The St. Louis Gateway Arch
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- Philip Rodgers
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1 The St. Louis Gateway Arch During the spring break trip to St. Louis, Missouri, I paid a visit to the St. Louis Gateway Arch. On the guided tour, I learned that the arch is 630 feet tall and the legs of the arch are 630 ft. apart. Being on spring break but still thinking about math, I look at the arch and immediately see a parabola. In a fervent effort to analyze this amazing curve, I realize that if I only knew three points, then I could determine the equation of the parabola. 1. Based on the information I learned in the tour, is it possible come up with three coordinates? If so, what might they be? Are these the only possible coordinates?
2 . Suppose you orient a coordinate grid with the bottom of one leg of the arch at the origin and the rest of the arch lying in the first quadrant so that the other leg is also on the x-axis. Sketch the arch in the first quadrant below and label as many coordinates as you can on the arch.
3 3. Using the general form of a quadratic function, y = a x + b x + c and the three coordinates labeled above, write a system of three equations to solve for the three unknowns, a,b and c. HINT: You will get c directly from one of the equations. Substitute this value into the other two. Use one of the other equations to find an equation for b in terms of a, and substitute this value for b into the last equation and solve for a. You now have numerical values for a and c use these to solve for b.
4 4. Using the values for a,b and c from question 3 write the quadratic equation for the Gateway Arch and sketch it.
5 5. Is there another way to orient the coordinate axes that might make finding the quadratic equation easier? Choose another orientation and find the quadratic equations using the new coordinate system. How are the two quadratic equations related?
6 6. The formula actually used in constructing the St. Louis Arch is displayed on the inside of the arch. It is the formula for a catenary curve, which is the shape a free hanging chain takes when held at both ends. Mathematically, the function that models such a curve is hyberbolic cosine. The formula used for the St. Louis Arch is y = 68.8 cosh(.01 x 1). Use your graphing calculator to graph this function. (If you have trouble find the cosh function, use the catalog feature of your calculator.) Is this graph close to the graph of either of the quadratic functions you found? Can you transform either of the functions (the quadratic or the hyberbolic cosine) in a way that preserves size (using only reflections and translations) to show that they model close to the same physical structure? December 0, 004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board.
7 Solutions 1. Based on the information I learned in the tour, is it possible come up with three coordinates? If so, what might they be? Are these the only possible coordinates? The students can orient the coordinate system in many different ways and get different sets of coordinates. Example: (0, 0), (315, 630) and (630, 0) or (-315, 0), (0, 630) and (315, 630). Suppose you orient a coordinate grid with the bottom of one leg of the arch at the origin and the rest of the arch lying in the first quadrant so that the other leg is also on the x-axis. Sketch the arch in the first quadrant below and label as many coordinates as you can on the arch.
8 3. Using the general form of a quadratic function, y = a x + b x + c and the three coordinates labeled above, write a system of three equations to solve for the three unknowns, a,b and c. HINT: You will get c directly from one of the equations. Substitute this value into the other two. Use one of the other equations to find an equation for b in terms of a, and substitute this value for b into the last equation and solve for a. You now have numerical values for a and c use these to solve for b. This creates an easy solution: Eqn 1: c = 0 Eqn : 315 a b + c = 630 Eqn 3: 630 a b + c = 0 Substitute c into Eqn and 3, a b = 630 a b = 0 From write b in terms of a, 315 a b = a b = = 315 a 315 Substitute in 3 and solve for a 630 a ( 315 a) = ( ) a = 630 a = = Solve for b b = 315 a b = 315 (.00635) = 4 Answer: a = b = 4 c = 0
9 4. Using the values for a,b and c from question 3 write the quadratic equation for the Gateway Arch and sketch it. y = x + 4 x
10 5. Is there another way to orient the coordinate axes that might make finding the quadratic equation easier? Choose another orientation and find the quadratic equations using the new coordinate system. How are the two quadratic equations related? One possibility is to center the legs around the y-axis and get the following three coordinates, (-315, 0), (0, 630) and (315, 0). By using knowledge about transformation of the parent function y=x we get the following solution: y = a x Plug in any value for y and x gives, 0 = a a = 315 = Leads to, y = x The two equations differ only in translation.
11 6. The formula actually used in constructing the St. Louis Arch is displayed on the inside of the arch. It is the formula for a catenary curve, which is the shape a free hanging chain takes when held at both ends. Mathematically, the function that models such a curve is hyberbolic cosine. The formula used for the St. Louis Arch is y = 68.8 cosh(.01 x 1). Use your graphing calculator to graph this function. (If you have trouble find the cosh function, use the catalog feature of your calculator.) Is this graph close to the graph of either of the quadratic functions you found? Can you transform either of the functions (the quadratic or the hyberbolic cosine) in a way that preserves size (using only reflections and translations) to show that they model close to the same physical structure? The catenary curve described by the formula is a reflection and a translation of the curve modeled with the quadratic. In order to see this, it can be manipulated by the parameters that will not change the size of our arch. The catenary curve can be transformed to y = 68.8 cosh (.01 x 3.15) This graph and the first quadratic model created are very close together. Some discrepancies occur because of where the tour guide s measurements are taken. We are told that the legs are 630 feet apart. This measurement is taken from the outside edge of both legs. A graphic of the actual arch with the published catenary curve overlaid shows the catenary curve does not follow the outer edge of the arch, but instead represents an internal structure. Still, the two curves are very close.
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