/*! 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 74 A bridge is to be built in the s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 74

A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

Short Answer

Expert verified
The height of the arch at its center is \( \frac{50}{3} \) feet.

Step by step solution

01

Identify the function describing the arch

Since the bridge is in the shape of a semielliptical arch, we can represent it using the equation of an ellipse. For an ellipse centered at the origin, the equation is: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) Here, \( a \) is the semi-major axis (half the span of the ellipse) and \( b \) is the semi-minor axis (the height of the arch at the center).
02

Set parameters based on the given span and height

The span of the bridge is 100 feet, so \( a = 50 \) feet. We need to find the value of \( b \). Given that the height of the arch at 40 feet from the center is 10 feet, we set \( x = 40 \) feet and \( y = 10 \) feet.
03

Substitute the given values into the ellipse equation

Using the ellipse equation: \( \frac{40^2}{50^2} + \frac{10^2}{b^2} = 1\) Simplify the left-hand side: \( \frac{1600}{2500} + \frac{100}{b^2} = 1\) Reduce fractions: \( \frac{16}{25} + \frac{100}{b^2} = 1\)
04

Solve for \( b\)

First, isolate \( \frac{100}{b^2} \): \( \frac{100}{b^2} = 1 - \frac{16}{25} \) Simplify the right-hand side: \( \frac{100}{b^2} = \frac{25}{25} - \frac{16}{25} \) \( \frac{100}{b^2} = \frac{9}{25} \) Next, solve for \( b^2 \): \( 100b^2 = 25 \times 100 \) \( b^2 = \frac{25 \times 100}{9} \) \( b^2 = \frac{2500}{9} \) Finally, take the square root to find \( b \): \( b = \frac{50}{3} \)
05

Determine the height of the arch at its center

In the context of the original problem, \( b \) represents the height of the arch at the center. Therefore, the height of the arch at its center is: \( H = \frac{50}{3} \) feet

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 91影视!

Key Concepts

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

ellipse equation
An ellipse is a 2D shape and its equation helps define it mathematically. When centered at the origin, the standard form is: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Here, \(a\) and \(b\) are constants that determine the ellipse's dimensions.
For a bridge arch, this equation is very useful since the arch can often be modeled as a part of an ellipse. Now, why does this equation matter for our bridge problem? Because it helps in calculating any point on the arch based on the span and height details!
semi-major axis
The semi-major axis, represented by \(a\), is half of the longest diameter of the ellipse.
In the problem, the bridge's span is given as 100 feet, making the total width of the ellipse 100 feet.
Thus, the semi-major axis is half of this span, so \(a=50\) feet.
Knowing the semi-major axis is crucial because it simplifies the ellipse equation, helping us solve for other variables needed to find the height of the arch.
semi-minor axis
The semi-minor axis, represented by \(b\), determines the height of the ellipse's vertex, perpendicular to the major axis.
In our exercise, \(b\) is the height of the arch at the center. To find \(b\), we use given points on the ellipse.
The problem states that at \(x = 40\) feet from the center, the height \(y\) is 10 feet. Plugging these values into the ellipse equation, we get:
\( \frac{40^2}{50^2} + \frac{10^2}{b^2} = 1 \).
Simplifying it helps us determine \(b\). This is how the semi-minor axis plays a key role in the problem.
solving for variables in algebra
In algebra, we solve for unknowns by isolating the variable of interest, i.e., arranging the equation such that the variable is on one side, and the known values are on the other.
For our bridge problem, we needed to find \(b\). First, we substituted the known values into the equation:
\( \frac{40^2}{50^2} + \frac{10^2}{b^2} = 1 \)
Simplifying, we found:
\( \frac{16}{25} + \frac{100}{b^2} = 1 \).
Next, we isolated \( \frac{100}{b^2} \):
\( \frac{100}{b^2} = \frac{9}{25} \).
This simplified to:
\(b^2 = \frac{2500}{9} \).
Finally, taking the square root, we obtained:
\( b = \frac{50}{3} \).
So, the algebraic steps help nail down the exact height of the arch!

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

Sonic Boom Aircraft such as fighter jets routinely go supersonic (faster than the speed of sound). An aircraft moving faster than the speed of sound produces a cone-shaped shock wave that "booms" as it trails the vehicle. The wave intersects the ground in the shape of one half of a hyperbola and the area over which the "boom" is audible is called the "boom carpet." If an aircraft creates a shock wave that intersects the ground in the shape of the hyperbola \(\frac{x^{2}}{484}-\frac{y^{2}}{100}=1\) (units in miles), how wide is the "boom carpet" 32 miles behind the aircraft?

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?

True or False The equation \(y^{2}=9+x^{2}\) is symmetric with respect to the \(x\) -axis, the \(y\) -axis, and the origin.

Find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. \(x^{2}-y^{2}=4\)

A racetrack is in the shape of an ellipse 100 feet long and 50 feet wide. What is the width 10 feet from a vertex?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation
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.