How Does the Quadratic Function Model Smooth Transitions in Highway Design?

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The quadratic function y = ax^2 + bx + c is essential in highway design as it creates smooth transitions between elevation changes. In this case, the initial gradient of 3% is represented by y = ax^2 + 0.3x + c, with specific elevation points provided. By substituting the known values of elevation at 200 feet and 1000 feet into the equation, two linear equations can be formed to solve for the coefficients a and c. The discussion highlights the simplicity of solving these equations, emphasizing the ease of applying quadratic functions in practical engineering scenarios. Overall, understanding this application aids in effective highway design.
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Homework Statement


In highway design, for civil engineers the quadratic function y= ax^2+bx +c is called the transition curve, because its properties provide a smooth transition between peaks and valleys. A road with an initial gradient of 3% can be represented by the formula y=ax^2+0.3x+c , where y is the elevation and x is the distance along the curve. Suppose the elevation of the road is 1105 feet at points 200 feet and 1000 feet along the curve. Find the equation of the transition curve.


Is this a theory of quadractic equation problem? Where I should use the roots and find the new equation?
 
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lionely said:

Homework Statement


In highway design, for civil engineers the quadratic function y= ax^2+bx +c is called the transition curve, because its properties provide a smooth transition between peaks and valleys. A road with an initial gradient of 3% can be represented by the formula y=ax^2+0.3x+c , where y is the elevation and x is the distance along the curve. Suppose the elevation of the road is 1105 feet at points 200 feet and 1000 feet along the curve. Find the equation of the transition curve.

Is this a theory of quadratic equation problem? Where I should use the roots and find the new equation?

Simply plug-in x=200 and y=1105 into y=ax^2+0.3x+c to get one equation.

Plug-in x=1000 and y=1105 into y=ax^2+0.3x+c to get another equation.

You now have two equations in two unknowns. solve them for a and c
 
Sigh really? =/
 
lionely said:
Sigh really? =/

Why "sigh"? You have two simple linear equations for a and c, and solving them is easy.
 
I said sigh because I didn't realize it was that easy. :(
 
Occam's razor!
 
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