Try the Free Math Solver or Scroll down to Tutorials!

Enter expression, e.g. (x^2-y^2)/(x-y)

Enter expression, e.g. x^2+5x+6

Enter expression, e.g. (x+1)^3

Enter a set of expressions, e.g. ab^2,a^2b

Enter equation to solve, e.g. 2x+3=4

Enter equation to graph, e.g. y=3x^2-1

Depdendent Variable

Number of equations to solve:

Equ. #1:

Equ. #2:

Equ. #3:

Equ. #4:

Equ. #5:

Equ. #6:

Equ. #7:

Equ. #8:

Equ. #9:

Solve for:

Enter inequality to solve, e.g. 2x+3>4

Enter inequality to graph, e.g. y<3x^2-1

Dependent Variable

Number of inequalities to solve:

Ineq. #1:

Ineq. #2:

Ineq. #3:

Ineq. #4:

Ineq. #5:

Ineq. #6:

Ineq. #7:

Ineq. #8:

Ineq. #9:

Solve for:

Math solver on your site

Please use this form if you would like
to have this math solver on your website,
free of charge.

Name:

Email:

Your Website:

Msg:

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. Using the general form of a quadratic function, y = a × x² + 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. Using the values for a,b and c from question 3 write the quadratic equation for the
Gateway Arch and sketch it.

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. 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?

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)

3. Using the general form of a quadratic function, y = a × x² + 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.

This creates an easy solution:

Eqn 1: c = 0
Eqn 2: 315² × a + 315×b + c = 630
Eqn 3: 630² × a + 630 ×b + c = 0

Substitute c into Eqn 2 and 3,

From 2 write b in terms of a,

Substitute in 3 and solve for a

Solve for b

Answer:
a = -.00635
b = 4
c = 0

4. Using the values for a,b and c from question 3 write the quadratic equation for the
Gateway Arch and sketch it.

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=x² we get the following solution:

Plug in any value for y and x gives,

Leads to,

The two equations differ only in translation.

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) + 700 .
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.