/*! 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 69 The reflector of a flashlight is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 69

The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 4 inches and a depth of 1 inch. How far from the vertex should the light bulb be placed?

Short Answer

Expert verified
The light bulb should be placed right at the base of the reflector which corresponds to the vertex of the parabolic surface, hence the height is 0 inch.

Step by step solution

01

Convert the given dimensions to the same units and determine the equation of the parabola

The diameter of the reflector is 4 inches, so the radius is half of the diameter which is 2 inches. The depth of the reflector is 1 inch. So, given the points (2,1) and (0,0), we can create the equation of the parabola. We start with the formula \(y=ax^2\), and substitute the point (2,1) into the equation which gives us \(1=a*2^2\). Solve for a to get the equation of the parabola.
02

Solve for a

Substitute \(x=2, y=1\) into the equation \(y=ax^2\) and solve for a which results in \(a =1/4\). This gives us the equation for the parabola \(y=1/4*x^2\).
03

Find the vertex of the parabola

The vertex of a parabola \(y=ax^2\) is always at (0, 0), so in this context, that's the base of the reflector of the flashlight. In the other words, the light bulb should be placed right on the base of the reflector, which is the vertex of the parabola, hence the height from the vertex to the flashlight bulb is 0 inch.

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!

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

Write an equation for the path of each of the following elliptical orbits. Then use a graphing utility to graph the two ellipses in the same viewing rectangle. Can you see why early astronomers had difficulty detecting that these orbits are ellipses rather than circles? \(\cdot\) Earth's orbit: Length of major axis: 186 million miles Length of minor axis: 185.8 million miles \(\cdot\) Mars's orbit: Length of major axis: 283.5 million miles Length of minor axis: 278.5 million miles

Use a graphing utility to graph the parabolas in Exercises 86鈥�87. Write the given equation as a quadratic equation in y and use the quadratic formula to solve for y. Enter each of the equations to produce the complete graph. $$ y^{2}+10 y-x+25=0 $$

If you are given the standard form of the equation of a parabola with vertex at the origin, explain how to determine if the parabola opens to the right, left, upward, or downward.

Use a graphing utility to graph the equation. Then answer the given question. $$ \begin{aligned} &r=\frac{3}{2+6 \cos \left(\theta+\frac{\pi}{3}\right)} ; \text { How does the graph differ from the }\\\ &\text { graph of } r=\frac{3}{2+6 \cos \theta} ? \end{aligned} $$

In Exercises \(51-60,\) convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$ 49 x^{2}+16 y^{2}+98 x-64 y-671=0 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete 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.