/*! 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 12 Let \(S=\\{(u, v): 0 \leq u \leq... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 12

Let \(S=\\{(u, v): 0 \leq u \leq 1\) \(0 \leq v \leq 1\\}\) be a unit square in the uv-plane. Find the image of \(S\) in the xy-plane under the following transformations. $$T: x=v \sin \pi u, y=v \cos \pi u$$

Short Answer

Expert verified
Answer: After the transformation \(T\), the image of the unit square S is a triangle with vertices at the points (0,0), (0,1), and (1,-1) in the xy-plane.

Step by step solution

01

1. Analyze the transformation T

The given transformation T is defined as: $$ T: x = v \sin \pi u, \quad y = v \cos \pi u $$ This transformation maps points in the uv-plane to points in the xy-plane using the sine and cosine functions. To find the image of the unit square S, we need to apply this transformation to each of the vertices of S.
02

2. List the vertices of the unit square S

The unit square S is defined by the following set of points in the uv-plane: $$ S = \{(u, v): 0 \leq u \leq 1, 0 \leq v \leq 1\} $$ The vertices of the unit square S are the points: $$ (0,0), \quad (0,1), \quad (1,0), \quad (1,1) $$
03

3. Apply the transformation T to each vertex

Next, we plug each vertex of S into the transformation T to get the corresponding points in the xy-plane: - For the vertex (0,0): $$ x = 0 \sin (\pi \cdot 0) = 0, \quad y = 0 \cos (\pi \cdot 0) = 0 $$ So, vertex (0,0) maps to point (0,0) in the xy-plane. - For the vertex (0,1): $$ x = 1 \sin (\pi \cdot 0) = 0, \quad y = 1 \cos (\pi \cdot 0) = 1 $$ So, vertex (0,1) maps to point (0,1) in the xy-plane. - For the vertex (1,0): $$ x = 0 \sin (\pi \cdot 1) = 0, \quad y = 0 \cos (\pi \cdot 1) = 0 $$ So, vertex (1,0) maps to point (0,0) in the xy-plane. - For the vertex (1,1): $$ x = 1 \sin (\pi \cdot 1) = 1, \quad y = 1 \cos (\pi \cdot 1) = -1 $$ So, vertex (1,1) maps to point (1,-1) in the xy-plane.
04

4. Describe the image of the unit square S in the xy-plane

The image of the unit square S under the transformation T is given by the following set of points in the xy-plane: $$ (0,0), \quad (0,1), \quad (0,0), \quad (1,-1) $$ This shape in the xy-plane is a triangle with vertices at the points (0,0), (0,1), and (1,-1).

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.

Unit Square
A unit square is a fundamental concept in geometry and mathematics. It refers to a square whose sides are each 1 unit long. This makes the total area of the unit square equal to 1 square unit.
In this exercise, the unit square is defined on the uv-plane, represented by the set of points \((u, v)\) where both \(u\) and \(v\) range from 0 to 1.
This results in four essential vertices that define the boundaries of the square: \( (0,0), (0,1), (1,0), ext{and} (1,1) \).

Understanding the unit square is crucial as it serves as the starting shape, which will undergo transformations. These transformations change its position or shape in a different coordinate system, in this case, the xy-plane.
Image in XY-plane
The concept of 'image in the XY-plane' refers to where points from another plane have been mapped to, after a transformation is applied. In this exercise, we are transforming the unit square from the uv-plane to another plane called the xy-plane using specific functions.
The transformation, given by \(T: x = v \sin \pi u, y = v \cos \pi u\), maps points from the uv-coordinate system to a new set of points in the xy-coordinate system.

To understand what happens during this transformation, we substitute the vertices of the unit square into the transformation equations:
  • The vertex \( (0,0) \) maps to \( (0,0) \) in the xy-plane.
  • The vertex \( (0,1) \) maps to \( (0,1) \) in the xy-plane.
  • The vertex \( (1,0) \) stays at \( (0,0) \) in the xy-plane.
  • The vertex \( (1,1) \) moves to \( (1,-1) \) in the xy-plane.

This results in a shape resembling a triangle in the xy-plane with vertices at \( (0,0), (0,1), ext{and} (1,-1) \).
Sine and Cosine Functions
The sine and cosine functions are two of the most fundamental trigonometric functions.
They describe the relationship between the angles and sides of a right triangle, but in this exercise, they are the components of the transformation of the unit square from the uv-plane to the xy-plane.

Here, the functions are used as part of a transformation:
  • \( x = v \sin \pi u \)
  • \( y = v \cos \pi u \)
These transformations indicate that each point \( (u,v) \) on the unit square is mapped using the sine and cosine of the angle \( \pi u \).

The periodic nature of sine and cosine functions, which oscillate between -1 and 1, is crucial. They help in stretching and rotating the image of the unit square in the xy-plane, resulting in changes such as the flip or rotation that yields the triangular shape of the final image.
Understanding these functions allows us to predict and explain how shapes will transform in different coordinate systems.

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

The following table gives the density (in units of \(\mathrm{g} / \mathrm{cm}^{2}\) ) at selected points of a thin semicircular plate of radius 3. Estimate the mass of the plate and explain your method. $$\begin{array}{|c|c|c|c|c|c|} \hline & \boldsymbol{\theta}=\mathbf{0} & \boldsymbol{\theta}=\boldsymbol{\pi} / \boldsymbol{4} & \boldsymbol{\theta}=\boldsymbol{\pi} / \boldsymbol{2} & \boldsymbol{\theta}=\boldsymbol{3} \pi / \boldsymbol{4} & \boldsymbol{\theta}=\boldsymbol{\pi} \\ \hline \boldsymbol{r}=\mathbf{1} & 2.0 & 2.1 & 2.2 & 2.3 & 2.4 \\ \hline \boldsymbol{r}=\mathbf{2} & 2.5 & 2.7 & 2.9 & 3.1 & 3.3 \\ \hline \boldsymbol{r}=\mathbf{3} & 3.2 & 3.4 & 3.5 & 3.6 & 3.7 \\ \hline \end{array}$$

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A tetrahedron is bounded by the coordinate planes and the plane \(x / a+y / a+z / a=1 .\) What are the coordinates of the center of mass?

Use integration to show that the circles \(r=2 a \cos \theta\) and \(r=2 a \sin \theta\) have the same area, which is \(\pi a^{2}\)

Changing order of integration If possible, write iterated integrals in cylindrical coordinates for the following regions in the specified orders. Sketch the region of integration. The solid outside the cylinder \(r=1\) and inside the sphere \(\rho=5\) for \(z \geq 0,\) in the orders \(d z d r d \theta, d r d z d \theta,\) and \(d \theta d z d r\)

Consider the thin constant-density plate \(\\{(r, \theta): 0

See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

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.