the arch of a concrete bridge is a semi ellipse having a span of 60 ft and a central height of 20 ft. if the roadway is 25 ft above the base. find at 10 ft intervals the distance from the arch to the roadway. (Im completely lost at what to find. Please help)

what you have is an ellipse with semi-axes of 30 and 20 ft, with the long axis horizontal. So, if the center is placed at (0,0) the equation is

x^2/30^2 + y^2/20^2 = 1

So, the upper arch is

y = 4/9 √(900-x^2)

Now just plug in your various values for x to find the height y.

To find the distance from the arch to the roadway at 10 ft intervals, we need to determine the equation of the semi-ellipse representing the arch.

The equation of a semi-ellipse with its major axis on the x-axis and center at the origin (0,0) can be expressed as:

x^2/a^2 + y^2/b^2 = 1

Where:
- "a" is the semi-major axis (half of the span)
- "b" is the semi-minor axis (half of the central height)

In our case, the span is 60 ft, so the semi-major axis (a) is 30 ft, and the central height is 20 ft, so the semi-minor axis (b) is 10 ft.

Plugging in these values, the equation becomes:

x^2/30^2 + y^2/10^2 = 1

Simplifying the equation further, we have:

x^2/900 + y^2/100 = 1

Now, to find the distance from the arch to the roadway, we need to calculate the y-coordinate for each x-coordinate at 10 ft intervals.

Starting from the top of the arch (where y = 0), we can substitute x = 0 into the equation to find the corresponding y-coordinate:

0^2/900 + y^2/100 = 1
y^2/100 = 1
y^2 = 100
y = ±10 ft

Since the roadway is 25 ft above the base, we consider only the positive y-coordinate:

y = 10 ft

Therefore, at x = 0, the distance from the arch to the roadway is 10 ft.

Next, we can substitute x = 10 ft into the equation:

(10^2)/900 + y^2/100 = 1
100/900 + y^2/100 = 1
y^2/100 = 1 - 100/900
y^2/100 = 1 - 1/9
y^2/100 = 8/9

Taking the square root of both sides:

y/10 = sqrt(8)/3
y = 10 * sqrt(8)/3

Therefore, at x = 10 ft, the distance from the arch to the roadway is 10 * sqrt(8)/3 ft.

You can repeat this process for x = 20 ft, x = 30 ft, and so on, to find the distance from the arch to the roadway at 10 ft intervals.

To find the distance from the arch to the roadway at 10 ft intervals, we can use a mathematical equation called the equation of the ellipse. Given the span of the bridge (60 ft) and the central height (20 ft), we can determine the equation of the ellipse in terms of x and y coordinates.

The general equation of an ellipse, centered at the origin, is:

(x^2 / a^2) + (y^2 / b^2) = 1

In this case, since the bridge is symmetric, the semi-major axis (a) is half of the span (60 ft / 2 = 30 ft), and the semi-minor axis (b) is the central height (20 ft).

Therefore, the equation becomes:

(x^2 / 30^2) + (y^2 / 20^2) = 1

To find the distance from the arch to the roadway, we need to solve for y when x is known. Given that the roadway is 25 ft above the base, we can substitute y with the value y = 20 + 25 = 45.

Substituting this value into the equation, we have:

(x^2 / 30^2) + (45^2 / 20^2) = 1

Simplifying the equation, we obtain:

(x^2 / 900) + (2025 / 400) = 1

Multiplying through by the least common denominator (LCD) of 900 and 400, we have:

4x^2 + 2025 * 9 = 900 * 400

4x^2 + 18225 = 360000

Rearranging the equation, we get:

4x^2 = 341775

x^2 = 341775 / 4

Taking the square root of both sides, we have:

x = √(341775) / 2

x ≈ 103.996 ft

Therefore, at 10 ft intervals, the distances from the arch to the roadway are approximately:

x = 0 ft → distance = 45 ft
x = 10 ft → distance ≈ 45 ft
x = 20 ft → distance ≈ 45 ft
...
x = 100 ft → distance ≈ 45 ft
x = 110 ft → distance ≈ 45 ft
...
x = 200 ft → distance ≈ 45 ft

So, the distance from the arch to the roadway remains approximately 45 ft at each of the 10 ft intervals.