The shape of the Gateway Arch in St. Louis. Missouri, can be approximately model

Question

The shape of the Gateway Arch in St. Louis. Missouri, can be approximately modeled with the equations of two parabolas: one parabola for the outer/upper surface, and one for the inner/lower surface. The height of the arch is 630 feet, and at ground level the outsides of the bases are 630 feet apart. The arch narrows as it rises. Therefore, the insides of the bases are only 540 feet apart at ground level, but the inside of the arch has a height of 615 feet. Suppose you are standing at the origin which is at ground level directly underneath the center of the arch. Find the vertex and .v-intercepts for both the outer and inner parabolas. Outer parabola: vertex: .(-intercepts:_ Inner parabola: vertex: .v-intercepts:_ Find the height above the ground of the focus and directrix for each parabola. Round your answers to the nearest tenth of a foot. Outer parabola: focus height: directrix height: _ Inner parabola: focus height: directrix height:_ Find the equation for both the outer and inner parabolas in standard form.

Explanation / Answer

16. (a) If the ground is assumed to be the X-Axis, the Axis of symmetry as the Y-Axis , the the origin is at the ground level, directly underneath the center of the arch. We know that the vertex of a parabola opening downwards, is its highest point. Further, both the outer and the inner parabolas in the arch open downwards. Since the height of the arch is 630 ft. and since the inside of the arch has a height of 615 ft., the vetex of the outer parabola is (0,630) and the vertex of the inner parabola is (0,615). We know that x-intercept is where y = 0. Hence, the x-intercepts of the outer parabola are ± 630/2 i.e. ± 315. The x-intercepts of the inner parabola are ± 540/2 i.e. ± 270.

(b). The equation of the outer parabola can be written as y = ax2 + 630 where a is negative and the equation of the inner parabola can be written as y = bx2 + 615, where b is negative . Further, we know that the outer parabola passes through the point (315,0), therefore, 0 = a(315)2 +630 or, a = -630/ 3152 = -2/315. Then the equation of the outer parabola changes to y = (-2/315)x2 + 630 or, y -630 = (-2/315)x2 or, x2 = – 315/2(y-630). Similarly, the equation of the inner parabola becomes y -615 =(-41/4860)x2 or, x2 = - 4860/41 (y-615) . We also know that when the equation of a parabola is (x-h)2 = m(y-k), then m = 4p, where p is the distance from the vertex to the focus and the directrix. In case of the outer parabola, p = 1/4(– 315/2) = -315/8 and In case of the inner parabola, p = 1/4 (-4860/41) = - 1215/41. Also, in the case of a parabola opening downwards, the focus is below the vertex and the directrix is above the vertex. Hence, the height above the ground of the focus of the outer parabola is 630- 315/8 = 4725/8 =590.625 ft = 590ft 7.5 inches. The height above the ground of the directrix of the outer parabola is 630 + 315/8 = 5355/8 = 669.375 ft = 669 ft 4.5 inches. Similarly, the height above the ground of the focus of the inner parabola is 615- 1215/41 = 24000/41 =585.366 ft = 585 ft 4.39 inches. The height above the ground of the directrix of the inner parabola is 615 + 1215/41 = 26430/41 = 644.634 ft = 644 ft 7. inches

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