/*! 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 72 A sealed-beam headlight is in th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 72

A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 1 inch from the vertex. If the depth is to be 2 inches, what is the diameter of the headlight at its opening?

Short Answer

Expert verified
The diameter of the headlight at its opening is approximately 5.66 inches.

Step by step solution

01

Understand the Problem

A paraboloid of revolution has a cross-section that is a parabola. The light bulb is at the focus of this parabola, 1 inch from the vertex. We need to find the diameter of the headlight at its 2-inch depth.
02

Define the Parabola Equation

The standard form of a parabola with vertex at the origin and focus at (0, f) is \[ y = \frac{x^2}{4f} \]. Since the focus is 1 inch from the vertex, we have \[ f = 1 \] inch. Thus, the equation of the parabola is \[ y = \frac{x^2}{4} \].
03

Substitute the Depth to Find x

At a depth of 2 inches, \[ y = 2 \]. Substitute y = 2 into the parabola equation \[ y = \frac{x^2}{4} \]:\[ 2 = \frac{x^2}{4} \].
04

Solve for x

Multiply both sides by 4:\[ 8 = x^2 \].Take the square root of both sides:\[ x = \pm \sqrt{8} \]. Simplify the square root:\[ x = \pm 2\sqrt{2} \].
05

Calculate the Diameter

The diameter is twice the absolute value of x:\[ \text{Diameter} = 2 \times 2\sqrt{2} = 4\sqrt{2} \approx 5.66 \text{ inches} \].

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.

parabola equation
A parabola is a symmetrical, open plane curve formed by the intersection of a cone with a plane parallel to its side. The simplest form of a parabola's equation is when its vertex is at the origin.
This standard form can be written as: \( y = \frac{x^2}{4f} \). Here, \( f \) represents the focus, the point where light (or sound) converges.
In the problem, the headlight's paraboloid has its focus 1 inch from the vertex. Plugging this value into our equation, f becomes 1, altering the equation to: \( y = \frac{x^2}{4} \). This equation will be pivotal in solving for the dimensions of the headlight.
focus of parabola
The focus of a parabola is significant because it denotes the point where light rays parallel to the axis of symmetry converge after reflection.
For our headlight problem, the light bulb's position at the focus ensures optimal illumination.
Given that the focus for this problem is 1 inch from the vertex, the coordinates become (0, 1). Using this value in the parabola's equation ensures accurate calculations of depths and widths. It also creates a direct link between the geometric shape of our paraboloid and its functional efficiency.
diameter calculation
Calculating the diameter at a specific depth in the paraboloid involves the use of the parabola equation. With the paraboloid's equation as \( y = \frac{x^2}{4} \), we can determine the width of the headlight at any given depth, \( y \).
The required depth is 2 inches, so substituting \( y = 2 \) into our equation: \( 2 = \frac{x^2}{4} \). This leads to \( x^2 = 8 \).
Solving for \( x \), we get \( x = \pm 2\sqrt{2} \). The diameter is the total distance across the opening of the paraboloid, which is twice the x-value: \( 2 \times 2\sqrt{2} = 4\sqrt{2} \approx 5.66 \) inches. This calculation ensures we correctly understand the physical dimensions.
algebraic solving techniques
Algebra is critical in solving geometry-related problems such as this headlight paraboloid. We'll go through some essential algebraic techniques step by step.
  • Setting up the equation: Start with the parabola's equation and insert known values. For our case: \( 2 = \frac{x^2}{4} \).
  • Isolating the variable: Multiply both sides by the denominator to simplify: \( 8 = x^2 \).
  • Solving for the variable: Take square roots of both sides to find \( x \): \( x = \pm \sqrt{8} \). Simplify to: \( x = \pm 2\sqrt{2} \).
  • Final calculation: Widen the solution to find the diameter as double the positive value of \( x \): \( 4\sqrt{2} \).
These steps highlight how algebra transforms geometric figures into solvable numerical forms, offering clear solutions to practical problems.

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

Parametric equations of four plane curves are given. Graph each of them, indicating the orientation. \(\begin{array}{ll}C_{1}: & x(t)=t, \quad y(t)=\sqrt{1-t^{2}} ; \quad-1 \leq t \leq 1 \\ C_{2}: & x(t)=\sin t, \quad y(t)=\cos t ; \quad 0 \leq t \leq 2 \pi \\\ C_{3}: & x(t)=\cos t, \quad y(t)=\sin t ; \quad 0 \leq t \leq 2 \pi \\\ C_{4}: & x(t)=\sqrt{1-t^{2}}, \quad y(t)=t ; \quad-1 \leq t \leq 1\end{array}\)

A radioactive substance has a half-life of 15 years. How long until there is \(40 \%\) of a sample remaining?

Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time the first person, standing at point \(A,\) hears the burst. One second later the second person, standing at point \(B,\) hears the burst. If the person at point \(B\) is due west of the person at point \(A\), and if the display is known to occur due north of the person at point \(A,\) where did the fireworks display occur?

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?

Lamp Shadow The light from a lamp creates a shadow on a wall with a hyperbolic border. Find the equation of the border if the distance between the vertices is 18 inches and the foci are 4 inches from the vertices. Assume the center of the hyperbola is at the origin.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.