A woman at a point A on the shore of a circular lake with radius r = 4km 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 the rate of 6 km/h and row a boat at 3 km/h. What is the shortest amount of time it would take her to reach point C?
I'm really confused with how i should approach this, i think it may have something to do with chords maybe. the last few questions have been on differentiation and implicit differentiation if that helps at all. thanks,
I'm really confused with how i should approach this, i think it may have something to do with chords maybe. the last few questions have been on differentiation and implicit differentiation if that helps at all. thanks,
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The distance she's going to walk is 2 * pi * 4 * (t / (2pi)) = 4 * t (we're dealing in radians here).
Distance = Speed * Time
D/S = T
4 * t / 6 = T[1]
(2t/3) = T[1]
Now, we can use the law of cosines to figure out the length of the chord
L^2 = 4^2 + 4^2 - 2 * 4 * 4 * cos(pi - t)
Can you see why it's cos(pi - t)? If you draw it out and label "t" as the angle between the starting point of the walk, the center, and the ending point of the walk, then you'd see that the total angle is pi radians
L^2 = 16 + 16 - 32 * (cos(pi)cos(t) + sin(pi)sin(t))
L^2 = 32 - 32 * (-1 * cos(t) + 0 * sin(t))
L^2 = 32 + 32 * cos(t)
L^2 = 16 * (2 + 2cos(t))
L = 4 * sqrt(2 * (1 + cos(t)))
L/3 = T[2]
T = T[1] + T[2]
T = (2/3) * t + 4 * sqrt(2 * (1 + cos(t))) / 3
T = (2/3) * (t + 2 * sqrt(2 + 2cos(t)))
Find dT/dt and set it to 0
dT/dt = (2/3) * (1 + 2 * (-2sin(t)) * (1/2) / sqrt(2 + 2cos(t))))
dT/dt = 0
0 = (2/3) * (1 - 2sin(t) / sqrt(2 + 2cos(t)))
0 = 1 - 2sin(t) / sqrt(2 + 2cos(t))
2 * sin(t) / sqrt(2 + 2cos(t)) = 1
2 * sin(t) = sqrt(2 + 2cos(t))
4sin(t)^2 = 2 + 2cos(t)
2sin(t)^2 = 1 + cos(t)
2 * (1 - cos(t)^2) = 1 + cos(t)
2 * (1 - cos(t)) * (1 + cos(t)) - (1 + cos(t)) = 0
(1 + cos(t)) * (2 * (1 - cos(t)) - 1) = 0
Distance = Speed * Time
D/S = T
4 * t / 6 = T[1]
(2t/3) = T[1]
Now, we can use the law of cosines to figure out the length of the chord
L^2 = 4^2 + 4^2 - 2 * 4 * 4 * cos(pi - t)
Can you see why it's cos(pi - t)? If you draw it out and label "t" as the angle between the starting point of the walk, the center, and the ending point of the walk, then you'd see that the total angle is pi radians
L^2 = 16 + 16 - 32 * (cos(pi)cos(t) + sin(pi)sin(t))
L^2 = 32 - 32 * (-1 * cos(t) + 0 * sin(t))
L^2 = 32 + 32 * cos(t)
L^2 = 16 * (2 + 2cos(t))
L = 4 * sqrt(2 * (1 + cos(t)))
L/3 = T[2]
T = T[1] + T[2]
T = (2/3) * t + 4 * sqrt(2 * (1 + cos(t))) / 3
T = (2/3) * (t + 2 * sqrt(2 + 2cos(t)))
Find dT/dt and set it to 0
dT/dt = (2/3) * (1 + 2 * (-2sin(t)) * (1/2) / sqrt(2 + 2cos(t))))
dT/dt = 0
0 = (2/3) * (1 - 2sin(t) / sqrt(2 + 2cos(t)))
0 = 1 - 2sin(t) / sqrt(2 + 2cos(t))
2 * sin(t) / sqrt(2 + 2cos(t)) = 1
2 * sin(t) = sqrt(2 + 2cos(t))
4sin(t)^2 = 2 + 2cos(t)
2sin(t)^2 = 1 + cos(t)
2 * (1 - cos(t)^2) = 1 + cos(t)
2 * (1 - cos(t)) * (1 + cos(t)) - (1 + cos(t)) = 0
(1 + cos(t)) * (2 * (1 - cos(t)) - 1) = 0
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