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)
8 answers
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.
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.
3 answers