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
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+%…
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+%…