/*! 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 4 Consider the mapping \(F: \mathb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 4

Consider the mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(3 y, 2 x) .\) Let \(S\) be the unit circle in \(\mathbf{R}^{2}\) that is, the solution set of \(x^{2}+y^{2}=1\) (a) Describe \(F(S)\). (b) Find \(F^{-1}(S)\)

Short Answer

Expert verified
(a) F(S) is an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). (b) F鈦宦�(S) is an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3).

Step by step solution

01

Apply the mapping F to the unit circle equation

We know the unit circle equation is x虏 + y虏 = 1. Since F maps (x, y) to (3y, 2x), let's substitute x = 2u and y = (1/3)v, where (u, v) are points on F(S). So the equation becomes: (2u)虏 + ((1/3)v)虏 = 1 Step 2: Simplify the equation
02

Simplifying the resulting equation

Now, we simplify the equation we just obtained to find the equation of F(S): (4u虏) + (1/9)v虏 = 1 This is the equation of an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). (a) Hence, F(S) is an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). Step 3: Finding F鈦宦�(S)
03

Find the inverse of the mapping F

To find F鈦宦�(S), we will first find the inverse mapping F鈦宦�, and then apply it to the unit circle equation. In the mapping F, the roles of x and y are switched, and they are multiplied by different constants. To invert the mapping, we need to switch back and divide by the corresponding multipliers. Let F鈦宦�(u,v) = (x, y): x = (1/2)v y = (1/3)u Step 4: Apply the inverse mapping to the unit circle equation
04

Applying F鈦宦� to the unit circle equation

Now that we have the F鈦宦� mapping, we will apply it to the unit circle equation (x虏 + y虏 = 1): ((1/2)v)虏 + ((1/3)u)虏 = 1 Step 5: Simplify the equation
05

Simplifying the resulting equation

Now, let's simplify the equation we got from the previous step to find the equation for F鈦宦�(S): (1/4)v虏 + (1/9)u虏 = 1 This is the equation of an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3). (b) Hence, F鈦宦�(S) is an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3).

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

Let \(A\) be an \(n \times n\) matrix with characteristic polynomial $$ f(t)=(-1)^{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{1} t+a_{0} . $$ Prove that \(f(0)=a_{0}=\operatorname{det}(A)\). Deduce that \(A\) is invertible if and only if \(a_{0} \neq 0\).

A subset \(X\) of a vector space \(V\) is said to be convex if the line segment \(L\) between any two points (vectors) \(P, Q \in X\) is contained in \(X\). (a) Show that the intersection of convex sets is convex; (b) suppose \(F: V \rightarrow U\) is linear and \(X\) is convex. Show that \(F(X)\) is convex.

Consider the zero mapping \(\mathbf{0}: V \rightarrow U\) defined by \(\mathbf{0}(v)=0, \forall v \in V .\) Find the kernel and the image of \(\mathbf{0}.\)

Show that the following mappings are not linear: (a) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=\left(x^{2}, y^{2}\right)\) (b) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+1, y+z)\) (c) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(x y, y)\) (d) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(|x|, y+z)\)

Let $A=\left(\begin{array}{rrr}1 & 1 & -3 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right)\(, let \)\mathrm{T}=\mathrm{L}_{A}\(, and let \)\mathrm{W}$ be the cyclic subspace of \(\mathrm{R}^{3}\) generated by \(e_{1}\). (a) Use Theorem \(5.21\) to compute the characteristic polynomial of \(\mathrm{T}_{\mathrm{W}}\). (b) Show that \(\left\\{e_{2}+\mathrm{W}\right\\}\) is a basis for \(\mathrm{R}^{3} / \mathrm{W}\), and use this fact to compute the characteristic polynomial of \(\overline{\mathrm{T}}\). (c) Use the results of (a) and (b) to find the characteristic polynomial of \(A\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Pure 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.