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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:\[ \text{Diameter} = 2 \times 2\sqrt{2} = 4\sqrt{2} \approx 5.66 \text{ inches} \].

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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.

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