/*! 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 42 Finding a Polar Equation In Exer... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 42

Finding a Polar Equation In Exercises \(39-44\) , find a polar equation for the conic with its focus at the pole and the given vertex or vertices. $$\begin{array}{ll}{\text { Conic }} & {\text { Vertex or Vertices }} \\\ {\text { Ellipse }} & {\left(2, \frac{\pi}{2}\right), \quad\left(4, \frac{3 \pi}{2}\right)}\end{array}$$

Short Answer

Expert verified
The polar equation for the ellipse with the given vertices is \(r = \frac{2}{1 - \frac{\sqrt{3}}{2}\cos{\theta}}\)

Step by step solution

01

Identify the vertices

The given vertices are \((2, \frac{\pi}{2})\) and \((4, \frac{3\pi}{2})\). The distance from pole to each vertex represents the semi-major axis (a) and semi-minor axis (b). Since the ellipse is symmetric about the polar axis, \(a = 4\) and \(b = 2\)
02

Calculate Eccentricity

For an ellipse, the eccentricity (\(e\)) can be calculated based on the semi-minor and semi-major axes: \(e^2 = 1 - \frac{b^2}{a^2}\). With \(a = 4\) and \(b = 2\), we have: \(e^2 = 1 - \frac{2^2}{4^2} = 1 - \frac{1}{4} = \frac{3}{4}\) Therefore, we have \(e = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\)
03

Formulate Polar Equation

Once the values of \(a\) and \(e\) are determined, we can plug these values into the polar equation of an ellipse : \(r = \frac{a(1-e^2)}{1-e\cos{\theta}}\). Thus we have: \(r = \frac{4(1-\frac{3}{4})}{1-\frac{\sqrt{3}}{2}\cos{\theta}} = \frac{4}{2 - \sqrt{3}\cos{\theta}}\)
04

Simplify the Polar Equation

Finally, we simplify the fraction to get the final polar equation: \(r = \frac{2}{1 - \frac{\sqrt{3}}{2}\cos{\theta}}\) This is the polar equation for the given ellipse with the focus at the pole and the specified vertices.

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.

Conic Sections
Conic sections are the curves formed by the intersection of a plane with a cone. These curves include circles, ellipses, parabolas, and hyperbolas. Each of these shapes has unique properties defined by their geometric and algebraic equations. An ellipse, which is the focus of our exercise, is one of these conic sections.

Ellipses have two main characteristics: they are closed, oval-shaped curves, and they have two foci. When the plane cuts through the cone in such a way that it doesn鈥檛 intersect the base, the resulting shape is an ellipse. The properties of these conic sections help us identify and derive their respective equations in different forms, such as polar coordinates.
Ellipse Properties
An ellipse is defined by several key properties. It is the set of all points for which the sum of the distances to two fixed points, called foci, is constant. This creates a smooth, oval-like shape.

Some key properties of ellipses include:
  • Vertices: These are the points on the ellipse that lie along the major axis, the longest diameter of the ellipse.
  • Axes: The major axis is the longest line that runs through the center, while the minor axis is perpendicular to it and shorter.
  • Center: The midpoint between the foci where the major and minor axes intersect.
By understanding these properties, we can describe the ellipse not only in Cartesian coordinates but also in polar form, as shown in the exercise.
Eccentricity
Eccentricity is a measure of how much an ellipse deviates from being a circle. It is denoted by the symbol \( e \). For a perfect circle, the eccentricity is 0. For an ellipse, the eccentricity lies between 0 and 1.

The formula for calculating the eccentricity of an ellipse is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

If \( e \) is closer to 0, the ellipse is more circular. If \( e \) is closer to 1, the ellipse is elongated. In our exercise, the eccentricity was determined to be \( \frac{\sqrt{3}}{2} \), showcasing a less circular, more elongated shape.
Semi-major Axis
The semi-major axis is one of the principal axes of an ellipse, acting as half of the longest diameter. It extends from the center to one of the vertices along the major axis.

In mathematical terms, it is the average of the maximum distance across an ellipse, measured through its foci. This crucial measurement determines other properties of the ellipse and plays a vital role in its equations.

Understanding the semi-major axis allows us to:
  • Identify the shape and scale of the ellipse.
  • Calculate the eccentricity with the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \).
  • Formulate polar equations for the ellipse.
In our exercise, the semi-major axis \( a \) is 4. This information is essential for constructing the ellipse's polar 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

Identifying and Sketching a Conic In Exercises \(13-22\) , find the eccentricity and the distance from the pole to the directrix of the conic. Then identify the conic and sketch its graph. Use a graphing utility to confirm your results. $$r=\frac{4}{1+\cos \theta}$$

Finding a Polar Equation Find a polar equation for the ellipse with the following characteristics. Focus: \((0,0)\) Eccentricity: \(e=\frac{1}{2}\) Directrix: \(r=4 \sec \theta\)

Finding the Area of a Polar Region Between Two Curves In Exercises \(37-44\) , use a graphing utility to graph the polar equations. Find the area of the given region analytically. Inside \(r=2 \cos \theta\) and outside \(r=1\)

Approximating Area Consider the circle \(r=8 \cos \theta\) (a) Find the area of the circle. (b) Complete the table for the areas \(A\) of the sectors of the circle between \(\theta=0\) and the values of \(\theta\) in the table. (c) Use the table in part (b) to approximate the values of \(\theta\) for which the sector of the circle composes \(\frac{1}{4}, \frac{1}{2}\) and \(\frac{3}{4}\) of the total area of the circle. (d) Use a graphing utility to approximate, to two decimal places, the angles \(\theta\) for which the sector of the circle composes \(\frac{1}{4}, \frac{1}{2},\) and \(\frac{3}{4}\) of the total area of the circle (e) Do the results of part (d) depend on the radius of the circle? Explain..

Area of a Region In Exercises \(55-58\) , use the integration capabilities of a graphing utility to approximate the area of the region bounded by the graph of the polar equation. $$r=\frac{2}{7-6 \sin \theta}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.