Road... Quadratic Formula

[MathBane]

When a road is being built, it usually has straight sections, all with the same grade, that must be linked to each other by curves. (By this we mean curves up and down rather than side to side, which would be another matter.) It's important that as the road changes from one grade to another, the rate of change of grade between the two be constant. The curve linking one grade to another grade is called a vertical curve.
Surveyors mark distances by means of stations that are 100 feet apart. To link a straight grade of g1 to a straight grade of g2, the elevations of the stations are given by the following equation.

\(\displaystyle \frac {g_2 - g_1}{2L}x^2+g_1x+E- \frac {g_1L}{2}\)

Here y is the elevation of the vertical curve in feet, g1 and g2 are percents, L is the length of the vertical curve in hundreds of feet, x is the number of the station, and E is the elevation in feet of the intersection where the two grades would meet. (See the figure shown below.) The station x = 0 is the very beginning of the vertical curve, so the station x = 0 lies where the straight section with grade g1 meets the vertical curve. The last station of the vertical curve is x = L, which lies where the vertical curve meets the straight section with grade g2.

Assume that the vertical curve you want to design goes over a slight rise, joining a straight section of grade 1.32% to a straight section of grade –1.74%. Assume that the length of the curve is to be 500 feet (so L = 5) and that the elevation of the intersection is 1010.62 feet.

(There are just so many words... I can't begin to make sense of it. All I know is L = 5 and E = 1010.62)

1. What is the equation for the vertical curve described above? Don't round the coefficients.


2. What are the elevations of the stations for the vertical curve? (Round your answers to two decimal places.)

Station Number | Elevation
0 | ___ft
1 | ___ft
2 | ___ft
3 | ___ft
4 | ___ft
5 | ___ft

 

[pitoten]

You are also given the grades as percents. Those percents are g1 and g2. Now that you have all the constants, you can build your
equation y = f(x), since you are given f(x) in the problem...then use the equation to solve the questions in (2).

This is a poorly worded problem, because a true equation must have some kind of = sign, greater/less than sign, proportional to etc...They mean to set the given equation = to y; that is, you should start off with y = the given equation with the correct numbers subbed in for L, g1, g2 etc...

 

[MathBane]

So... the formula should be:

\(\displaystyle y = \frac {g_2 - g_1}{2L}x^2+g_1x+E- \frac {g_1L}{2}\)

L = 5
E = 1010.62
g1 = 1.32 ?
g2 = -1.74 ?

\(\displaystyle y = \frac {-1.74 - 1.32}{2(5)}x^2+1.32x+1010.62- \frac {1.32(5)}{2}\)

Simplified, it yields:

\(\displaystyle y = -7.65x^2+1.32x+1007.32\)

Does it matter what is g1 or g2? I tried both ways but I got it wrong. Am I missing some steps?

 

[Deleted member 4993]

So... the formula should be:

\(\displaystyle y = \frac {g_2 - g_1}{2L}x^2+g_1x+E- \frac {g_1L}{2}\)

L = 5
E = 1010.62
g1 = 1.32 ? g1 = 1.32% = 0.0132
g2 = -1.74 ? g2 = 1.74% = 0.0174

\(\displaystyle y = \frac {-1.74 - 1.32}{2(5)}x^2+1.32x+1010.62- \frac {1.32(5)}{2}\)

Simplified, it yields:

\(\displaystyle y = -7.65x^2+1.32x+1007.32\)

Does it matter what is g1 or g2? I tried both ways but I got it wrong. Am I missing some steps?

 

[MathBane]

Ugh, I tried it every which way... but for some reason I can't get it to work. I switched g1 and g2 values... but it was still wrong. I must be making the same mistake every time.

 

[Denis]

\(\displaystyle y = \frac {-1.74 - 1.32}{2(5)}x^2+1.32x+1010.62- \frac {1.32(5)}{2}\)
Simplified, it yields:
\(\displaystyle y = -7.65x^2+1.32x+1007.32\)

Your -7.65 should be -.306 ; how did you ever manage to get -7.65? :shock:

 

[MathBane]

Your -7.65 should be -.306 ; how did you ever manage to get -7.65? :shock:

I got that by doing -1.74-1.32 first... then dividing the answer by 2(5), but that was before it was pointed out to me that I forgot to change the percents into decimals.

 

[Deleted member 4993]

Your -7.65 should be -.306 ; how did you ever manage to get -7.65? :shock:

I got that by doing -1.74-1.32 (= -3.06) first... then dividing the answer by 2(5) [ = -3.06/10 = -0.306 not -7.65]but that was before it was pointed out to me that I forgot to change the percents into decimals.

 

[MathBane]

Blast. All this time I was doing /2(5). That must mean to the calculator to divide by two then multiply by five. In fact, it does. Ugh. At least I learned something to not do on the final.