Need help, Math C homework. please.?
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Need help, Math C homework. please.?

[From: ] [author: ] [Date: 14-04-10] [Hit: ]
3.0075327775562348888060714294033 It looks like walking around will take the shortest amount of time.Makes sense, too.The length of the perimeter is 1.57 times the length of the longest distance between the 2 sides and she walks 2 times as fast as she rows.......

1 + cos(t) = 0
cos(t) = -1
t = pi

2 * (1 - cos(t)) - 1 = 0
2 * (1 - cos(t)) = 1
1 - cos(t) = 1/2
1/2 = cos(t)
t = pi/3

This gives us 2 possible angles to check

T = (2/3) * (t + 2 * sqrt(2 + 2cos(t)))
T = (2/3) * (pi + 2 * sqrt(2 + 2 * cos(pi))) , (2/3) * (pi/3 + 2 * sqrt(2 + 2 * cos(pi/3)))
T = (2/3) * (pi + 2 * sqrt(2 - 2)) , (2/3) * (pi/3 + 2 * sqrt(2 + 1))
T = (2/3) * (pi + 2 * 0) , (2/3) * (1/3) * (pi + 6 * sqrt(3))
T = (2pi/3) , (2/9) * (pi + 6 * sqrt(3))
T = 2.0943951023931954923084289221863 , 3.0075327775562348888060714294033

It looks like walking around will take the shortest amount of time. Makes sense, too. The length of the perimeter is 1.57 times the length of the longest distance between the 2 sides and she walks 2 times as fast as she rows. It'd be more advantageous for her to just walk it, even though it's a longer combined distance.

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let the woman row part way at an angle Θ to the *diameter*,
and walk along the arc for the rest of the way.

since the triangle on the diameter is right angled,
angle subtended at the centre by the arc walked will be 2Θ

distance rowed is 2rcosΘ = 8cosΘ
and length of the arc walked = r•2Θ = 8Θ, Θ in radians,
so T(Θ) = 8cosΘ/3 + 8Θ/6
= (8/3)cosΘ + (4/3)Θ

T'(Θ) = -(8/3)sinΘ + (4/3)
but T"(Θ) = -(8/3)cosΘ, so setting T'(Θ) to 0 will give a MAXIMUM
thus minimum time has to be at one of the 2 extrema (Θ=0 or pi/2)
T(0) = 8/3 = 2.667
T(π/2) = (2/3)π = 2.094 hrs
so least time = 2.094 hrs, walking the whole way
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