/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Arch in St. Louis is often mista... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Arch in St. Louis is often mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 630 feet high and 630 feet wide at its base. (a) Find the equation of a parabola with the same dimensions. Let \(x\) equal the horizontal distance from the center of the arch. (b) The table below gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). $$ \begin{array}{|cc|} \hline \text { Width (ft) } & \text { Height (ft) } \\ \hline 567 & 100 \\ 478 & 312.5 \\ 308 & 525 \\ \hline \end{array} $$ (c) Do the data support the notion that the Arch is in the shape of a parabola?

Short Answer

Expert verified
The equation of the parabola is \( y = \frac{1}{157.5} x^2 \). The Arch's data does not support it being a parabola.

Step by step solution

01

Understand the problem

The goal is to find the equation of a parabola that mimics the dimensions of the St. Louis Arch and compare its height at specific widths to see if the Arch could be parabolic.
02

Set up the parabola equation

The parabola's vertex is at the origin (0,0), and its equation is in the form of \(y = ax^2\). Given the width is 630 feet at the base and the maximum height is 630 feet, you have coordinates \(x = \frac{630}{2} = 315, y = 630\).
03

Solve for 'a'

Substitute the known point (315, 630) into the equation \(y = ax^2\) to find 'a': \[ 630 = a(315)^2 \]
04

Simplify and find 'a'

Calculate \(a\) as follows: \[ 630 = a(99225) \] \[ a = \frac{630}{99225} = \frac{1}{157.5}\]
05

Write the final parabolic equation

The equation of the parabola is \[ y = \frac{1}{157.5} x^2 \]
06

Calculate heights at given widths

Substitute the widths (567, 478, and 308 feet) into the parabolic equation to find the corresponding heights. \[ y(567) = \frac{1}{157.5} (567)^2 = 2041.21 \] \[ y(478) = \frac{1}{157.5} (478)^2 = 1450.16 \] \[ y(308) = \frac{1}{157.5} (308)^2 = 602.32 \]
07

Compare heights with actual data

Compare the calculated heights for the parabola to the heights given for the Arch in the provided table: \[ \begin{array}{|cc|} \hline \text { Width (ft) } & \text { Height (ft) } \ \hline \567 & 100 \ \478 & 312.5 \ \308 & 525 \ \hline \end{array} \]
08

Analyze the data

The calculated heights for the parabola (2041.21, 1450.16, 602.32) are significantly higher than the given heights of the Arch (100, 312.5, 525), indicating that the Arch is not parabolic.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

parabola
A parabola is a U-shaped curve that can open either upwards or downwards. Its equation is typically written as \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The simplest form, \( y = ax^2 \), has its vertex (the highest or lowest point) at the origin (0,0).

To find the equation of a parabola when given specific points, you substitute the known x and y coordinates into the equation and solve for \( a \), \( b \), or \( c \).

In our exercise, we only needed \( y = ax^2 \) because the parabola is centered at the origin with no linear or constant term.
St. Louis Arch
The St. Louis Arch is a famous monument constructed to commemorate the westward expansion of the United States. It stands at 630 feet tall and 630 feet wide, which leads many to mistakenly believe it is parabolic.

However, the actual shape of the Arch is a catenary curve, not a parabola. A catenary curve is the shape that a flexible chain or cable assumes when supported at its ends and acted on by gravity. It has a more complex equation than a simple parabolic curve, resulting in the unique and iconic shape of the Arch.
quadratic formula
The quadratic formula is used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula helps find the values of \( x \) where the parabola crosses the x-axis (the roots).

In the context of the exercise, solving for the parabola’s equation primarily involved substituting values and solving for the constant \( a \). Although the quadratic formula wasn't directly used, it's important to understand how it applies to general quadratic equations.
catenary curve
A catenary curve is mathematically described by the hyperbolic cosine function: \( y = a \cosh(\frac{x}{a}) \), where \( a \) is a constant determining the curve’s shape.

Unlike a parabola, a catenary captures the shape a chain naturally forms, creating a stronger and more stable structure. This property makes it an ideal shape for the St. Louis Arch, designed to withstand environmental forces like wind and gravity.

Understanding the difference between parabolic and catenary curves explains why the actual height measurements of the Arch at specific widths don't match those predicted by a simple parabolic equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph each function. Be sure to label any intercepts. [Hint: Notice that each function is half a hyperbola.] \(f(x)=-\sqrt{9+9 x^{2}}\)

Jim, standing at one focus of a whispering gallery, is 6 feet from the nearest wall. His friend is standing at the other focus, 100 feet away. What is the length of this whispering gallery? How high is its elliptical ceiling at the center?

Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in \(1672 .\) The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) and the focal length (distance from the vertex to the focus) of the parabola is \(6,\) find the equation of the parabola.

Use a graphing utility to graph the plane curve defined by the given parametric equations. \(x(t)=4 \sin t+2 \sin (2 t)\) \(y(t)=4 \cos t+2 \cos (2 t)\)

Show that the graph of an equation of the form $$A x^{2}+C y^{2}+D x+E y+F=0 \quad A \neq 0, C \neq 0$$ where \(A\) and \(C\) are of the same sign, (a) is an ellipse if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F\) is the same sign as \(A\). (b) is a point if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F=0\). (c) contains no points if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F\) is of opposite sign to \(A\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.