The Gateway Arch in St. Louis, Missouri is not a parabola but a shape known as a catenary. The name is given to the shaoe formed by the Graph of the hyperbolic cosine (cosh). The arch has a height of 625 feet andna span of 600 feet. The hyperbolic cosine is defined as:
Cosh x= 1/2 e^x + 1/2 e^-x
•on the same set of axis graph f(x)=1/2e^x, g(x)=1/2e^-x,and h(x)=(f+g)(x)
•create a parabolic model that matches the arch at the vertex and ground level
•the equation that gives shape of the arch is c(x)=693.8597-68.7672cosh(0.0100333x). Define x and c(x) in terms of the situation
•rewrite the gateway arch equation into exponential function form.
Thanks ahead of time!
you can do all the graphing stuff. Lots of online places for that. Given the definition of cosh(x), you have the arch as
c(x)=693.8597-68.7672(e^0.0100333x+e^-0.0100333x)/2.
The vertex of the arch is at
c(0) = 693.8597-65.7672 = 828.0925
the arch meets the ground where c(x) = 0, or x = ±299.226
So, the parabola whose vertex is where the arch's is is
y = -0.007750x^2 + 693.8957
oops. wrong vertex. Should be (0,628.0925)
y = -0.007015x^2+628.0925
To graph the functions f(x) = 1/2e^x, g(x) = 1/2e^-x, and h(x) = (f+g)(x) on the same set of axes:
1. Choose a set of x-values, such as -4, -2, 0, 2, and 4.
2. Calculate the corresponding y-values for each function using the given formulas.
- For f(x), substitute the x-values into the formula f(x) = 1/2e^x.
- For g(x), substitute the x-values into the formula g(x) = 1/2e^-x.
- For h(x), sum the corresponding y-values of f(x) and g(x) at each x-value.
3. Plot the points for each function on the graph.
4. Connect the points with smooth curves to represent the functions f(x), g(x), and h(x).
Now, to create a parabolic model that matches the arch at the vertex and ground level, you need to find the equation of a parabola that fits these criteria. The vertex form of a parabola is y = a(x-h)^2 + k, where (h, k) represents the vertex coordinates.
In this case, the vertex is at the ground level, so the y-coordinate (k) is 0. The x-coordinate (h) of the vertex can be found by solving the equation c(x) = 693.8597 - 68.7672cosh(0.0100333x) = 0. Set c(x) equal to 0 and solve for x to find the value of x at the vertex.
Next, rewrite the equation c(x) = 693.8597 - 68.7672cosh(0.0100333x) into exponential function form.
The hyperbolic cosine function cosh(x) can be rewritten using the exponential function e^x as follows:
cosh(x) = (e^x + e^-x) / 2
Substitute this expression into the equation for c(x):
c(x) = 693.8597 - 68.7672 * (e^0.0100333x + e^-0.0100333x) / 2
Simplify as needed.
Note: Some slight variations or approximations may arise due to rounding or decimal precision.
To answer your question, let's break it down step by step:
1. Graphing the Functions:
To graph the functions f(x) = 1/2e^x, g(x) = 1/2e^-x, and h(x) = (f+g)(x), you can use a graphing calculator or any software that can plot graphs. Simply input the functions and the desired range for x-values.
2. Creating a Parabolic Model:
To create a parabolic model that matches the arch at the vertex and ground level, you need to find the equation of a parabola that fits these points. Start by identifying these points from the given information: vertex (highest point) and ground level.
The vertex of the arch is the point (0, 625) since the height is 625 feet, and the ground level is at (300, 0) (assuming the center of the arch is at x = 300).
Using the vertex form of a parabola, the equation will be of the form:
y = a(x - h)^2 + k, where (h, k) is the vertex and a represents the curvature.
To find 'a', we need another point on the parabola. Let's use the ground level point (300, 0):
0 = a(300 - 300)^2 + 625
0 = a(0) + 625
0 = 625
Since this does not work out, it suggests that a normal parabola does not match the shape of the arch. Therefore, a single parabolic model cannot match the arch at the vertex and ground level.
3. Equation in Terms of the Situation:
The equation given to represent the shape of the arch is c(x) = 693.8597 - 68.7672cosh(0.0100333x).
Here, x represents the horizontal distance along the arch, and c(x) represents the vertical displacement (height) from the ground level. The arch's maximum height is 625 feet, so the ground level is at c(x) = 0.
4. Rewriting the Equation in Exponential Function Form:
The given equation for the shape of the arch, c(x) = 693.8597 - 68.7672cosh(0.0100333x), can be rewritten in exponential function form. To do this, recall the definition of hyperbolic cosine (cosh) as:
cosh(x) = (1/2)e^x + (1/2)e^(-x).
Substituting this definition, we can rewrite the equation as:
c(x) = 693.8597 - 68.7672[(1/2)e^(0.0100333x) + (1/2)e^(-0.0100333x)].
Simplifying further, we get:
c(x) = 693.8597 - 34.3836e^(0.0100333x) - 34.3836e^(-0.0100333x).
Thus, the equation for the shape of the Gateway Arch in exponential function form is c(x) = 693.8597 - 34.3836e^(0.0100333x) - 34.3836e^(-0.0100333x).
I hope this helps clarify the steps and answer your question. If you have any further queries, feel free to ask!