long derivative word problem

[KNS]

A bridge must span a linear distance of 500 feet from points we will call C and D. The difference in elevation between C and D is 20 feet with D being the higher point. The bridge must connect smoothly with the existing roadways at points C and D. At the C side of the swamp the roadway has a 7% grade (slope of the roadway is 0.07) and at the D side the grade is 4% (slope of –0.04). C is at swamp level. The coordinate system should have the origin at C.

This is what I have done.

f(x)=ax^3+bx^2+cx+d d=0

f'(x)=3ax^2+2bx+c and f'(0)=3a(0)^2+2b(0)+c=0.07 so c=0.07

f'(x)=3ax^2+2bx+0.07

Now I have solved the equation where the variable equals something in terms of b.

f'(500)=3a(500)^2+2b(500)+0.07=-0.04 a=(-0.11-1,000b)/750,000

The problem I have is when I solve this equation for b it just cancels itself out. But the problem has to equal a number that when put back into the original equation it equals 20. Please help, I am so lost and I am most definatly going spare.

 

[stapel]

From what you've done, I will guess that some information is missing from the problem statement; nothing in the exercise says anything about the roadway being modelled with a cubic equation...? But assuming that this is what you're supposed to do, you appear to have made a good start:

You have some cubic for the bridge, which must "mesh" smoothly with the roadway on either end. This means that it must, at C and D, have the required slopes. You have:

. . . . .f(x) = ax³ + bx² + cx + d

...with f(C) = f(0) = 0 and f(D) = f(500) = 20. Then:

. . . . .f(0) = d = 0

...so d = 0:

. . . . .f(x) = ax³ + bx² + cx

...and:

. . . . .125 000 000 a + 250 000 b + 500 c = 20 *

Turning now to the ends, you have:

. . . . .f'(x) = 3ax² + 2bx + c

...with f'(C) = f'(0) = 0.07 and f'(D) = f'(500) = -0.4. Then:

. . . . .f'(0) = c = 0.07

. . . . .f'(500) = 750 000 a + 1 000 b + 0.07 = -0.04 **

Plugging the now-known value for "c" into *, we get, with **, the new system:

. . . . .125 000 000 a + 250 000 b + 35 = 20
. . . . .750 000 a + 1 000 b + 0.07 = -0.04 **

Rearranging a bit, we get:

. . . . .125 000 000 a + 250 000 b = -15
. . . . .75 000 000 a + 100 000 b = -4

Multiplying the second equation by -2.5, we get:

. . . . .125 000 000 a + 250 000 b = -15
. . . . .-187 500 000 a - 250 000 b = 10

Solve for "a"; back-solve for "b".

Hope that helps at bit!

Eliz.

 

[galactus]

You just have some algebra troubles somewhere.

How does the high end have a negative slope?. Shouldn't that be +4% going out?.

\(\displaystyle \L\\0=a(0)^{3}+b(0)^{2}+c(0)+d\), d=0

\(\displaystyle \L\\20=a(500)^{3}+b(500)^{2}+c(500)\)

\(\displaystyle \L\\y'=3a(500)^{2}+2b(500)+c=\frac{1}{25}\)

\(\displaystyle \L\\y'=3a(0)^{2}+2b(0)+c=\frac{7}{100}\), c=7/100

Solve the system and you get:

\(\displaystyle \L\\y=\frac{3}{25000000}x^{3}-\frac{3}{25000}x^{2}+\frac{7}{100}x\)

\(\displaystyle \L\\y'=\frac{9}{25000000}x^{2}-\frac{3}{12500}x+\frac{7}{100}\)

 

[KNS]

Thank you

galactus,
The derivitive at point D (500,20) is -0.04 because it is a bridge and if graphed it would be concave down.

galactus and stapel,
Thank you for your help. I have figured it out, and I feel a little silly that I missed it. Thank you again.

kristine

 

[galactus]

In my experience, cubics are not used in real-world design of vertical curves in highway and bridge construction; They are based on a parabola. That certainly doesn't mean they're not implemented in some circles, though.

We have a starting point known as the PVC (Point of Curvature), PVI (Point of Intersection), PVT (Point of Tangent).

The elevation of the PVC is known and from it we can calculate the elevation of the PVI by extending the tangent line grade.

Knowing the elevations of the endpoints, the PVC and PVT, we find the midpoint by
\(\displaystyle \L\\B=\frac{PVT+PVC}{2}\)

We subtract this from the PVI elevation and divide by 2:

\(\displaystyle \L\\\frac{B-PVI}{2}\)

This gives us \(\displaystyle \L\\y_{m}\) which is equal to a in \(\displaystyle \L\\y=ax^{2}\)

Now, we use the equation of the parabola to find the amounts to subtract or add to the tangent line elevations to find the elevations of the points on the curve.

Using the diagram , here’s an example to illustrate:

http://img170.imageshack.us/img170/933/ ... 007cr6.png


The curve length L=500, G1= grade in is 0.5% and G2=grade out is +1.86%

We use \(\displaystyle y=ax^{2}\)

x=the increment divided by the length of half the curve. L/2=250
Say, we go 50 feet, then (50/250)=1/5 and we have \(\displaystyle (\frac{1}{5})^{2}\cdot{1.475}=0.06\). Since this is an under vertical or concave up curve, we add this to the tangent line elevation at that point.

Station     Tangent elev.     Curve elev.    Calculation for yn  

15+00      594.41                594.41         -------------------
15+50      594.16                594.22         (1/5)^2*1.475=0.06
16+00      593.91                594.15         (2/5)^2*1.475=0.24
16+50      593.66                594.19         (3/5)^2*1.475=0.53
17+00      593.41                594.35         (4/5)^2*1.475=0.94
17+50(PVI) 593.16              594.64       (5/5)^2*1.475=1.475
18+00       594.09                595.03         same as 17+00
18+50       595.02               595.55         same as 16+50
19+00       595.95                596.19         same as 16+00
19+50       596.88                596.94         same as 15+50
20+00(PVT)597.81               597.81         ---------------------

The low or high point of the curve can be obtained by

\(\displaystyle \L\\\frac{L\cdot{G1}}{G1-G2}\)

In this case it is x=105.93. Which is 15+00+105.93=16+05.93
The elevation is then \(\displaystyle (\frac{105.93}{250})^{2}\cdot{1.475}=0.26\)

594.41-.005*105.93=593.88; 593.88+0.26=593.14

We could also find the equation of the parabola and go that route:

\(\displaystyle \L\\y=\frac{59}{2500000}x^{2}-\frac{1}{200}x+594.41\)

This is actually easier than the method used above, but what I outlined above is how it is normally done.

 

[KNS]

Thank you for your help, my problem specifically askes for a cubic equation but I did find your solution very intersting. Do you work in construction or engineering??

 

[galactus]

Yes, I worked as a survey party chief for about 15 years. I laid out many horizontal and vertical curves for road construction.