Highway parabola

[dono_van24]

Roads are often designed with parabolic surfaces so that the water can flow from the road. Define the road to be 36 feet wide. The center of the road is .5 feet higher than the road at the sides.

1. Set up a coordinate system and sketch the parabolic curve. Define the coordinate system so that the center of the road occurs at x=0. Define three points.



















2. Solve for a,b, and c so that y = ax2 + bx + c

a) Solve for b

b) Find c by letting x = 0.


c) Replace b and c and solve for a


d) Define a quadratic function which describes the highway surface.


3. How far from the center of the road is the surface .2 feet above the horizontal?

 

[dono_van24]

Sorry I forgot to post any work...well I haven't done any yet because I am completely lost on this!

 

[dono_van24]

so I believe that my 3 points are (-18,0),(0,0.5),(18,0), but I cannot figure out how to solve for b? or any of the rest

 

[galactus]

You have a parabola that is concave down, passes through the points (18,0), (-18,0), and has vertex at (0,1/2)

You can use the form of a parabola:

\(\displaystyle y=a(x-h)^{2}+k\)

(h,k) are the coordinates of the vertex.

Sub in the given x,y,h, and k and solve for a.

\(\displaystyle 0=a(18-0)^{2}+\frac{1}{2}\)

\(\displaystyle 0=a\cdot 8^{2}+\frac{1}{2}\)

\(\displaystyle a=\frac{-1}{648}\)

The parabola has equation \(\displaystyle y=\frac{-1}{648}x^{2}+\frac{1}{2}\)

See how to appraoch it now?.