91Ó°ÊÓ

For a function to be classified as continuously differentiable, it must meet certain criteria. Specifically, the function's derivatives must not only exist but also be continuous. In the context of transformations, this means

• The function must be smooth without any abrupt changes or breaks in its curve.
• The derivatives of the function must be calculated and should not have any discontinuity.

This smoothness guarantees that the transformation behaves predictably when moving from one set of variables to another. In our exercise, verifying that the function \( T(u, v) = (x, y) \) is continuously differentiable is the first step in ensuring it qualifies as a \( C^1 \) transformation. This classification helps in understanding the behavior of transformations across different regions.

The Jacobian matrix is a crucial tool for understanding transformations. It captures information about how changes in input affect changes in output within the transformation. Mathematically, the Jacobian matrix for a transformation \( T(u, v) = (x, y) \) is constructed using the partial derivatives:\[J_{T} = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}\]For the given problem, we calculated:\[J_{T} = \begin{pmatrix} 1 & -1 \1 & 1 \end{pmatrix}\]

• \( \frac{\partial x}{\partial u} = 1 \)
• \( \frac{\partial x}{\partial v} = -1 \)
• \( \frac{\partial y}{\partial u} = 1 \)
• \( \frac{\partial y}{\partial v} = 1 \)

The determinant of this matrix, which is non-zero, confirms \( T \) has continuous derivatives, supporting its classification as a \( C^1 \) transformation. The Jacobian also offers insights into how areas in the input space are stretched or compressed in the output space.

The image of a transformation refers to the set of all points that can be obtained by applying the transformation to every point in the original region. In our exercise:

• The original region \( S \) is the unit square \( \{ (u, v) \mid 0 \leq u \leq 1, 0 \leq v \leq 1 \} \).
• Applying the transformation \( T(u, v) = (u - v, u + v) \) results in a new set of points in the \( (x, y) \) space.

Specifically, the function \( T \) maps the four vertices of \( S \) to points:

• \( (0, 0) \rightarrow (0, 0) \)
• \( (1, 0) \rightarrow (1, 1) \)
• \( (0, 1) \rightarrow (-1, 1) \)
• \( (1, 1) \rightarrow (0, 2) \)

The resulting image, \( R \), forms a parallelogram defined by these vertices in the \( (x, y) \) plane.

Region mapping describes how the shape and size of a region change under a transformation. Using our specific transformation \( T \):

• The original region \( S \) is a square.
• Through the transformation \( T(u, v) = (x, y) = (u-v, u+v) \), the square becomes a different shape.

This exercise specifically results in a parallelogram. To find the new shape, we calculated the images of the vertices of the square. After transformation, these vertices form the corners of the parallelogram:

• \( (0, 0), (1, 1), (-1, 1), (0, 2) \)

Connecting these points gives a clear visualization of the region \( R \). Understanding how transformations map regions helps in visualizing geometric transformations and analyzing how areas are altered by mathematical functions.