Let's break down the problem step by step: ### Problem Restatement - The pool is circular, lifeguard is at point \(A\) on the circumference. - The drowning person is at the **diametrically opposite** point \(B\). - Lifeguard can **run** along the edge **3 times** as fast as he can **swim**. - He can run from \(A\) to some point \(C\) on the edge, then swim straight from \(C\) to \(B\). - **Question:** What is the optimal path (how much to run, how much to swim) to minimize time? --- ## 1. Assign Variables Let: - Radius of the pool = \( r \) - Running speed = \( v \) - Swimming speed = \( u \), with \( v = 3u \) Let: - Distance along circumference from \(A\) to \(C\) = \( x \) (in radians; arc length = \( x r \)) - Length of swim from \(C\) to \(B\) = ? (to be calculated) \(A\) and \(B\) are diametrically opposite, so the angle subtended at the center is \(\pi\). Let \(C\) be at an angle \(x\) from \(A\) (in direction from \(A\) to \(B\), so \(0 \leq x \leq \pi\)). --- ## 2. Express Distances - **Run:** Arc \(A \to C\): distance \( = r x \) - **Swim:** Chord \(C \to B\): Angle subtended at the center from \(C\) to \(B\) is \( \pi - x \). Chord length formula: \[ \text{Chord length} = 2r \sin\left(\frac{\theta}{2}\right) \] Here, \( \theta = \pi - x \): \[ \text{Swim distance} = 2r \sin\left(\frac{\pi - x}{2}\right) \] --- ## 3. Express Total Time - **Run time:** \( \frac{r x}{v} \) - **Swim time:** \( \frac{2r \sin\left(\frac{\pi - x}{2}\right)}{u} \) - \( v = 3u \) So, \[ \text{Total time} = \frac{r x}{3u} + \frac{2r \sin\left(\frac{\pi - x}{2}\right)}{u} \] --- ## 4. Simplify Expression Divide by \( r/u \) to get: \[ T(x) = \frac{x}{3} + 2 \sin\left(\frac{\pi - x}{2}\right) \] We need to **minimize \( T(x) \) for \( 0 \le x \le \pi \)**. --- ## 5. Find Minimum Take derivative with respect to \( x \): \[ T(x) = \frac{x}{3} + 2 \sin\left(\frac{\pi - x}{2}\right) \] Let’s find the derivative: \[ T'(x) = \frac{1}{3} + 2 \cdot \cos\left(\frac{\pi - x}{2}\right) \cdot \left(-\frac{1}{2}\right) \] The derivative of \( \sin(f(x)) \) is \( \cos(f(x)) \cdot f'(x) \). Here, \( f(x) = \frac{\pi - x}{2} \Rightarrow f'(x) = -\frac{1}{2} \). So, \[ T'(x) = \frac{1}{3} - \cos\left(\frac{\pi - x}{2}\right) \] Set derivative to zero: \[ \frac{1}{3} - \cos\left(\frac{\pi - x}{2}\right) = 0 \implies \cos\left(\frac{\pi - x}{2}\right) = \frac{1}{3} \] --- ## 6. Solve for \( x \) \[ \frac{\pi - x}{2} = \arccos\left(\frac{1}{3}\right) \] \[ \pi - x = 2 \arccos\left(\frac{1}{3}\right) \] \[ x = \pi - 2 \arccos\left(\frac{1}{3}\right) \] --- ## 7. Find Run and Swim Distances - **Run distance:** \( r x = r \left[ \pi - 2 \arccos\left(\frac{1}{3}\right) \right] \) - **Swim distance:** \( 2r \sin\left( \frac{\pi - x}{2} \right) = 2r \sin\left( \arccos\left( \frac{1}{3} \right) \right) \) Let’s compute \( \sin(\arccos(1/3)) \): If \( \theta = \arccos(1/3) \), then \( \cos \theta = 1/3 \), so \[ \sin \theta = \sqrt{1 - (1/3)^2} = \sqrt{1 - 1/9} = \sqrt{8/9} = \frac{2\sqrt{2}}{3} \] So: - **Swim distance:** \( 2r \cdot \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}r \) Now, for the run distance: First, find \( \arccos(1/3) \) in radians. - \( \arccos(1/3) \approx 1.23096 \) radians Therefore, \[ x = \pi - 2 \times 1.23096 \approx 3.1416 - 2.46192 = 0.67968 \text{ radians} \] - **Run distance:** \( r x \approx r \times 0.68 \) radians --- ## **Final Numeric Answers (in terms of radius \(r\)):** - **Run distance:** \( r x = r \left[ \pi - 2 \arccos\left(\frac{1}{3}\right) \right] \approx 0.68 r \) - **Swim distance:** \( \frac{4\sqrt{2}}{3} r \approx 1.89 r \) --- ### **Summary Table** | Step | Formula | Numeric value (in \( r \)) | |--------|---------------------------------------------------|----------------------------| | Run | \( r \left[\pi - 2\arccos(\frac{1}{3})\right] \) | \( 0.68 r \) | | Swim | \( 2r \sin(\arccos(\frac{1}{3})) = \frac{4\sqrt{2}}{3} r \) | \( 1.89 r \) | --- ### **Explanation** The lifeguard should **run a short distance along the edge (\(0.68 r\))** and then **swim the straight line (\(1.89 r\))** to the person. This is faster than either running all the way (\(\pi r \approx 3.14 r\)) or swimming straight (\(2r\)), because running is much faster than swimming. --- ## **Final Answer** **The lifeguard should run a distance of \( r \left[\pi - 2\arccos(\frac{1}{3})\right] \approx 0.68 r \) along the edge, then swim the straight-line distance of \( \frac{4\sqrt{2}}{3} r \approx 1.89 r \) across the pool to the person.**