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You want to get from a point a on the straight shore of the beach to a buoy which is 60 meters out in the water from a point b on the shore. b is 70 meters from you down the shore. if you can swim at a speed of 3 meters per second and run at a speed of 5 meters per second, at what point along the shore, x meters from b, should you stop running and start swimming if you want to reach the buoy in the least time possible?

Respuesta :

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The time t for a constant speed is given by:
t = distance/ velocity

Let's call the distance to run on the beach x₁ and the remaining distance to swim x₂, then the equation becomes:
[tex]t = \frac{x_1}{5} + \frac{x_2}{3} [/tex]

The distance x₂ depends on x₁:
[tex]x_2 = \sqrt{(70 - x_1)^2 + 60^2} [/tex]

So: [tex]t = \frac{x_1}{5} + \frac{ \sqrt{(70 - x_1)^2 + 60^2} }{3} [/tex]

Find the minimum by setting the derivative to zero: [tex] \frac{dt}{dx_1} = 0[/tex]
[tex] \frac{dt}{dx_1} = \frac{1}{5} - \frac{1}{3} \frac{70 - x_1} {\sqrt{70 - x_1)^2 + 60^2} } = 0 [/tex]
[tex]70 - x_1 = \frac{3}{5} \sqrt{(70 - x_1)^2 + 60^2} \\ \\ (70 - x_1)^2 = \frac{9}{25}((70 - x_1)^2 + 60^2) \\ \\ \frac{16}{25}(70 - x_1)^2 = \frac{9}{25}60^2 \\ \\ \frac{4}{5}(70 - x_1) = \frac{3}{5}60 \\ \\ x_1 = 25[/tex]

You should stop running 70m - 25m = 45m from b.

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