SOLUTION: The arch beneath a bridge is ​semi-​elliptical, and a​ one-way roadway passes under the arch. The width of the roadway is 30 feet and the height of the arch over

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The arch beneath a bridge is ​semi-​elliptical, and a​ one-way roadway passes under the arch. The width of the roadway is 30 feet and the height of the arch over       Log On


   

Question 1116130: The arch beneath a bridge is ​semi-​elliptical, and a​ one-way roadway passes under the arch. The width of the roadway is 30 feet and the height of the arch over the center of the roadway is 12 feet. Two trucks plan to use this road. They are both 10 feet wide. Truck 1 has an overall height of 11 feet and Truck 2 has an overall height of 10 feet. Draw a rough sketch of the situation and determine which of the trucks can pass under the bridge.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This is what the ground, the road, and the arch beneath the bridge look like. .
This is what the two trucks going under the bridge, down the middle of the roadway would look like.
and .
A graph of the semi-elliptical arch on x-y coordinates representing length in feet, would look like this.
. The equation is x%5E2%2F15%5E2%2By%5E2%2F12%5E2=1 for x%3E=0 only.

So, the height in feet, over the roadway at a distance abs%28x%29=5 feet from the center of the roadway is
.

NOTE: To make a good sketch, I would draw the roadway as a line,
I would mark the center of the roadway
and the two foci of the ellipse at a focal distance
c=sqrt%2815%5E2-12%5E2%29=9 from the center.
Then I would put pins at a the foci;
would wrap around them a loop of string with total length
2%2A15%2B2%2A8=48 .
I would stretch the loop tight with the tip of a pencil and would move it in an arc keeping the string tight.
That would make the sum of the distances from the foci to the tip of the pencil
equal to the 2%2A15 major axis of the ellipse.