/*! 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 43 Use set-builder notation to desc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 43

Use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. The region inside the left half of the circle \(r=6 \sin (\theta)\).

Short Answer

Expert verified
The region in set-builder notation is \( S = \{ (r, \theta) \,|\, 0 \leq r \leq 6 \sin(\theta), \, \frac{\pi}{2} \leq \theta \leq \pi \} \).

Step by step solution

01

Understand the Circle Equation

The given equation \( r = 6 \sin(\theta) \) represents a circle in polar coordinates. This is a circle centered at \((3, \frac{\pi}{2})\) with a radius of 3. The full circle is described by rotating through angles from \(\theta = 0\) to \(\theta = \pi\).
02

Determine the Region

The specific region we are interested in is the 'left half' of this circle. In polar coordinates, the left portion corresponds to \( \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \). However, the original instruction refers to the left 'half', which suggests that \( \frac{\pi}{2} \leq \theta \leq \pi \) provides the bounds for the sector in question. The inequality \( 0 \leq r \leq 6 \sin(\theta) \) describes that our region is the area inside the arc and the line \(\theta = \pi\).
03

Use Set-Builder Notation

In set-builder notation, we express the region as the set of all points \((r, \theta)\) that satisfy the specific conditions we've analyzed. Thus, the region is: \[ S = \{ (r, \theta) \,|\, 0 \leq r \leq 6 \sin(\theta), \, \frac{\pi}{2} \leq \theta \leq \pi \} \]. This describes all points in our specified left half of the circle.

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.

Set-Builder Notation
Set-builder notation is a mathematical way to describe a set of elements that satisfy certain conditions. It is a precise method often used in algebra and calculus to define regions or groups of numbers. In this case, we use it to describe a polar region.
Here's how it works:
  • The notation encapsulates the idea of filtering a set of all possible inputs, keeping only those that meet particular conditions.
  • In a polar coordinate system, points are expressed in terms of \(r\) for radial distance and \(\theta\) for the angle.
  • The conditions that bound the region define the set. For example, in this exercise, we define the bounds for \(r\) and \(\theta\).
The expression \( S = \{ (r, \theta) \,|\, 0 \leq r \leq 6 \sin(\theta), \, \frac{\pi}{2} \leq \theta \leq \pi \} \) succinctly captures all the points that fall within the specified sector of the circle.
Circle Equation
In polar coordinates, the equation \( r = 6 \sin(\theta) \) describes a circle. Understanding this notation is crucial for defining regions accurately. Here's a breakdown of how this circle appears and what its parameters mean:
  • The center of the circle in the polar system is located at \( (3, \frac{\pi}{2}) \), indicating it鈥檚 centered along the positive y-axis when visualized in Cartesian terms.
  • The radius of the circle is 3. This gives a direct relation between angles, \(\sin(\theta)\), and distance \(r\).
  • The function \( r = 6 \sin(\theta) \) produces a complete circle as \(\theta\) ranges from 0 to \(\pi\).
It is important to note that this doesn't describe a standard circle centered at the origin but rather one influenced by the sine function, making it a bit more challenging to visualize at first.
Bounding Curves
Bounding curves are the limits that define the extent of a region on a graph. In polar coordinates, these are crucial to determining the area of interest:
  • For this circle described as \( r = 6 \sin(\theta) \), the bounding curve includes all radii \( r \) up to the circle's edge.
  • In this scenario, the boundary curve for the left half is determined by the segment where \( \frac{\pi}{2} \leq \theta \leq \pi \).
  • Bounding curves in this context define not just the extent of \(r\), but also the angular sweep, effectively slicing out a 'half' segment of the full shape.
Understanding these boundary constraints lets us describe the area accurately using set-builder notation, ensuring we capture all relevant points properly within those limits.
Polar Region Description
Describing a region in polar coordinates involves understanding both the radial and angular bounds. In this exercise, we focused on the left half of a circle:
  • The region is delineated by its position within the circle, where \( 0 \leq r \leq 6 \sin(\theta) \) holds true.
  • Angular specification is crucial: this sector is defined from \( \frac{\pi}{2} \leq \theta \leq \pi \), ensuring only the desired portion is included in the description.
  • This bounded area creates a sector, challenging visual understanding compared to traditional Cartesian systems, which necessitates careful translation of conditions into mathematical language.
By carefully setting and combining these parameters, one can fully describe the area they wish to study, model, or utilize, all contained succinctly within the set-builder notation.

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

Convert the equation from polar coordinates into rectangular coordinates. \(r=2 \sec (\theta) \tan (\theta)\)

Use the pair of vectors \(\vec{v}\) and \(\vec{w}\) to find the following quantities.\(\bullet \vec{v} \cdot \vec{w}\) \( \bullet \operatorname{proj}_{\vec{w}}(\vec{v})\) \( \bullet\)The angle \(\theta\) (in degrees) between \(\vec{v}\) and \(\vec{w}\) \( \bullet\) \(\vec{q}=\vec{v}-\operatorname{proj}_{\vec{w}}(\vec{v})\) (Show that \(\left.\vec{q} \cdot \vec{w}=0 .\right)\) $$ \vec{v}=\left\langle\frac{1}{2},-\frac{\sqrt{3}}{2}\right\rangle \text { and } \vec{w}=\left\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right\rangle $$

Find a parametric description for the given oriented curve. the curve \(x=y^{2}-9\) from (0,3) to (-5,-2) . (Shift the parameter so \(t=0\) corresponds to \((0,3) .)\)

Convert the equation from polar coordinates into rectangular coordinates. \(r=4 \cos (\theta)\)

Given any natural number \(n \geq 2,\) the \(n\) complex \(n^{\text {th }}\) roots of the number \(z=1\) are called the \(n^{\text {th }}\) Roots of Unity. In the following exercises, assume that \(n\) is a fixed, but arbitrary, natural number such that \(n \geq 2\). (a) Show that \(w=1\) is an \(n^{\text {th }}\) root of unity. (b) Show that if both \(w\), and \(w_{k}\) are \(n^{\text {th }}\) roots of unity then so is their product \(w, w_{k}\) - (c) Show that if \(w_{j}\) is an \(n^{\text {th }}\) root of unity then there exists another \(n^{\text {th }}\) root of unity \(w_{j}\) such that \(w_{y} w_{y^{\prime}}=1 .\) Hint: If \(w_{j}=\operatorname{cis}(\theta)\) let \(w_{y^{\prime}}=\operatorname{cis}(2 \pi-\theta) .\) You'll need to verify that \(w_{y^{\prime}}=\operatorname{cis}(2 \pi-\theta)\) is indeed an \(n^{\text {th }}\) root of unity.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.