Grace S.

asked • 10/31/19# I don't even know where to start

A bridge with a supporting parabolic arch spans 60 feet with a 30-ft wide road passing underneath the bridge. In order to have a minimum clearance of 16 feet at either edge of the roadway, what is the maximum clearance?

## 2 Answers By Expert Tutors

William W. answered • 10/31/19

Math and science made easy - learn from a retired engineer

I'm thinking the picture looks like this:

Use the vertex form of a parabola y = a(x - 30)^{2} + k

Using the sketch, the parabola goes through the points (0, 0) and (15, 16) and the vertex is at (30, k) plug those in and get 2 equations in 2 unknowns to solve for k,

Barbara F. answered • 10/31/19

Physics and Engineering Tutor, with specialization in advanced topics

We know that the standard equation form for a parabola is:

(x-h)^{2} = 4p(y-k)

-where (h, k) correspond to the vertex of the parabola

-If we place the vertex at x =0, then h=0 and k will be the maximum y-value

To solve for k, we need to find points that we can plug into the equation that will give us an idea of what is p.

To do this, we need to divide 60 and 30 by 2 to get the points which are symmetric about the x-axis:

point 1: (0, k) <--- the vertex

point 2/3: (30, 0) and (-30, 0) <-- the ends of the arch that touch the ground

point 4/5: (15, 16) and (-15, 16) <-- the minimum clearance at the edge of the roadway

Now let's plug in what we know to our equation:

(x-h)^{2} = 4p(y-k)

(x-0)^{2} = 4p(y-k)

Since we have two unknowns, we need to create two equations by plugging point 2 and 4 into our parabolic equation:

i) (30-0)^{2} = 4p(0-k)

ii) (15-0)^{2} = 4p(16-k)

i) (30)^{2} = 4p(-k)

ii) (15)^{2} = 4p(16-k)

i) (30)^{2} = -4pk

ii) (15)^{2} = 64p-4pk

subtract i-ii:

(30)^{2 }- (15)^{2} = -4pk -64p+4k

(30)^{2 }- (15)^{2} = -64p

p = ((30)^{2 }- (15)^{2})/64 = -10.5

Plug p back into equation i to get k:

(30)^{2} = -4pk

(30)^{2} = -4(-10.5)k

k = (30)^{2}/(-4(-10.5)) = 21.4

k= 21.4ft.

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Paul M.

10/31/19