach a destination.
• Scenario:
  • Person at point A needs to reach point B.
  • River width: 3 km.
  • Distance along the opposite bank: 8 km.
  • Speeds: Rowing at 6 km/h, Running at 8 km/h.

Methodology

1. Objective: Minimize travel time

   • Options for crossing:
     • Row directly to B
     • Row to some point D between B and C, then run to B
   • Define variables:
     • CD = s (distance along the opposite bank)
   • Use Pythagorean Theorem:
     • AC² + CD² = AD²
     • AC = 3 km, CD = s
     • AD = ( \sqrt{9 + s^2} )

2. Calculate Time

   • Rowing Time: ( \frac{AD}{6} )
   • Running Time: ( \frac{8 - s}{8} )
   • Total Time Function: [ T(s) = \frac{\sqrt{9 + s^2}}{6} + \frac{8 - s}{8} ]

3. Optimization Using Derivatives

   • Differentiate Time Function:
     • Use chain rule and derivative of a square root function.
   • Set derivative to zero to find critical points.
   • Solve for s:
     • ( \frac{2s}{2\sqrt{9 + s^2}} - \frac{1}{8} = 0 )
     • Simplify and solve: [ 16s^2 = 81(9 + s^2) ]
       • Solve quadratic equation for s.
     • Accept positive solution since distance must be positive.

4. Solution

   • Optimal s: ( \frac{9}{\sqrt{7}} )
     • Approximate s ≈ 3.402 km.
   • Calculate total time and verify with calculations.
   • Result: Optimal path ensures shortest travel time.

Conclusion

• Calculated optimal landing point ensures minimal travel time from A to B across the river.
• Key takeaway: Use derivative optimization techniques to solve real-world problems effectively.

Thank you for attending the lecture!