I was the one that gave you the above solution.
What part of it don't you understand?
Up to what point an you follow it ?
I placed the ellipse on the gird so that the centre of the roadway is at (0,0)
Since the far side of the road is at (30,0)
wouldn't there be 10 ft intervals at
0, 10ft, 20ft, and 30 ft?
Because of the symmetry I did not do the -10, -20, and -30 since we could get the same result as 10, 20, 30.
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.
6 answers
Since i need to find the distance from the arch to the roadway. Shouldnt the distance from 0-30 increase? In your case it decreases..
the distance form the arch to the roadway is the y value , the vertical, the up-and-down.
at x = 0 , y = 20 ---> middle of road
at x = 10, y = 18.856
it would be the same height at x = -10
of course it has to decrease, it would be highest in the middle.
at x = 0 , y = 20 ---> middle of road
at x = 10, y = 18.856
it would be the same height at x = -10
of course it has to decrease, it would be highest in the middle.
But in the question, it says that the roadway is above the arch. Not below
Ok, you are right, I mis-read the question.
I was visualizing the arch above the road, with the road as an underpass.
no big deal, easily fixed.
so we can still use my equation etc.
except we will make the road as y = 25 ft
so at x = 0, the road = 25-20 or 5 ft above
at x = 10 the road = 25 - 18.856 = 6.144 ft
etc
calculate the y above in my equation, and subtract it from 25
Notice now it would be increasing.
I was visualizing the arch above the road, with the road as an underpass.
no big deal, easily fixed.
so we can still use my equation etc.
except we will make the road as y = 25 ft
so at x = 0, the road = 25-20 or 5 ft above
at x = 10 the road = 25 - 18.856 = 6.144 ft
etc
calculate the y above in my equation, and subtract it from 25
Notice now it would be increasing.
Ahhh. THANK YOU SO MUCH!! This greatly helped me
8 answers