Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the image of the rectangle with the given corners and find the Jacobian of the transformation. $$ x=2 u+3 v, y=u-v ;(0,0),(3,0),(3,1),(0,1) $$

Step-by-Step Solution

Answer

The image is (0,0), (6,3), (9,2), (3,-1); the Jacobian determinant is -5.

Step 1: Define the Transformation

The given transformation equations are \( x = 2u + 3v \) and \( y = u - v \). We need to find how these transform the given rectangle's corners.

Step 2: Transform Corner Points

Transform each corner point from \((u, v)\) to \((x, y)\):- For \((u, v) = (0,0)\), \( x = 2(0) + 3(0) = 0 \), \( y = 0 - 0 = 0 \). Thus the image is \((0, 0)\).- For \((u, v) = (3,0)\), \( x = 2(3) + 3(0) = 6 \), \( y = 3 - 0 = 3 \). Thus the image is \((6, 3)\).- For \((u, v) = (3,1)\), \( x = 2(3) + 3(1) = 9 \), \( y = 3 - 1 = 2 \). Thus the image is \((9, 2)\).- For \((u, v) = (0,1)\), \( x = 2(0) + 3(1) = 3 \), \( y = 0 - 1 = -1 \). Thus the image is \((3, -1)\).

Step 3: State the Image Rectangle

The image points of the corners of the rectangle are \((0,0), (6,3), (9,2), (3,-1)\).

Step 4: Find the Jacobian Matrix

The Jacobian matrix \( J \) of the transformation is derived from the partial derivatives:\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\]Calculate the partial derivatives:\(\frac{\partial x}{\partial u} = 2, \quad \frac{\partial x}{\partial v} = 3 \\frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1\)Thus,\[ J = \begin{bmatrix} 2 & 3 \ 1 & -1 \end{bmatrix} \]

Step 5: Calculate the Determinant of the Jacobian

The determinant of the Jacobian \( J \) is given by:\[det(J) = \left| \begin{matrix} 2 & 3 \ 1 & -1 \end{matrix} \right| = (2)(-1) - (3)(1) = -2 - 3 = -5\]This determinant represents the factor by which area is scaled by the transformation.

Key Concepts

Jacobian matrix

The Jacobian matrix plays a crucial role in understanding coordinate transformations, such as converting a rectangle's position from one coordinate system to another. It consists of the partial derivatives of the transformation equations. In this exercise, the transformation equations are given by \( x = 2u + 3v \) and \( y = u - v \). The Jacobian matrix \( J \) is then formed by calculating the partial derivatives of \( x \) and \( y \) with respect to the variables \( u \) and \( v \). Let's break it down:- Partial derivative of \( x \) with respect to \( u \) is \( \frac{\partial x}{\partial u} = 2 \).- Partial derivative of \( x \) with respect to \( v \) is \( \frac{\partial x}{\partial v} = 3 \).- Partial derivative of \( y \) with respect to \( u \) is \( \frac{\partial y}{\partial u} = 1 \).- Partial derivative of \( y \) with respect to \( v \) is \( \frac{\partial y}{\partial v} = -1 \).Combining these, the Jacobian matrix is:\[J = \begin{bmatrix} 2 & 3 \ 1 & -1 \end{bmatrix}\]

Determinant of Jacobian

The determinant of the Jacobian matrix provides valuable information about the transformation's effect on areas and orientation. It equates to the factor by which the area is scaled when transforming a region from one coordinate space to another. In our example, the Jacobian matrix \( J \) is \[\begin{bmatrix} 2 & 3 \ 1 & -1 \end{bmatrix}\]We calculate the determinant of this matrix as follows:- Multiply the top-left and bottom-right elements: \( 2 \times (-1) = -2 \)- Multiply the top-right and bottom-left elements: \( 3 \times 1 = 3 \)- Subtract the second product from the first: \( -2 - 3 = -5 \)Therefore, the determinant of the Jacobian, \( det(J) \), is \(-5\). This negative value indicates not only the scaling factor of 5 but also that the transformation involves a change in orientation.

Transformation of coordinates

Transformation of coordinates involves altering the system used to define a point or region in space. In this exercise, we observe how a rectangle with specific corners in \( (u, v) \) coordinates transforms into a new shape in \( (x, y) \) coordinates using the given transformation functions \( x = 2u + 3v \) and \( y = u - v \). Transformation of coordinates allows us to:- Map points from one space to another.- Understand how shapes and sizes alter in different coordinate systems.For each corner of the rectangle, we calculate new \( x \) and \( y \) values:- \( (u, v) = (0,0) \) becomes \( (0,0) \) in \( (x, y) \) coordinates.- \( (u, v) = (3,0) \) becomes \( (6,3) \).- \( (u, v) = (3,1) \) becomes \( (9,2) \).- \( (u, v) = (0,1) \) becomes \( (3,-1) \).This transformation modifies the shape and scale of the original rectangle.

Partial derivatives

In calculus, partial derivatives represent how a function changes as one of the variables changes, while other variables remain fixed. They are essential in constructing the Jacobian matrix, indicating how small changes in the \( u \) and \( v \) coordinates affect the \( x \) and \( y \) coordinates. Here's how we apply it to this transformation:- The partial derivative of \( x = 2u + 3v \) with respect to \( u \) is \( \frac{\partial x}{\partial u} = 2 \), showing how \( x \) changes with \( u \).- The partial derivative of \( x = 2u + 3v \) with respect to \( v \) is \( \frac{\partial x}{\partial v} = 3 \), reflecting the influence of \( v \) on \( x \).- For \( y = u - v \), \( \frac{\partial y}{\partial u} = 1 \) and \( \frac{\partial y}{\partial v} = -1 \), detailing how \( y \) shifts with changes in \( u \) and \( v \) respectively.These derivatives are the components of the Jacobian matrix, aiding in quantifying the transformation's effect on coordinates.