A coral reef is 6 miles due north of its closest point along a straight shoreline. A visitor is staying in a tent that is 9 miles east of th

Question

A coral reef is 6 miles due north of its closest point along a straight shoreline. A visitor is staying in a tent that is 9 miles east of that point. The visitor is planning to go from the tent to the coral reef. Suppose the visitor runs at a rate of 4 mph and swims at a rate of 2 mph. How far should the visitor run to minimize the time it takes to reach the coral reef? Round to two decimal places.

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3 weeks 2021-09-09T06:46:33+00:00 1 Answer 0

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    2021-09-09T06:47:40+00:00

    Answer:

    • 5.54 miles

    Explanation:

    1. Assume she runs up to a point that is x miles to the east of the coral reef:

    • distance running = 9-x\text{ }miles
    • distance swimming = \sqrt{x^2+6^2}=\sqrt{x^2+36}\text{ }miles

    2. Times:

    time= distance / speed

    • time running = \dfrac{9-x}{4}
    • time swimming = \dfrac{\sqrt{x^2+36} }{2}

    3. Total time

    • time running + time swimming:  

                  T=\dfrac{9-x}{4}+\dfrac{\sqrt{x^2+36} }{2}

    4. Minimize time

    The minimum time is when the derivative of the function of time is zero.

    Find the derivative of T(x), using rule of the sum, the derivative of the sum is the sum of the derivatives, and the chain rule:

             \dfrac{dT}{dx}=\dfrac{-1}{4}+\dfrac{1}{2}\cdot \dfrac{1}{2}\dfrac{2x}{\sqrt{x^2+36}}=0

    Solve for x:

              \dfrac{2x}{\sqrt{x^2+36}}=1\\ \\ \\ 2x=\sqrt{x^2+36\\ } \\ \\ (2x)^2=x^2+36\\ \\ \\ 4x^2=x^2+36\\ \\ \\ 3x^2=36\\ \\ \\ x^2=12\\ \\ \\ x=2\sqrt{3}

    5. Distance the visitor should run to minimize the time

    Hence, the visitor should run 9-2\sqrt{3}\approx 5.54  miles.

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