A man is on the shore of a circular lake with a radius of 3km at point P and wants to cross the lake directly through the middle to reach point Q. he can canoe across the lake at a speed of 4km/h or he can wak along the shore of the lake at a speef of 6km/h. At what angle should he row in order to minimize the time required to reach point Q?

Answer 1

If I read that correctly, point Q is directly across from point P along a diameter PQ.
let the centre be O and let the point A be such that the time to row along PA and then walk AQ is a minimum.
Let the central angle of the sector QOA be θ, then the central angle of triangle POA is π-θ sector QOAtriangle POA be θ, then in then central angle will be
recall cos(π-θ) = cosπcosθ + sinπsinθ
= -1(cosθ) + 0(sinθ) = -cosθ

arclength AQ = rθ = 3θ
PA^2 = 9+9-2(3)(3)cos(π-θ)
= 18 + 18cosθ
PA = (18+18cosθ)^(1/2)

time = (18+18cosθ)^(1/2)/4 + 3θ/6
= (1/8)(18+18cosθ)^(-1/2)*(-18sinθ dθ/dt) + (1/2)dθ/dt
= (18+18cosθ)^(-1/2)*(-18sinθ dθ/dt) + 4dθ/dt
= (18+18cosθ)^(-1/2)*(-18sinθ ) + 4
18sinθ/√(18+18cosθ) = 4
9sinθ/√(18+18cosθ) = 2
square both sides:
81sin^2 /(18+18cosθ) = 4
72 + 72cosθ = 81(1-cos^2 θ
81cos^2 θ + 72cosθ - 9 = 0
9cos^2 θ + 8cosθ - 1 = 0
(9cosθ - 1)(cosθ + 1) = 0
cosθ = 1/9 or cosθ = -1
θ = appr 83.62 ° or θ = 180° <---- this one would be rowing straight across

now you can find every angle in the diagram.
Let me know how you answered it.

I should really write these out first, better check my algebra.

Answer 2

Hmmm. The central angle QOA is twice the inscribed angle QPA=θ
That makes PA = 2r cosθ = 6cosθ
so the total distance is 6cosθ(water) + 6θ (land)
The total time required is thus
T = 6cosθ/4 + 6θ/6 = 3/2 cosθ + θ
dT/dθ = -3/2 sinθ + 1
dT/dθ = 0 when sinθ = 2/3
θ = 41.8°