Problem 12
Question
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. $$T: x=v \sin \pi u, y=v \cos \pi u$$
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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.
2Step 2: 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:
$$
(0,0), \quad (0,1), \quad (1,0), \quad (1,1)
$$
3Step 3: 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.
4Step 4: 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).
Key Concepts
Unit SquareImage in XY-planeSine and Cosine Functions
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.
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:
This results in a shape resembling a triangle in the xy-plane with vertices at \( (0,0), (0,1), ext{and} (1,-1) \).
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:
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.
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:
- \( x = v \sin \pi u \)
- \( y = v \cos \pi u \)
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.
Other exercises in this chapter
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