Thread: Parabola word problem

Hey guys thanks in advance. The surface of a roadway over a bridge follows the parabolic curve with vertex at the middle of the bridge. The span of the bridge is 400m. The roadway is 16m higher in the middle than at the endpoints. how far above the end supports is a point 50m from the middle? 150 from middle?

Hey bearjr32.

You need to figure out the equation of the parabola which has the form f(x) ax^2 + bx + c = 0.

Hint: You have f(-200) = f(200) = 0 if x = 0 is the middle of the bridge and you also have f(0) = 16.

Can you use this to solve for a, b, and c and get the height at those arbitrary end-points?

Chiro
To be honest I am 30 and haven't done anything but very basic algebra and trig since I graduated. I'm struggling even getting that far.. Really I am struggling with this whole parabola concept. I get the basic formulas and the focus and all that but after that I am lost. That's what I get for delaying my college education!

No worries, I'll explain it a bit further.

A parabola has the form of y = f(x) = ax^2 + bx + c. It has a minimum or a maximum (depending on if it is concave up or down) and it is symmetric about this value (for symmetry think about that if you drew a line parallel to the y axis at the maximum or minimum, the graph would be a reflection around that line).

From the information you have given there is a maximum at 16m at the center of the bridge.

What I have done is chosen for the middle of the bridge to be at the y-axis (or at x = 0). Since the height is 16m higher than the end points we set f(0) = 16.

We also know that the span is 400m and because of the symmetry, we know that f(-x) = f(x) and since the span is 200m, we know that f(-200) = f(200) = 0.

If you graphed it, it would look like a half football with the highest point at x = 0 and the end points at x = -200 and x = 200.

Since the function of a parabola is always f(x) = ax^2 + bx + c and since we have three pieces of independent information for f(x), we can solve for the parameters a, b, and c which we have to work out.

So lets look at the pieces of information available:

f(0) = 16 means a*(0)^2 + b*(0) + c = 16 which means c = 16 (Substituting x = 0 into f(x)))
f(-200) = 0 means a*(-200)^2 + b*(-200) + c = 40000a - 200b + 16 = 0
f(+200) = 0 means a*(200)^2 + b*(200) + c = 40000a + 200b + 16 = 0

So we have c but we need to figure out a. Remember that we have two equations:

40000a - 200b + 16 = 0 (Equation 1)
40000a + 200b + 16 = 0 (Equation 2)

Adding both equations we get Equation 3:

80000a + 32 = 0 which means
a = -32/80000.

To get b we use Equation 4 = Equation 2 - Equation 1 which leaves us with:

400b = 0 which means b = 0

You could have substituted a into the equation to get b as well.

So now we have the equation for our bridge that goes from x = -200 to x = 200 with the center at x = 0 which is

f(x) = -32/80000*x^2 + 16

Now if you want the height of the bridge relative to that of the end-points, you simply substitute in a value of x (remembering that the start of the bridge starts at x = -200) and get f(x) which is the height.

It's amazing when someone shows it to me how simple it seems. Do you mind checking me on this next one? The shape of a wire hanging between two poles closely approximates a parabola. Find the equation of a wire that is suspended between two poles 40m apart and whose lowest point is 10m below the level of the insulator. I came up with the formula

F(x)=1/40x^2

Show us your working if you don't mind.

I did the three equations again
F(x)=ax^2+bx+c

using the lowest point as origin it gives the first coordinate (0,0) then I know that (-20,10) and (20,10) are the endpoints
F(0)=a(0)^2+b(0)+c=0 C=0
F(-20)=400a-40b+0=10
F(20)=400a+40b+0=10

adding the bottom two together gives me 800a=20 or a=1/40
substituting this into an equation gives me 10+40b=10 or b=0

knowing b and c are equal to zero I am left with F(x)=1/40(x^2)

That looks pretty good.