\"A semi elliptic archway has a height of 16 feet in the center and a width of 4
ID: 1914642 • Letter: #
Question
"A semi elliptic archway has a height of 16 feet in the center and a width of 40 feet. Determine if a truck 13 feet high and 12 feet wide can drive under the archway without going over the road's centerline. Show all work including the height at the point you are checking." Ok. I know how to start it but something is wrong. Where does the 12 feet wide come into play? (x^2/20^2) + (y^2/16^2) = 1 (13^2/20^2) + (y^2/16^2) = 1 (13^2/20^2) + (y^2/16^2) - (13^2/20^2) = 1 - (13^2/20^2) (y^2/16^2) = (1 - (13^2/20^2)) (y^2/16^2) (16^2) = (1 - (13^2/20^2)) (16^2) y^2 = 147.84 y = 12.16 So now what?...Explanation / Answer
If the truck cannot cross the centerline and is 12 ft wide and 13 ft high, the height of the arch 12 feet from the centerline must be greater than or equal to 13 feet. If the coordinate system is defined such that y is the vertical height of the ellipse and x is the horizontal width with the point (0,0) at the center of the roadway, 20 is the semi-major axis and 16 is the semi-minor axis, then the equation of the ellipse is: (x/20)² + (y/16)² = 1 y = 16v[1-(x/20)²] (ignore the negative value since we are only interested in height above the roadway) At a point 12 ft from the centerline, the height is: y = 16v[1-(12/20)²] = 12.8 feet Since the truck is 13 feet high, and the arch is only 12.8 feet high at a point 12 feet from the centerline, the truck will have to cross the centerline to pass under the arch.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.