To access a new pumping station, Oil Company Z needs to construct a roadway acro

Question

To access a new pumping station, Oil Company Z needs to construct a roadway across a swamp. Their construction crew says it would be easiest to span the distance with a bridge in the shape of a simple arc. The 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. A diagram appears below. The coordinate system should have the origin at C.

1. Use a cubic equation to model the bridge span and find a formula which is smooth at both C and D.

2. We need information about the height of the span above the swamp. We
have a 30 foot wide barge carrying a crane that has a height above the water of 21 feet. Will the
crane fit under the bridge (allow 0.5 foot of vertical space between the top of the crane and the
bottom of the bridge for clearance), and if so, where do we need to dredge a channel to allow it
to pass under? Assume that the top of the crane can be located anywhere above the deck of the
barge. Base your answer on the cubic model you found for the bridge.

* I have been attempting this problem since it was given to us and every equation that I come up with does not work, so I need to know what I am doing so wrong! We were told that we would have a bonus question similar to this on our final so I need to know how to do this! Please show steps and explanations to this problem so I know what I am doing.

Explanation / Answer

As the slope is .07 at 0, and f(0) = 0, the equation has constant = 0 and linear term .07x

Then, y = .07x + ax^2 + bx^3

As f(500) = 20, .20 = .07*500 + a*500^2 + b*500^3

The slope is -.04 at 500, so -.04 = f'(500) = .07 + 2ax + 3bx^2 = .07+2*a * 500 + 3*b*500^2

20 = 35 + 250000 a + 125000000b

-.04 = .07 + 1000 a + 750000 b

Then, 250000 a + 125000000b = -15

1000 a + 750000 b = -.11

We then solve for a and b

Multiply the bottom equation by 250

250000 a + 187500000 b = -27.5

If we subtract this from the top equation, we get

-62500000 b = -12.5

b = -0.0000002

1000 a + 750000 b = -.11

a = (-.11 - 750000 b)/1000 = (-.11 +  0.0000002 *750000))/1000 = 0.00004

Then, the equation of the curve is y = .07x + 0.00004 x^2 -0.0000002 x^3

We can find the maximum height by finding where f'(x) = 0 and determining f(x) there

f'(x) = .07 + 2ax + 3bx^2 = .07 + 2* 0.00004 x + 3 * -0.0000002 x^2, or

.07 + 0.00008 x -0.0000006 x^2

If we multiply by -1, we get

0.0000006 x^2 - 0.00008 x - .07 = 0

From the form of the equation, we know that there will be a positive and negative solution, and we will discard the negative solution.

Then, using the quadratic formula,

x = ( 0.00008 + sqrt( 0.00008^2 + 4 * .07 * 0.0000006))/(2*0.0000006) = 414.676883630352

Note that 414 is between 0 and 500 and much closer to 500.

The height there is

.07x + 0.00004 x^2 -0.0000002 x^3 =

.07*414.676883630352 + 0.00004*414.676883630352^2 -0.0000002*414.676883630352^3 =

21.6443468069815

This is above 21.5 feet, so the crane will fit.

Since the crane may be placed in various places, we have options. As it is 30 ft. wide, the obvious place to dredge is 15 feet on either side of the x where the bridge has maximum height.

Dredge from 414.676883630352-15 to 414.676883630352 + 15, or from

399.676883630352 to 429.676883630352

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