Respuesta :
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.
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.