Can anyone help me solve this question please A new bridge i

Can anyone help me solve this question please?

A new bridge is to be constructed over the East River in New York City. The space between the supports needs to be 1050 ft and the height at the center of the arch needs to be 350 ft. Two structural possibilities exist; the arch could be in the shape of a parabola or the shape of a semiellipse. Consider that an empty tanker needs a 280 ft clearance to pass beneath the bridge. The width of the channel must be determined to verify that the tanker can pass through the bridge. a. Determine the parabola with these characteristics. Then, determine the width of the channel that the tanker can pass through. b. Determine the semiellipse with these characteristics. Then, determine the width of the channel that the tanker can pass through. c. If the river were to flood and rise 10 ft, how would the clearances of the two plans be affected? How does this affect your decision as to which design to choose? Explain.

Solution

A.

Parabola has x-intercepts x = 525, x = 525 (the distance between x-intercepts = 1050)

f(x) = a (x + 525) (x 525)

Parabola has vertex = (0, 350) ----> height of arch = 350

f(0) = 350
a (0 + 525) (0 525) = 350
275625 a = 350
a = 14/11025

f(x) = 14/11025 (x + 525) (x 525)
f(x) = 14/11025 (x² 275625)  

B.

Ellipse is centred at (0,0) with vertices at (525,0), (525,0) and co-vertices at (0,350) and (0,350), although we are not interested in graph below x-axis.

Equation of ellipse: (x²/525² ) + (y²/350²) = 1

To find equation of semi-ellipse, we need to solve for y, with y >= 0

y²/350² = 1 x²/525²
y² = 350² 350²x²/525²
y² = 122500 4x²/9
y² = 4/9 275625 x²
y = 2/3 (275625 x²)

g(x) = 2/3 (275625 x²)

C) To find width of channels, we need to find x-values for each equations where height (y-value) = 280

Parabola:

f(x) = 280
14/11025 (x² 275625) = 280
x² 275625 = 280 * 11025/14
x² 275625 = 220500
x² = 55125
x = 1055, 1055

Width of channel = (1055) (1055) = 2105 = 469.57 ft

Semi-ellipse:

g(x) = 280
2/3 (275625 x²) = 280
(275625 x²) = 420
275625 x² = 176400
x² = 99225
x = 315, 315

Width of channel = (315) (315) = 630 ft