A problem to minimize (optimize) the time taken to walk from one point to another is presented. An analytical method, using derivatives and other calculus concepts and theorems, is developed in order to find an analytical solution to the problem.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
You decide to walk from point A (see figure below) to point C. To the south of the road through BC, the terrain is difficult and you can only walk at \( 3 \; \text{km/hr} \). However, along the road BC you can walk at \( 5 \; \text{km/hr} \). The distance from point A to the road is \( 5 \; \text{km} \). The distance from B to C is \( 10 \; \text{km} \). What path do you have to follow in order to arrive at point C in the shortest (minimum) time possible?
Show Solution to the Problem
Let us suppose that you follow the path from A to P and from P to C along straight lines. Let distance BP be equal to \( x \). Using the Pythagorean theorem, we can express the distances AP and PC:
We now find the time \( t_1 \) to walk distance AP (\( \text{time} = \text{distance} / \text{speed} \)):
\[ t_1 = \frac{\sqrt{5^2 + x^2}}{3} \]The time \( t_2 \) to walk distance PC is given by:
\[ t_2 = \frac{10 - x}{5} \]The total time \( t \) is found by adding \( t_1 \) and \( t_2 \):
\[ t(x) = \frac{\sqrt{5^2 + x^2}}{3} + \frac{10 - x}{5} \]The domain of the function \( t(x) \) is the closed interval \( [0, 10] \). For values of \( x \) outside this interval (where point P is to the left of B or to the right of C), the total time will increase.
To find the value of \( x \) that minimizes \( t \), we calculate the first derivative \( \dfrac{dt}{dx} \):
\[ \frac{dt}{dx} = \frac{\frac{x}{3}}{\sqrt{5^2 + x^2}} - \frac{1}{5} \]Setting \( \dfrac{dt}{dx} = 0 \) to find critical points:
\[ \frac{x}{3\sqrt{25 + x^2}} - \frac{1}{5} = 0 \implies \frac{x}{3\sqrt{25 + x^2}} = \frac{1}{5} \]Cross-multiplying:
\[ 5x = 3\sqrt{25 + x^2} \]Squaring both sides:
\[ (5x)^2 = 9(25 + x^2) \implies 25x^2 = 225 + 9x^2 \]Grouping like terms:
\[ 16x^2 = 225 \implies x^2 = \frac{225}{16} \]Solving for \( x \) (\( x > 0 \)):
\[ x = \sqrt{\frac{225}{16}} = \frac{15}{4} = 3.75 \; \text{km} \]
The first derivative \( \dfrac{dt}{dx} \) is negative for \( x < 3.75 \), zero at \( x = 3.75 \), and positive for \( x > 3.75 \). Evaluating \( t(x) \) at the endpoints gives \( t(0) \approx 3.6 \; \text{hrs} \) and \( t(10) \approx 3.7 \; \text{hrs} \), whereas at \( x = 3.75 \), \( t(3.75) \approx 3.3 \; \text{hrs} \) (the minimum).
Conclusion: You must walk to point P such that \( BP = 3.75 \; \text{km} \) and then proceed along the road to C to arrive in the shortest possible time.
Exercises
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Exercise 1: Solve the same problem as above but with the modified parameters shown in the diagram below.
Figure 4. Diagram for practice exercise Show Solution to Exercise 1
Following the same analytical derivative method with the exercise parameters:
\[ x = 6.26 \; \text{km} \quad (\text{rounded to 2 decimal places}) \]