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The shoreline of a lake is a circle with diameter 3 km. Peter stands at point E and wants to reach the diametrically opposite point W. He intends to jog along the north shore to a point P and then swim the straight line distance to W. If he swims at a rate of 3 km/h and jogs at a rate of 24 km/h. How far should he jog in order to arrive at point W in the least amount of time?

The problem involves minimizing the time taken by Peter to jog from point \( E \) to point \( P \) along the shore of the lake and then swim from \( P \) to point \( W \). The lake has a circular shoreline with a diameter of 3 km, and Peter needs to reach the diametrically opposite point on the circle.

### Let's break down the problem:

1. **Given:**
   - Diameter of the circle = 3 km, so the radius \( r \) of the circle = 1.5 km.
   - The total distance \( EW \) = 3 km.
   - Speed of jogging = 24 km/h.
   - Speed of swimming = 3 km/h.

2. **Objective:**
   - Minimize the total time taken by Peter to jog from \( E \) to \( P \) and then swim from \( P \) to \( W \).

3. **Approach:**
   - Let the angle between line segment \( WP \) and the line \( EW \) be \( \theta \).
   - The length of the arc \( EP \) is \( r \theta = 1.5\theta \) (since \( r = 1.5 \)).
   - The distance \( PW \) is the straight line from \( P \) to \( W \). By the law of cosines:
     \[
     PW = \sqrt{3^2 + 3^2 - 2 \cdot 3 \cdot 3 \cdot \cos(\theta)} = 3 \sqrt{2(1 - \cos(\theta))} = 6 \sin(\frac{\theta}{2})
     \]

4. **Total time taken:**
   - Time to jog \( = \frac{1.5\theta}{24} \)
   - Time to swim \( = \frac{6\sin(\frac{\theta}{2})}{3} = 2\sin(\frac{\theta}{2}) \)
   - Total time \( T(\theta) \) is given by:
     \[
     T(\theta) = \frac{1.5\theta}{24} + 2\sin(\frac{\theta}{2})
     \]

### Finding the Minimum Time:
To minimize the total time, we differentiate \( T(\theta) \) with respect to \( \theta \) and set it equal to zero. Then, we solve for \( \theta \) to determine the optimal angle, and hence, the distance to jog.

### Solution via Derivative:
Let's compute it now.

The derivative of the time function \( T(\theta) \) does not yield any real solutions within the valid range \( 0 \leq \theta \leq \pi \). This suggests that there might be no critical points in this interval, meaning the minimum time likely occurs at the boundary values of \( \theta \).

To determine the optimal strategy for Peter:

1. **If \( \theta = 0 \)**: Peter jogs directly from \( E \) to \( W \) without swimming.
2. **If \( \theta = \pi \)**: Peter jogs along half the circumference (to the point diametrically opposite), then swims directly across the diameter.

By evaluating the total time at these boundary values, we can determine which strategy is faster. Let's calculate the time for both scenarios.The calculations yield the following results:

1. **If \( \theta = 0 \)** (Peter jogs directly from \( E \) to \( W \) without swimming):
   - Total time: \( 0 \) hours (since no movement occurs).

2. **If \( \theta = \pi \)** (Peter jogs half the circumference, then swims):
   - Total time: \( 0.0625\pi + 2 \approx 2.197 \) hours.

Since Peter's goal is to minimize time, the fastest strategy is to **jog directly along the shore from point \( E \) to \( W \)**, which takes no time at all since he's already at point \( E \).

Would you like to explore any further details or ask questions?

### Related Questions:
1. What is the total time if Peter jogs half the circle and then swims?
2. How does the speed of swimming impact the optimal strategy?
3. What if Peter’s swimming speed was higher than his jogging speed?
4. How would the optimal path change if the lake were elliptical instead of circular?
5. What is the relationship between the jog distance and swim distance for different angles \( \theta \)?

### Tip:
When solving optimization problems involving time, always consider boundary values if no critical points exist within the interval.

Explore how Peter optimizes his route around a circular lake to reach the diametrically opposite point. Understand the application of the Law of Cosines and optimization techniques in minimizing travel time. Includes detailed calculations and strategy analysis.

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