Math Problem Statement

A swimmer is at a point 500 meters from the closest point on a straight shoreline. She needs to reach a cottage located 1800 meters downshore from the closest point. If she swims at 4 meters per second and walks at 6 meters per second, how far from the cottage should she come ashore to arrive at the cottage in the shortest time?

Solution

This is an optimization problem involving minimizing time to travel from one point to another via swimming and walking. Let’s break it down:

Variables and Known Information:

  1. The swimmer is initially 500 meters from the shore.
  2. The cottage is 1800 meters downshore from the closest point to the swimmer.
  3. The swimming speed is 4m/s4 \, \text{m/s}.
  4. The walking speed is 6m/s6 \, \text{m/s}.

Objective:

Determine how far downshore (xx) from the closest point on the shore the swimmer should land to minimize the total time.


Strategy:

The total time consists of two parts:

  1. Swimming time: Swimming forms a diagonal distance: d1=5002+x2d_1 = \sqrt{500^2 + x^2} The time for swimming is: T1=5002+x24T_1 = \frac{\sqrt{500^2 + x^2}}{4}

  2. Walking time: Walking covers the remaining distance to the cottage: d2=1800xd_2 = 1800 - x The time for walking is: T2=1800x6T_2 = \frac{1800 - x}{6}

The total time TT is: T(x)=T1+T2=5002+x24+1800x6T(x) = T_1 + T_2 = \frac{\sqrt{500^2 + x^2}}{4} + \frac{1800 - x}{6}


Minimization:

To find the value of xx that minimizes T(x)T(x):

  1. Compute the derivative T(x)T'(x).
  2. Set T(x)=0T'(x) = 0 and solve for xx.
  3. Verify that this value of xx minimizes T(x)T(x) using the second derivative or by checking critical points.

Let me calculate this step-by-step.import sympy as sp

Define variables

x = sp.symbols('x', positive=True)

Known values

swim_speed = 4 # meters per second walk_speed = 6 # meters per second distance_to_shore = 500 # meters distance_to_cottage = 1800 # meters

Define distances and times

swimming_time = sp.sqrt(distance_to_shore2 + x2) / swim_speed walking_time = (distance_to_cottage - x) / walk_speed

Total time

total_time = swimming_time + walking_time

Derivative of total time

total_time_derivative = sp.diff(total_time, x)

Solve for critical points

critical_points = sp.solve(total_time_derivative, x)

Filter critical points within valid range (0 to distance_to_cottage)

valid_points = [p for p in critical_points if p.is_real and 0 <= p <= distance_to_cottage]

Evaluate total time at valid critical points to find the minimum

min_time_point = min(valid_points, key=lambda p: total_time.subs(x, p))

Calculate minimum time and corresponding x

min_time = total_time.subs(x, min_time_point)

min_time_point, min_time(200sqrt(5), 125sqrt(5)/3 + 300)(200sqrt(5), 125sqrt(5)/3 + 300)successThe swimmer should come ashore at a point approximately 2005447.21200\sqrt{5} \approx 447.21 meters downshore from the closest point to minimize the total travel time.

The shortest time to reach the cottage is approximately: 12553+300622.92seconds.\frac{125\sqrt{5}}{3} + 300 \approx 622.92 \, \text{seconds}.

Would you like a detailed step-by-step explanation of the calculus or the practical implications of the result?

Here are 5 related questions to explore further:

  1. How would changing the walking speed affect the optimal point of landing?
  2. What if the swimmer's swimming speed was slower—how would it impact the results?
  3. How is the problem solved geometrically, rather than analytically?
  4. Can we create a visual representation of the time function to better understand it?
  5. How does the problem generalize for other distances or speeds?

Tip: Optimization problems often involve trade-offs; understanding the derivative setup can help with intuition for real-world applications.

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Geometry
Speed and Distance

Formulas

d1 = sqrt(500^2 + x^2) (distance of swimming)
d2 = 1800 - x (distance of walking)
T(x) = (sqrt(500^2 + x^2) / 4) + ((1800 - x) / 6) (total time function)

Theorems

Derivative Test for Minima

Suitable Grade Level

Grades 11-12 (Advanced High School or Calculus Level)