if anyone could help me with this, i would really appreciate it!
In order to build a highway, it is necessary to fill in a section of a valley where the grades (slopes) of the sides are 9% and 6%. The top of the filled region will have the shape of a parabolic arc that is tangent between the two slopes at the points A and B. The horizontal distance between the points A and B are 1,000 feet.
a) Find a quadratic function y = ax^2 + bx + c, -500 < x < 500, that describes the top of the filled region. (less than or equal to <)
please!
In order to build a highway, it is necessary to fill in a section of a valley where the grades (slopes) of the sides are 9% and 6%. The top of the filled region will have the shape of a parabolic arc that is tangent between the two slopes at the points A and B. The horizontal distance between the points A and B are 1,000 feet.
a) Find a quadratic function y = ax^2 + bx + c, -500 < x < 500, that describes the top of the filled region. (less than or equal to <)
please!
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You haven't specified where the coordinate system is placed, so I assume that the origin is at the bottom of the valley and the x-axis is horizontal.
Since the slopes of the left and right sides are 9% and 6%, the end points of the parabolic arc are:
Left: (-500, 9% of 500) = (-500, 45)
Right: (500, 6% of 500) = (500, 30)
These two points are on the parabola, so substituting into its equation y = ax^2 + bx + c, we get:
45 = 25000a - 500b + c
30 = 25000a + 500b + c
Subtracting, we get 15 = -1000b, so b = -15/1000
Now, the slope of the tangent at the right end point is 2ax + b, with x = 500. Since this slope must be 6% or 6/100, we have:
2a (500) + b = 6/100.
Substituting b = -15/1000 and solving for a yields: a = 75/1,000,000
Using either of the first two equations, we now derive c = 35625/1000
Thus, the solution is a = 75/1,000,000, b = 15/1,000, c = 35,625/1,000
P.S: This system is overspecified. We never used the fact that the slope of the tangent on the left end point is -9%. Although this happens to be true here, if it were changed, say to 10%, the solution may no longer exist. Basically, two points on a parabola and the slope of the tangent at one of these suffices to reconstruct the parabola uniquely. Here, we are also given the slope of the tangent at the second point.
Since the slopes of the left and right sides are 9% and 6%, the end points of the parabolic arc are:
Left: (-500, 9% of 500) = (-500, 45)
Right: (500, 6% of 500) = (500, 30)
These two points are on the parabola, so substituting into its equation y = ax^2 + bx + c, we get:
45 = 25000a - 500b + c
30 = 25000a + 500b + c
Subtracting, we get 15 = -1000b, so b = -15/1000
Now, the slope of the tangent at the right end point is 2ax + b, with x = 500. Since this slope must be 6% or 6/100, we have:
2a (500) + b = 6/100.
Substituting b = -15/1000 and solving for a yields: a = 75/1,000,000
Using either of the first two equations, we now derive c = 35625/1000
Thus, the solution is a = 75/1,000,000, b = 15/1,000, c = 35,625/1,000
P.S: This system is overspecified. We never used the fact that the slope of the tangent on the left end point is -9%. Although this happens to be true here, if it were changed, say to 10%, the solution may no longer exist. Basically, two points on a parabola and the slope of the tangent at one of these suffices to reconstruct the parabola uniquely. Here, we are also given the slope of the tangent at the second point.