Asked by Shannon
A woman at point A on the shore of a circular lake with radius 2.5mi wants to arrive at the point C diametrically opposite A on the other side of the lake
in the shortest possible time. She can walk at a rate of 3.25 mi/hr and row at a rate of 2 mi/hr.
Find shortest time.
in the shortest possible time. She can walk at a rate of 3.25 mi/hr and row at a rate of 2 mi/hr.
Find shortest time.
Answers
Answered by
Steve
If she walks to subtend an angle of θ, she walks 2.5θ miles
The distance to row across is thus
d^2 = 2(2.5^2)(1+cosθ)
time elapsed is
t = 2.5θ/3.25 + 2.5√2(1+cosθ)/2
= .7692θ + 1.7678√(1+cosθ)
dt/dθ = .7629 - .8839sinθ/√(1+cosθ)
we want dt/dθ=0. That happens when θ=1.31
So, she walks 2.5*1.31 = 3.275 mi
and she rows 3.965 mi
time = 3.275/3.25 + 3.965/2 = 3 hours
The distance to row across is thus
d^2 = 2(2.5^2)(1+cosθ)
time elapsed is
t = 2.5θ/3.25 + 2.5√2(1+cosθ)/2
= .7692θ + 1.7678√(1+cosθ)
dt/dθ = .7629 - .8839sinθ/√(1+cosθ)
we want dt/dθ=0. That happens when θ=1.31
So, she walks 2.5*1.31 = 3.275 mi
and she rows 3.965 mi
time = 3.275/3.25 + 3.965/2 = 3 hours
Answered by
Marta
For anyone looking at the response, I think it should be 1-cosθ, not 1+cosθ.
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