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Question
A woman at a point A on the shore of a circular lake with a radius of 3km 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 6km/h and row a boat at 3km/h. How should she proceed?
Diagram 1: http://img7.imageshack.us/img7/3416/diagram1m.jpg
Hint: Suppose she sets out at an angle theta from the diameter line joining A and C. You need to determine the angle theta which will minimise the time taken. In the diagram below some additional angles are provided (make sure you can see how those angles were determined), and the rowing distance is denoted D, while the walking distance is denoted L. Remember that speed is equal to the distance traveled divided by the time taken.
Diagram 2: http://img155.imageshack.us/img155/2099/diagram2.jpg
Lets say she rows the boat first and the angle between the direction she rows and the direction she goes on shore is Θ. This makes that angle in the center of the circle and the place she's going to 2Θ. The distance on the shore, when the angle is in radians is r(2Θ), or 2rΘ.
The distance she went in the water is the hypoteneuse of a isosceles triangle given by
dē = 2r(r-cos(2Θ)). This makes d = √(2r[r-cos(2Θ)]).
The total time is then t(Θ) = √(2r[r-cos(2Θ)]) / 3 + r(2Θ) / 6,
which is t(Θ) = (√(2r[r-cos(2Θ)]) + rΘ)/3.
To find the answer, we need t'(Θ).
That is seen to be t'(Θ) = (1/2){1/√(2r[r-cos(2Θ)])}*(2sinΘ) + r]/3.
Setting it to 0 gives (1/2){1/√(2r[r-cos(2Θ)])}*(2sinΘ) + r]/3 = 0.
Multiplying by 6 gives {1/√(2r[r-cos(2Θ)])}*(2sinΘ) + 2r = 0.
Subtracting 2r from both sides gives {1/√(2r[r-cos(2Θ)])}*(2sinΘ) = -2r.
Multiplying both sides by {1/√(2r[r-cos(2Θ)])} gives 2sinΘ = -2r*√(2r[r-cos(2Θ)]).
Dividing by 2 and then squaring gives sinēΘ = -rē*(2r[r-cos(2Θ)]).
Multiplying out gives sinēΘ = -2(r^4) + 2(r^3)cos(2Θ).
The cos(2Θ) could be converted to 1 - 2sinēΘ, giving sinēΘ = -2(r^4) + 2(r^3)(1 - 2sinēΘ).
This could be multipled out, the terms with sinēΘ could be moved to one side,
sinēΘ could be facotred out, and then the coefficient in front of it divided by.
Taking the squareroot of both sides, you will get an expression for sinΘ.
Taking the inverse sin() of both sides gives Θ.
By the way, I can't find either file.
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Scott A Wilson
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