A man launches his boat from point A on a bank of a straight river, 1 km wide, and wants to reach point B, 1 km downstream on the opposite bank, as quickly as possible (see the figure below). He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.)
you dont have to necessarily give me the answer, you can just explain how to set it all up.
thanks!!
you dont have to necessarily give me the answer, you can just explain how to set it all up.
thanks!!
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From D to B on the opposite bank is (1 - x) km.
From A to D on the opposite bank is √(1 + x^2) km.
The time required to go from A to D, rowing, is √(1 + x^2)/6, and the time required to go from D to B, running, is (1 - x)/8.
The total time T(x), then, is √(1 + x^2)/6 + (1 - x)/8.
At this point, I would differentiate T(x) and set it = 0.
Or, using technology, I would graph T(x) and find the minimum value.
I find that x ≈ 1.134 km minimizes the time, which is 0.235 hr.
From A to D on the opposite bank is √(1 + x^2) km.
The time required to go from A to D, rowing, is √(1 + x^2)/6, and the time required to go from D to B, running, is (1 - x)/8.
The total time T(x), then, is √(1 + x^2)/6 + (1 - x)/8.
At this point, I would differentiate T(x) and set it = 0.
Or, using technology, I would graph T(x) and find the minimum value.
I find that x ≈ 1.134 km minimizes the time, which is 0.235 hr.
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A boat
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C ...D ...B
we shall solve in general form & then substitute numerical values
let AC = w
CD = x
CB = y
time ratio/km water:land = k (4/3)
angle CAD = z
dist. over water = w•sec z,
dist. over land = y - w•tan z
time T ∞ k•sec z +(y - tan z)
dT/dz = k•secz•tan z - sec^2 z
= (ksin z - 1) / cos^2 z
for minima, sin z = 1/k
which yields x = w /√(k^2 - 1) = 1/sqrt((4/3)^2-1) = 1.134 km > y
this means he should row directly to B
----------------- ------------------ ---------------
|\
|-\
|--\
|----\
|-----\------
C ...D ...B
we shall solve in general form & then substitute numerical values
let AC = w
CD = x
CB = y
time ratio/km water:land = k (4/3)
angle CAD = z
dist. over water = w•sec z,
dist. over land = y - w•tan z
time T ∞ k•sec z +(y - tan z)
dT/dz = k•secz•tan z - sec^2 z
= (ksin z - 1) / cos^2 z
for minima, sin z = 1/k
which yields x = w /√(k^2 - 1) = 1/sqrt((4/3)^2-1) = 1.134 km > y
this means he should row directly to B
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