Respuesta :
Answer:
The route that take the least amount of time is the one that she swimms 1.25 miles across the river, to go out of the river on the other shore, 0.25 miles horizontally from the starting point, where she walks 76.25 miles to the campground.
x=0.25 (see figure attached).
Step-by-step explanation:
We have to minimize the time to make the route to the camp.
The time can be expressed as the sum of the time walking and the time swimming:
[tex]t=t_w+t_s=\dfrac{d_w}{v_w}+\dfrac{d_s}{v_s}=\dfrac{d_w}{55}+\dfrac{d_s}{33}[/tex]
The distance swimming can be expressed using the Pithagoras theorem, using an auxiliary variable x (see picture attached):
[tex]d_s^2=x^2+w^2=x^2+1\\\\d_s=\sqrt{x^2+1}[/tex]
Using the same auxiliary variable x, the distance walking can be expressed as:
[tex]d_w=L-x=77-x[/tex]
Then, we can minimize t in function of x by deriving t and equal to 0:
[tex]t=\dfrac{d_w}{55}+\dfrac{d_s}{33}=\dfrac{77-x}{55}+\dfrac{\sqrt{x^2+1}}{33}\\\\\\\dfrac{dt}{dx}=-\dfrac{1}{55}+\dfrac{1}{33}\dfrac{x}{\sqrt{x^2+1}}=0\\\\\\\dfrac{x}{\sqrt{x^2+1}}=\dfrac{33}{55}\\\\\\55x=33\sqrt{x^2+1}\\\\55^2x^2=33^2(x^2+1)=33^2x^2+33^2\\\\3025x^2=1089x^2+1089\\\\(3025-1089)x^2=1089\\\\1936x^2=1089\\\\x=\sqrt{\dfrac{1089}{1936}}=\sqrt{0.5625}=0.75[/tex]
The distance swimming is then:
[tex]d_s=\sqrt{x^2+1}=\sqrt{0.5625+1}=\sqrt{1.5625}=1.25[/tex]
The distance walking is:
[tex]d_w=77-0.75=76.25[/tex]
Then, the route that take the least amount of time is the one that she swimms 1.25 miles across the river, to go out of the river on the other shore, 0.25 miles horizontally from the starting point, where she walks 76.25 miles to the campground.
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