You are a lifeguard and spot a drowning child 40 meters along the shore and 50 meters from the shore to the child. You run along the shore for a while and then jump into the water, swimming from there directly to the child. You can run at a rate of 3 meters per second and swim at a rate of 0.8 meters per second. How far along the shore should you run before jumping into the water in order get to the child in the least amount of time?

Hint: Label the distance from the origin where you jump in the water x. Can you come up with a formula in terms of x for the distance you swim in the water? You are minimizing time so you need to come up with a formula for the total time to go along the shore plus the time swimming)

Round your answer to three decimal places

Let's label the distance you run along the shore before jumping into the water as \( x \). The total distance you run along the shore is \( x \) meters, and the remaining distance along the shore to the point directly opposite the child is \( 40 - x \) meters.

The distance you swim to reach the child forms a right triangle with legs \( 50 \) meters (the distance from the shore to the child) and \( 40 - x \) meters (the horizontal distance from where you jump in to the child's position). The hypotenuse of this triangle, which is the distance you swim, can be expressed as:

\[ \text{Distance you swim} = \sqrt{(40 - x)^2 + 50^2} \]

We need to minimize the total time it takes to reach the child. The total time \( T \) is the sum of the time running and the time swimming.

The time running along the shore is:

\[ T_{\text{run}} = \frac{x}{3} \]

The time swimming is:

\[ T_{\text{swim}} = \frac{\sqrt{(40 - x)^2 + 2500}}{0.8} \]

The total time \( T \) as a function of \( x \) is:

\[ T(x) = \frac{x}{3} + \frac{\sqrt{(40 - x)^2 + 2500}}{0.8} \]

To find the value of \( x \) that minimizes \( T(x) \), we take the derivative of \( T(x) \) with respect to \( x \) and set it to zero. 

Let's compute the derivative of \( T(x) \):

\[ T'(x) = \frac{d}{dx} \left( \frac{x}{3} + \frac{\sqrt{(40 - x)^2 + 2500}}{0.8} \right) \]

\[ T'(x) = \frac{1}{3} + \frac{d}{dx} \left( \frac{\sqrt{(40 - x)^2 + 2500}}{0.8} \right) \]

Using the chain rule for the second term:

\[ \frac{d}{dx} \left( \frac{\sqrt{(40 - x)^2 + 2500}}{0.8} \right) = \frac{1}{0.8} \cdot \frac{1}{2\sqrt{(40 - x)^2 + 2500}} \cdot (-2)(40 - x) \]

\[ = \frac{- (40 - x)}{0.8 \sqrt{(40 - x)^2 + 2500}} \]

Thus,

\[ T'(x) = \frac{1}{3} - \frac{(40 - x)}{0.8 \sqrt{(40 - x)^2 + 2500}} \]

Setting \( T'(x) = 0 \):

\[ \frac{1}{3} = \frac{(40 - x)}{0.8 \sqrt{(40 - x)^2 + 2500}} \]

Multiplying both sides by \( 0.8 \sqrt{(40 - x)^2 + 2500} \):

\[ 0.8 \sqrt{(40 - x)^2 + 2500} = 3(40 - x) \]

Squaring both sides to eliminate the square root:

\[ 0.64 ((40 - x)^2 + 2500) = 9(40 - x)^2 \]

\[ 0.64 (40 - x)^2 + 1600 = 9(40 - x)^2 \]

\[ 1600 = 8.36 (40 - x)^2 \]

\[ (40 - x)^2 = \frac{1600}{8.36} \]

\[ (40 - x)^2 = 191.3907 \]

\[ 40 - x = \sqrt{191.3907} \]

\[ 40 - x \approx 13.835 \]

\[ x \approx 40 - 13.835 \]

\[ x \approx 26.165 \]

Thus, the distance you should run along the shore before jumping into the water to reach the child in the least amount of time is approximately \( 26.165 \) meters.

Would you like more details or have any questions?

Here are five questions you might want to ask next:
1. Can you explain how the chain rule was applied in this solution?
2. What is the significance of setting the derivative equal to zero?
3. How would the solution change if the running and swimming speeds were different?
4. Can you provide a step-by-step explanation of how to compute the final distance run?
5. How do you interpret the physical meaning of the solution in this problem?

Tip: When solving optimization problems, always verify your critical points by considering the context and possible constraints.

Learn how to solve an optimization problem involving a lifeguard rescuing a drowning child. Understand the application of the distance formula and derivative rules to find the optimal distance to run along the shore before swimming. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation