Problem 3 - Thomas Calculus

Question

\begin{equation} \begin{array}{l}{\text { a. Solve the system }} \\ {u=3 x+2 y, \quad v=x+4 y} \\\ {\text { for } x \text { and } y \text { in terms of } u \text { and } v . \text { Then find the value of the }} \\ {\text { Jacobian } \partial(x, y) / \partial(u, v) .} \\ {\text { b. Find the image under the transformation } u=3 x+2 y} \\ {v=x+4 y \text { of the triangular region in the } x y \text { -plane bounded }}\\\\{\text { by the } x \text { -axis, the } y \text { -axis, and the line } x+y=1 . \text { Sketch the }} \\ {\text { transformed region in the } u v \text { -plane. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The Jacobian determinant is 0.10, and the transformed triangle in uv-plane has vertices at (0,0), (3,1), and (2,4).

Step 1: Express x and y in terms of u and v

Given the equations:\[\begin{aligned}u & = 3x + 2y \v &= x + 4y \end{aligned}\]we need to solve this system for \(x\) and \(y\) in terms of \(u\) and \(v\). First, rewrite the system in matrix form:\[\begin{bmatrix}3 & 2 \ 1 & 4\end{bmatrix} \begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}u \ v\end{bmatrix}\]To solve for \(x\) and \(y\), we find the inverse of the coefficient matrix, if it exists.

Step 2: Calculate the inverse of the matrix

The matrix to invert is \(\begin{bmatrix}3 & 2 \ 1 & 4\end{bmatrix}\). Compute its determinant:\[det = 3 \cdot 4 - 1 \cdot 2 = 12 - 2 = 10\]Since the determinant is non-zero, the inverse exists. The inverse is:\[\frac{1}{10}\begin{bmatrix}4 & -2 \ -1 & 3\end{bmatrix} = \begin{bmatrix}0.4 & -0.2 \ -0.1 & 0.3\end{bmatrix}\]

Step 3: Express x and y using the inverse matrix

Multiply the inverse matrix by the column vector \([u \, v]^T\):\[\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}0.4 & -0.2 \ -0.1 & 0.3\end{bmatrix} \begin{bmatrix}u \ v\end{bmatrix}\]\[x = 0.4u - 0.2v\]\[y = -0.1u + 0.3v\]

Step 4: Find the Jacobian determinant

The Jacobian \(\frac{\partial(x, y)}{\partial(u, v)}\) is the determinant of the matrix formed by the partial derivatives of \(x\) and \(y\):\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}\]Calculate the partial derivatives:\[\frac{\partial x}{\partial u} = 0.4, \quad \frac{\partial x}{\partial v} = -0.2\]\[\frac{\partial y}{\partial u} = -0.1, \quad \frac{\partial y}{\partial v} = 0.3\]The Jacobian is:\[J = \begin{vmatrix} 0.4 & -0.2 \ -0.1 & 0.3 \end{vmatrix} = (0.4)(0.3) - (-0.2)(-0.1) = 0.12 - 0.02 = 0.10\]

Step 5: Identify the triangular region in the xy-plane

In the \(xy\)-plane, we have the triangular region defined by the points where \(x = 0\), \(y = 0\), and \(x + y = 1\). This is a triangle with vertices at \( (0,0) \), \( (1,0) \), and \( (0,1) \).

Step 6: Transform the triangular region to uv-plane

Using the transformation equations, calculate the points in the \(uv\)-plane:- For \((0,0)\): \[u = 3(0) + 2(0) = 0, \quad v = 0 + 4(0) = 0\]- For \((1,0)\): \[u = 3(1) + 2(0) = 3, \quad v = 1 + 4(0) = 1\]- For \((0,1)\): \[u = 3(0) + 2(1) = 2, \quad v = 0 + 4(1) = 4\]The transformed triangle in the \(uv\)-plane has vertices at \((0,0)\), \((3,1)\), and \((2,4)\).

Step 7: Sketch the transformed region

The transformed triangular region in the \(uv\)-plane would have its vertices at \((0,0)\), \((3,1)\), and \((2,4)\). Draw lines connecting these points to form a triangle. The region is a non-standard triangle due to the scaling and translation of the original triangle defined by the linear transformation.

Key Concepts

Jacobian Determinant

The Jacobian determinant is an essential tool in vector calculus. It represents how a function transforms volume near a given point. In our context, it's about understanding how the transformation involving variables shifts.
The transformation equations given were:

• \(u = 3x + 2y\)
• \(v = x + 4y\)

To find this Jacobian, partial derivatives of the new variables with respect to the old variables must be calculated. This is laid out as a matrix with rows and columns:

• \(\frac{\partial x}{\partial u}, \frac{\partial x}{\partial v}\)
• \(\frac{\partial y}{\partial u}, \frac{\partial y}{\partial v}\)

For this problem, the determinant of that matrix came out to 0.10. In simpler terms, the Jacobian determinant helps us determine the rate at which area transforms under a coordinate change. Here, a determinant of 0.10 conveys how areas shrink under this specific transformation.

Inverse Matrix

An inverse matrix plays a pivotal role in solving systems of linear equations. When you have a set of equations like we did here, written as a matrix equation, you often seek an inverse.
Here, the system \(\begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} u \ v \end{bmatrix}\) defines the transformation. To revert or solve for original terms, we need its inverse.
The steps involve:

• Calculating the determinant.
• Ensuring it's non-zero, signaling that an inverse exists.
• Constructing the inverse using specific formulas, like rearranging matrix terms adjusted by the determinant.

Finding this inverse, like \(\begin{bmatrix} 0.4 & -0.2 \-0.1 & 0.3 \end{bmatrix}\), translates into finding back-transformed values of \(x\) and \(y\) when provided \(u\) and \(v\). The ability to switch back-and-forth between these forms is crucial in linear transtformation scenarios.

Matrix Algebra

Matrix algebra streamlines computations in linear transformations. Given a transformation described through equations, using matrices aids in representing and solving them efficiently.
In this exercise, the initial formulation \(\begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}\) acts as the transformation matrix. It organizes coefficients systematically in compact notation.

• Matrix multiplication bundles related equations.
• The determinant informs about matrix invertibility.
• Inverse matrices solve for variables succinctly.

These matrices simplify complex calculations. They allow handling transformations systematically, leveraging properties like associativity and distributivity found in regular arithmetic but extended to higher dimensions. Such algebra becomes crucial, especially in scenarios of multi-variable transformations.

Coordinate Transformation

Coordinate transformation is the conversion between two different systems of coordinates. In our case, transitioning from \((x, y)\) to \((u, v)\).
It entails using equations that relate one set of coordinates to another. This lets us understand the impact of transformations on geometric objects like regions or shapes.

• The triangle in the \(xy\)-plane had vertices \((0, 0)\), \((1, 0)\), and \((0, 1)\).
• Solving the equations transformed its vertices to \((0, 0), (3, 1), (2, 4)\) in the \(uv\)-plane.

Graphically, what started simple can transform unpredictably, like stretching or squeezing. Identifying this transformed area tells us about the scaled or changed nature of such a region under linear transformations. Understanding transformations is vital in scenarios involving shifts, rotations, or scalings in various applications.