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Question 1042427: A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal. If the parkway is centered, how high must the vertex of the arch be in order to give a minimum clearance of 5 meters over the parkway. (Solve using the equation of parabola)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The arc(parabola) curves downward and without loss of generality we can construct the parabola with x intercepts at (0,0) and (30,0).
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The x value for the vertex of the parabola is 15 and we are given two more points on the parabola, namely, (5,5) and (25,5) which are the horizontal ends of the parkway.
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The standard form of a parabola is
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y = ax^2 +bx +c where a, b, c are real numbers
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our c is 0 since one of our parabola's x intercepts is the origin (0,0), therefore
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y = -ax^2 +bx
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note that a < 0 because our parabola curves downward
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we have the following two equations in a and b
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1) -a(5)^2 +5b = 5
2) -a(25)^2 +25b = 5
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Let's work with equation 1)
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-25a +5b = 5
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divide both sides of = by 5
-5a + b = 1
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b = 5a +1
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now substitute for b in equation 2)
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-625a +25(5a+1) = 5
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-625a +125a +25 = 5
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-500a = -20
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a = 1/25
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substitute for a in 1)
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b = 6/5
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the equation of our parabola with one x intercept at the origin is
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y = -(x^2/25) +(6x/5)
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now if x = 15 then
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y = -(225/25) + (90/5) = -9 + 18 = 9
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The height of the vertex of the arc must be 9 meters
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