Chapter 14: Problem 9
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 $$T: x=u^{2}-v^{2}, y=2 u v$$
Short Answer
Expert verified
Answer: The image of the unit square in the xy-plane is a parallelogram.
Step by step solution
01
Analyzing the transformation
The transformation is given by:
- \(x = u^2 - v^2\)
- \(y = 2uv\)
Let's apply this transformation to each corner of the unit square.
02
Finding the corner points after the transformation
The unit square in the uv-plane has its corner points at (0, 0), (0, 1), (1, 0) and (1, 1).
1. For the point (0, 0):
- \(x = 0^2 - 0^2 = 0\)
- \(y = 2(0)(0) = 0\)
The transformed point is (0, 0).
2. For the point (0, 1):
- \(x = 0^2 - 1^2 = -1\)
- \(y = 2(0)(1) = 0\)
The transformed point is (-1, 0).
3. For the point (1, 0):
- \(x = 1^2 - 0^2 = 1\)
- \(y = 2(1)(0) = 0\)
The transformed point is (1, 0).
4. For the point (1, 1):
- \(x = 1^2 - 1^2 = 0\)
- \(y = 2(1)(1) = 2\)
The transformed point is (0, 2).
So, the corner points of the image of \(S\) in the xy-plane are (0, 0), (-1, 0), (1, 0) and (0, 2).
03
Drawing the transformed image in the xy-plane
Using the transformed corner points, we can sketch the image of the unit square in the xy-plane. The image is a parallelogram with points: (0, 0), (-1, 0), (1, 0) and (0, 2).
To summarize, the image of the unit square \(S\) in the xy-plane under the transformation \(T: x = u^2 - v^2, y = 2uv\) is a parallelogram with corner points (0, 0), (-1, 0), (1, 0) and (0, 2).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Square
A unit square is a specific type of geometric shape characterized by having each of its sides equal to 1 unit. This means the area of the unit square is precisely 1 square unit. In the context of coordinate planes, such as the uv-plane in this exercise, a unit square can be defined by the set of points where each coordinate lies between 0 and 1 inclusive. Thus, the boundaries of the unit square in the uv-plane are:
- 0 ≤ u ≤ 1
- 0 ≤ v ≤ 1
Image in Plane
The image of a shape upon transformation onto a different plane is essentially how the shape appears after it undergoes a specified transformation. When discussing the transformation of coordinates, the focus is on how the positions of the points in one plane (uv-plane in this instance) reappear in another plane (xy-plane here).
For the given transformation:
For the given transformation:
- \(x = u^2 - v^2\)
- \(y = 2uv\)
Corner Points
Corner points are the vertices of a shape that indicate its size and position in a coordinate plane. For transformations, they are critical since changing these points redefines the entire shape in its new context. In our exercise, the unit square has corner points at:
- (0, 0)
- (0, 1)
- (1, 0)
- (1, 1)
- (0, 0) remains (0, 0) because both coordinates are zero.
- (0, 1) transforms to (-1, 0) as \(x = 0^2 - 1^2 = -1\) and \(y = 2 \times 0 \times 1 = 0\).
- (1, 0) transforms to (1, 0) since \(x = 1^2 - 0^2 = 1\) and \(y = 2 \times 1 \times 0 = 0\).
- (1, 1) changes to (0, 2) as \(x = 1^2 - 1^2 = 0\) and \(y = 2 \times 1 \times 1 = 2\).
Parallelogram
A parallelogram is a four-sided figure in which opposite sides are parallel, and it can be viewed as a skewed rectangle. When we transform a unit square on the uv-plane, what results often is not another square. In this specific transformation, the image turns into a parallelogram.
The transformed corner points, which help define the shape of this parallelogram in the xy-plane, are (0, 0), (-1, 0), (1, 0), and (0, 2).
Here's how they connect:
Here's how they connect:
- Points (0, 0) and (-1, 0) form one side, and (0, 0) and (1, 0) form another.
- The segment from (-1, 0) to (0, 2) and from (1, 0) to (0, 2) creates the skewed edges.