The question:

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 avertical curve.

Surveyors mark distances by means ofstationsthat are 100 feet apart. To link a straight grade of $\displaystyle g_1$ to a straight grade of $\displaystyle g_2$, the elevations of the stations are given by the following equation.

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

Here $\displaystyle y$ is the elevation of the vertical curve in feet, $\displaystyle g_1$ and $\displaystyle g_2$ are percents,$\displaystyle L$ is the length of the vertical curve in hundreds of feet,$\displaystyle x$ is the number of the station, and $\displaystyle E$ is the elevation in feet of the intersection where the two grades would meet. (See the figure shown below.) The station $\displaystyle x = 0$ is the very beginning of the vertical curve, so the station $\displaystyle x = 0$ lies where the straight section with grade $\displaystyle g_1$ meets the vertical curve. The last station of the vertical curve is $\displaystyle x = L$, which lies where the vertical curve meets the straight section with grade $\displaystyle g_2$.

Assume that the vertical curve you want to design goes over a slight rise, joining a straight section of grade 1.34% to a straight section of grade –1.77%. Assume that the length of the curve is to be 500 feet (soL= 5) and that the elevation of the intersection is 1070.64 feet.

(a) What phrase in the first paragraph of this exercise assures you that a quadratic model is appropriate?

(I think it's: The last station of the vertical curve isx=L, which lies where the vertical curve meets the straight section with gradeg2.)

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

(For the formula, I'm getting:

$\displaystyle y= \frac {-0.0177-0.0134}{2(5)}x^2+0.0134x+1070.64- \frac {0.0134(5)}{2}$

after cleaning it up I got:

$\displaystyle y = -0.07775x^2+0.0134x+535.2865$

I'm sure this is not right.)

(c) 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

(d) Where is the highest point of the road on the vertical curve? (Give the distance along the vertical curve and the elevation. Round your answers to two decimal places.)

We find the maximum at [____] feet along the vertical curve with an elevation of= [____] feet.E

(I assume that answers to (c) and (d) should come easy after the right formula is found, unless I need to take some extra steps...)