# calculus

You are standing at the edge of a slow moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground which is one mile from the point directly across the river where you start your swim. What route will take the least amount of time?

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1. so I know you will use
T=dr/rr +dL/rL

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2. I cannot tell what your variables represent
My sketch has the distance from the point directly across his starting point to the end of his swim as x
and the distance to be walked is then 1 -x
let the length of his swim by y
in the right-angled triangle:
x^2 + 1 = y^2
y = (1+x^2)^(1/2)

time for the swim = (1/2)(x^2 + 1)^(1/2)
time for the walk = (1/3)(1-x)

Time = (1/2)(x^2 + 1)^(1/2) + (1/3)(1-x)
d(time)/dx = (1/4)(x^2 + 1)^(-1/2) (2x) - 1/3 = 0 for min time

x/( 2√(x^2 + 1) ) = 1/3
square both sides
x^2/(4x^2 + 4) = 1/9
9x^2 = 4x^2 + 4
5x^2 = 4
x = 2/√5 = .894

He should aim for a spot .894 miles down from directly across, or .106 miles from the campground

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