Solving for the Height and Width of St. Louis Arch: A Calculus Problem

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In summary, the St. Louis arch was constructed using a hyperbolic cosine function, with cross sections in the shape of equilateral triangles. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x, where x represents the distance along the arch's path and ranges from -299.2239 to 299.2239. For each value of x, the area of the cross-sectional triangle is given by A = 125.1406 cosh .0100333x. To determine the height of the highest cross-sectional triangle, x is set to 0 and the equation becomes 693.8597-68.7672 cosh (0) =
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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)
 
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Can you please make it more sparse, to make it more readable? Like breaking it into paragraphs.
 
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St. Louis Arch:

The St. Louis arch was constructed using the hyperbolic cosine function. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x. -299.2239<=x<=299.2239. (1) 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 of the cross sectional triangle is A = 125.1406 cosh .0100333x. (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)
 
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Can you please make it more sparse, to make it more readable? Like breaking it into paragraphs.
 
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gjh said:
St. Louis Arch:

The St. Louis arch was constructed using the hyperbolic cosine function. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x. -299.2239<=x<=299.2239. (1) 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 of the cross sectional triangle is A = 125.1406 cosh .0100333x. (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.)

gjh said:
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.

gjh said:
hen 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)
 

1. How do you use calculus to solve for the height and width of the St. Louis Arch?

Calculus is a branch of mathematics that deals with rates of change and accumulation. In this problem, we can use the concepts of derivatives and integrals to find the height and width of the St. Louis Arch.

2. What information do you need to solve this problem?

To solve for the height and width of the St. Louis Arch, we need the measurements of the base and the midpoint of the arch. We also need to know the shape of the arch, which is a catenary curve.

3. Can this problem be solved using other mathematical methods?

Yes, there are other mathematical methods that can be used to solve this problem, such as geometry or trigonometry. However, calculus is the most efficient and accurate method for solving this type of problem.

4. Is it possible to solve for the height and width of the St. Louis Arch without knowing the exact shape of the arch?

No, the shape of the arch is a crucial piece of information needed to solve this problem. Without it, we cannot accurately determine the height and width of the arch using calculus.

5. What are the real-world applications of solving for the height and width of the St. Louis Arch?

Solving for the height and width of the St. Louis Arch has practical applications in architecture, engineering, and construction. It can also be used to analyze and design other structures with similar shapes, such as bridges and tunnels.

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