Problem 10 - Advanced Engineering Mathematics

Question

In Problems \(7-10\), find the Jacobian of the transformation \(T\) from the \(u v\) -plane to the \(x y\) -plane. $$ u=\frac{2 x}{x^{2}+y^{2}}, v=\frac{-2 y}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

The Jacobian determinant is \(-\frac{8}{(x^2+y^2)^2}\).

Step 1: Understanding the Transformation

The transformation given is a mapping from variables \((x, y)\) to \((u, v)\). Here, \(u\) and \(v\) are expressed in terms of \(x\) and \(y\) according to the formulas given: \(u = \frac{2x}{x^2 + y^2}\) and \(v = \frac{-2y}{x^2 + y^2}\). The goal is to determine the Jacobian matrix of this transformation, \(T\).

Step 2: Calculating Partial Derivatives

To find the Jacobian matrix, we must compute the partial derivatives of \(u\) and \(v\) with respect to \(x\) and \(y\). We need:\[ \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \text{ and } \frac{\partial v}{\partial y} \]

Step 3: Computing \(\frac{\partial u}{\partial x}\)

Apply the quotient rule: If \(u = \frac{2x}{x^2 + y^2}\), then:\[ \frac{\partial u}{\partial x} = \frac{(2)(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2) - 4x^2}{(x^2+y^2)^2} = \frac{2y^2 - 2x^2}{(x^2+y^2)^2} \]

Step 4: Computing \(\frac{\partial u}{\partial y}\)

Similarly, using the quotient rule for \(u = \frac{2x}{x^2 + y^2}\):\[ \frac{\partial u}{\partial y} = \frac{0 - 2x(2y)}{(x^2 + y^2)^2} = \frac{-4xy}{(x^2 + y^2)^2} \]

Step 5: Computing \(\frac{\partial v}{\partial x}\)

For \(v = \frac{-2y}{x^2 + y^2}\), applying the quotient rule gives:\[ \frac{\partial v}{\partial x} = \frac{0 + 2y(2x)}{(x^2 + y^2)^2} = \frac{4xy}{(x^2 + y^2)^2} \]

Step 6: Computing \(\frac{\partial v}{\partial y}\)

Using the quotient rule for \(v = \frac{-2y}{x^2 + y^2}\):\[ \frac{\partial v}{\partial y} = \frac{-2(x^2 + y^2) + 4y^2}{(x^2 + y^2)^2} = \frac{-2x^2 + 2y^2}{(x^2 + y^2)^2} \]

Step 7: Forming the Jacobian Matrix

The Jacobian matrix \(J\) for the transformation \(T\) is\[J = \begin{bmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix}= \begin{bmatrix}\frac{2y^2 - 2x^2}{(x^2+y^2)^2} & \frac{-4xy}{(x^2 + y^2)^2} \\frac{4xy}{(x^2 + y^2)^2} & \frac{-2x^2 + 2y^2}{(x^2 + y^2)^2}\end{bmatrix}\]

Step 8: Calculating the Determinant of the Jacobian

The determinant of the Jacobian matrix \(J\) is found by:\[ \text{det}(J) = \left(\frac{2y^2 - 2x^2}{(x^2+y^2)^2}\right) \left(\frac{-2x^2 + 2y^2}{(x^2+y^2)^2}\right) - \left(\frac{-4xy}{(x^2 + y^2)^2}\right) \left(\frac{4xy}{(x^2 + y^2)^2}\right) \]Expanding each term and simplifying results in:\[ \frac{(2y^2-2x^2)(-2x^2+2y^2) - (4xy)(4xy)}{(x^2+y^2)^4} = \frac{-8x^4 - 8y^4}{(x^2+y^2)^4} = -\frac{8(x^2+y^2)^2}{(x^2+y^2)^4} = -\frac{8}{(x^2+y^2)^2} \]

Key Concepts

Transformation

In mathematics, a transformation refers to a process that changes the position, size, or shape of a figure in a systematic way. When dealing with coordinate systems, transformations can map points from one plane to another, which is precisely what the problem at hand requires us to do. Here, we're looking at a transformation from the \((x, y)\) plane to the \((u, v)\) plane.

The given transformation is described by the equations:

• \( u = \frac{2x}{x^2 + y^2} \)
• \( v = \frac{-2y}{x^2 + y^2} \)

These equations take the coordinates \((x, y)\) and transform them into new coordinates \((u, v)\). Understanding this transformation is crucial when wanting to determine how certain areas or shapes change as they move from one set of axes to the other. It involves deep understanding of relationships between different coordinate spaces and how algebraic expressions can modify geometric entities.

Partial Derivatives

To find the Jacobian matrix of a transformation, you need to calculate partial derivatives. Partial derivatives are derivatives taken of multivariable functions with respect to one variable, while keeping other variables constant. They are essential in multivariable calculus and help us understand how the function changes.

For the given transformation equations involving \( u \) and \( v \), we calculate the partial derivatives with respect to \( x \) and \( y \):

• \( \frac{\partial u}{\partial x} = \frac{(2)(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2+y^2)^2} \)
• \( \frac{\partial u}{\partial y} = \frac{0 - 2x(2y)}{(x^2 + y^2)^2} = \frac{-4xy}{(x^2 + y^2)^2} \)
• \( \frac{\partial v}{\partial x} = \frac{0 + 2y(2x)}{(x^2 + y^2)^2} = \frac{4xy}{(x^2 + y^2)^2} \)
• \( \frac{\partial v}{\partial y} = \frac{-2(x^2 + y^2) + 4y^2}{(x^2 + y^2)^2} = \frac{-2x^2 + 2y^2}{(x^2 + y^2)^2} \)

Understanding these partial derivatives allows us to assemble them into the Jacobian matrix, which is a core tool in dealing with transformations from one plane to another.

Determinant

The determinant is a key value related to square matrices that gives us essential information about the linear transformation represented by the matrix. For a Jacobian matrix, the determinant can tell us how the transformation scales areas near a certain point.

For the described exercise, the Jacobian matrix \( J \) is:\[J = \begin{bmatrix}\frac{2y^2 - 2x^2}{(x^2+y^2)^2} & \frac{-4xy}{(x^2 + y^2)^2} \\frac{4xy}{(x^2 + y^2)^2} & \frac{-2x^2 + 2y^2}{(x^2+y^2)^2}\end{bmatrix}\]The determinant of the Jacobian matrix is calculated as:\[\text{det}(J) = \left(\frac{2y^2 - 2x^2}{(x^2+y^2)^2}\right) \left(\frac{-2x^2 + 2y^2}{(x^2+y^2)^2}\right) - \left(\frac{-4xy}{(x^2 + y^2)^2}\right) \left(\frac{4xy}{(x^2 + y^2)^2}\right)\]This determinant simplifies to:\[-\frac{8}{(x^2+y^2)^2}\]A determinant of zero would indicate that the transformation squishes everything along at least one dimension, meaning it is not invertible at that point. For non-zero determinants, it provides insight into how the transformation stretches or shrinks areas around each point, and negative sign signifies change in orientation.