Word problem please help...Thank you
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Word problem please help...Thank you

[From: ] [author: ] [Date: 12-11-06] [Hit: ]
how wide is the surface of the water in the ditch?~ 14,so when the canal is 10√2 ft (approximately 14.http://www.wolframalpha.com/input/?......
A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch?

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y = ax^2
10=100a
a = 1/10
Equation is y = x^2/10
5 = x^2/10
x = sqrt(50)
Width = 2sqrt(50) = 10sqrt(2)
~ 14,14 ft

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y = a(x/2)^2

with y = depth and x = width

we know 10 = a(20/2)^2
=> 10 = a100
=> a = 1/10

so y = [(x/2)^2]/10

so 5 = [(x/2)^2]/10
=> 50 = (x/2)^2
=> 5√2 = x/2
=> 10√2 = x

so when the canal is 10√2 ft (approximately 14.142 ft (3dp) ) wide when it is 5 feet deep

if you want to see a nice graph of the canal's parabola let x/2 = u => y = u^2/10
http://www.wolframalpha.com/input/?i=5+%…
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