Question

Semielliptical Arch Bridge An arch for a bridge over a highway is in the form of half an ellipse. The top of the arch is 20 feet above the ground level (the major axis). The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each 4 feet wide. What should the span of the bridge be (the length of its major axis) if the height 28 feet from the center is to be 13 feet

Answer 1

Answer:

the span of the bridge is 73.7 feet

Step-by-step explanation:

The equation of an ellipse with a vertical major axis(i.e major axis parallel to y axis) is given by:

$\frac{(x-h)^{2} }{b^{2} } + \frac{(y-k)^{2} }{a^{2} }= 1$ a>b

where (h,k) are the coordinates of the center of the ellipse, a is the length of the major axis and b is the length of the minor axis

For this problem, the center of the ellipse (h,k) = (0,0)

Therefore:

$\frac{(x)^{2} }{b^{2} } + \frac{(y)^{2} }{a^{2} }= 1$

The top of the arch is 20 feet above the ground level (the major axis), therefore a=20

length of the major axis = 2a= 2*20 = 40

$\frac{x^{2} }{b^{2} } + \frac{y^{2} }{20^{2} }= 1\\\frac{x^{2} }{b^{2} } + \frac{y^{2} }{400 }= 1$

The coordinates of the ellipse (x,y) = (28,13)

$\frac{28^{2} }{b^{2} } + \frac{13^{2} }{400 }= 1$

$\frac{28^{2} }{b^{2} } + 0.4225= 1$

$\frac{784 }{b^{2} } = 0.5775$

b² ≅ 1358

b≅36.85

Length of minor axis (2b) = 73.7 feet.

the span of the bridge is 73.7 feet