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 which to solve/find)
Make a sketch of the semi-ellipse with centre (0,0)
I assume you know the basic equation of an ellipse to be
x^2/a^2 + y^2/b^2 = 1, where 2a is the major axis , and 2b is the minor axis
in our case, a = 30, and b = 20
equation:
x^2/900 + y^2/400 = 1
so we need the values of y for x = 0,10,20,30
x^2/900 + y^2/400 = 1
times 3600
4x^2 + 9y^2 = 3600
y^2 = (3600 - 4x^2)/9
when x = 0
y^2= 3600/9 = 400, y = √400 = 20
when x = 10
y^2 =(3600-400)/9 = 3200/9 , y = 18.856
you do the last two
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 which to solve/find)
Igot this answer:
"
Make a sketch of the semi-ellipse with centre (0,0)
I assume you know the basic equation of an ellipse to be
x^2/a^2 + y^2/b^2 = 1, where 2a is the major axis , and 2b is the minor axis
in our case, a = 30, and b = 20
equation:
x^2/900 + y^2/400 = 1
so we need the values of y for x = 0,10,20,30
x^2/900 + y^2/400 = 1
times 3600
4x^2 + 9y^2 = 3600
y^2 = (3600 - 4x^2)/9
when x = 0
y^2= 3600/9 = 400, y = √400 = 20
when x = 10
y^2 =(3600-400)/9 = 3200/9 , y = 18.856
you do the last two "
BUT I DON'T STILL GET HOW TO find at 10 ft intervals the distance from the arch to the roadway.
Since the roadway is at (0,25) in the center, I think that the 25 ft here would be subtracted from the stuff you got from x. So when x = 0, then (25-20) = 5 so (0,5). Then when x=10, (25 - 18.86) = 6.14 so (10, 6.14). The y increases as x is.
Since the roadway is at (0,25) in the center, I think that the 25 ft here would be subtracted from the stuff you got from x. So when x = 0, then (25-20) = 5 so (0,5). Then when x=10, (25 - 18.86) = 6.14 so (10, 6.14). The y increases as x is. Correct me if I'm wrong. Lol.
Well, sir/madam, whoever u are. Then the intervals would have coordinates (0,5), (10, 6.14), (20,10.09), and (30,25)?
That's the same with what I've solved. Since it's a span of 60ft/2 then 30ft to the right, then the roadway connecting to/touching the end of the arch imo (more-like in my computation) is (30, 25). Lol.
Isn't that a bit too much of increase in y sir/madam? From 13.94 or 14 to 25 real quick?
Idk man/dude/kid. Try doin' it yerself. You'll see how'd I end up there. Try graphing too and check. As I've said b4, 'Correct me if I'm wrong. Lol.'. Lol. Peace out. G'day.
To find the distance from the arch to the roadway at 10 ft intervals, we need to determine the equation of the semi-ellipse and then substitute different values to find the corresponding distances.
Let's start by visualizing the problem. We have a semi-ellipse which represents the arch of the concrete bridge. The span of the semi-ellipse is 60 ft, which means the width of the arch from one end to the other is 60 ft. The central height of the semi-ellipse is 20 ft, representing the highest point of the arch.
To find the equation of this semi-ellipse, we can use the standard form of the equation for a vertical ellipse:
(x-h)²/a² + (y-k)²/b² = 1,
where (h, k) is the center of the ellipse, and 'a' and 'b' are the semi-major and semi-minor axes, respectively.
Since the semi-ellipse is symmetrical, the center will be at the midpoint of the span, which in this case is (30, 20). Thus, the equation becomes:
(x-30)²/a² + (y-20)²/b² = 1.
To find 'a' and 'b', we can use the following information:
- The span is 60 ft, which means the total width from one end to the other is 2a. So, 2a = 60, which implies a = 30.
- The central height is 20 ft, so 'b' is the distance from the center (30, 20) to either the top or bottom of the semi-ellipse. We can use the Pythagorean theorem to find 'b':
b² = a² - h²
b² = 30² - 20²
b² = 900 - 400
b² = 500
b ≈ 22.36 ft.
Therefore, the equation of the semi-ellipse becomes:
(x-30)²/30² + (y-20)²/22.36² = 1.
Now, we need to find the distance from the arch to the roadway at 10 ft intervals from the center (30, 20). We can do this by substituting 'y' with '25' and solving for 'x' in the equation of the semi-ellipse.
(30-x)²/30² + (25-20)²/22.36² = 1.
Now, we can solve this equation for different values of 'x' to find the corresponding values of 'y', which will give us the distance from the arch to the roadway.
For example:
- When 'x' is 10, we substitute it into the equation:
(30-10)²/30² + (25-20)²/22.36² = 1.
Simplifying and solving the equation will give us the value of 'y', which represents the distance from the arch to the roadway when 'x' is 10.
Repeat this process at 10 ft intervals until you reach the other end of the span at 'x' = 60.