- #1

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- 10

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- Homework Statement
- The St. Louis Arch was constructed using the hyperbolic cosine function. The equation used for construction was y=693.8597-68.7672 cosh 0.0100333x, -299.2239<=x<=299.2239, x and y in feet. Cross sections of the arch are equilateral triangles, and (x,y) traces the path of the centers of mass of the cross-sectional triangles. For each value of x, the Area A of the cross-sectional is A = 125.1406 cosh 0.0100333x.

- Relevant Equations
- y=693.8597-68.7672 cosh 0.0100333x, -299.2239<=x<=299.2239, x and y in feet. (1)

Area A of the cross-sectional is A = 125.1406 cosh 0.0100333 (2)

(a) How high above the ground is the center of the highest triangle? (At ground level, y=0)

Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925

(b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of the triangle and the center of mass of the triangle is located at two-thirds the height of the triangle.)

From equation (2) above, I found the area A of the highest cross-sectional triangle as A = 125.1406 cosh (0) = 125.1406. Then used A = 125.1406 = sqrt (3) c^2 to solve for c and got c=8.5. Then found the height of the highest cross-section triangle as h = tan (60) x (8.5) = 14.72. The height of the arch is then 625.0925 + (1/3) 14.72 = 625.0925 +4.0975 =630 feet. This checks with Arch height on web, so believe (a) and (b) are correct.

(c) How wide is the arch at ground level? Here's where I ran into problems.

I reasoned that the center of mass equation is satisfied for y=0 -> 0 = 693.8597 - 68.7672 cosh 0.0100333 x -> 68.7672 cosh .0100333x = 693.8587 -> cosh .0100333x =693.8587/68.7672 = 10.09. Then I took the inverse cosh of both sides of the equation and got x=2.9941. The substituted back into Area function A = 125.1406 cosh (0.0100333(2.9941) = 125.1406 cosh (1.0045) = 125.1971. Solving for c, 125.71 = sqrt (3) c^2 -> c=8.5. If I add that to 299.2239, I get 307.7239. Web lists the width as 630 feet, so half width is 315. I'm off by over 7 feet. Any help appreciated.

Note: This problem is from Calculus, 8th ed, by Larson et al, that I'm using for home study)

Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925

(b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of the triangle and the center of mass of the triangle is located at two-thirds the height of the triangle.)

From equation (2) above, I found the area A of the highest cross-sectional triangle as A = 125.1406 cosh (0) = 125.1406. Then used A = 125.1406 = sqrt (3) c^2 to solve for c and got c=8.5. Then found the height of the highest cross-section triangle as h = tan (60) x (8.5) = 14.72. The height of the arch is then 625.0925 + (1/3) 14.72 = 625.0925 +4.0975 =630 feet. This checks with Arch height on web, so believe (a) and (b) are correct.

(c) How wide is the arch at ground level? Here's where I ran into problems.

I reasoned that the center of mass equation is satisfied for y=0 -> 0 = 693.8597 - 68.7672 cosh 0.0100333 x -> 68.7672 cosh .0100333x = 693.8587 -> cosh .0100333x =693.8587/68.7672 = 10.09. Then I took the inverse cosh of both sides of the equation and got x=2.9941. The substituted back into Area function A = 125.1406 cosh (0.0100333(2.9941) = 125.1406 cosh (1.0045) = 125.1971. Solving for c, 125.71 = sqrt (3) c^2 -> c=8.5. If I add that to 299.2239, I get 307.7239. Web lists the width as 630 feet, so half width is 315. I'm off by over 7 feet. Any help appreciated.

Note: This problem is from Calculus, 8th ed, by Larson et al, that I'm using for home study)