Problem 24 - Calculus Early Transcendentals

Question

Solve the following relations for \(x\) and \(y,\) and compute the Jacobian \(J(u, v)\) $$u=x y, v=x$$

Step-by-Step Solution

Answer

Question: Find the Jacobian of the transformation from (x, y) to (u, v) for the following relations: u = xy and v = x. Answer: The Jacobian, J(u,v), of the given transformation is: J(u,v) = -1/v.

Step 1: Solve the given relations for x and y

From the given relations, we have: $$u=xy \quad \text{and} \quad v=x$$ Since \(v=x\), we can rewrite the first relation as: $$u=xy = vy$$ Now solve for \(y\), we get: $$y=\frac{u}{v}$$ We have found the expressions for \(x\) and \(y\) in terms of \(u\) and \(v\): $$x = v \quad \text{and} \quad y=\frac{u}{v}$$

Step 2: Compute the Jacobian

To compute the Jacobian \(J(u,v)\), we first find the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\): $$\frac{\partial x}{\partial u} = 0 \quad \text{and} \quad \frac{\partial x}{\partial v} = 1$$ $$\frac{\partial y}{\partial u} = \frac{1}{v} \quad \text{and} \quad \frac{\partial y}{\partial v} = -\frac{u}{v^2}$$ Now, the Jacobian is given by the determinant of the matrix formed by these partial derivatives: $$J(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ \frac{1}{v} & -\frac{u}{v^2} \end{vmatrix}$$ $$J(u,v)=\left(0\right)\left(-\frac{u}{v^2}\right) - \left(1\right)\left(\frac{1}{v}\right) = -\frac{1}{v}$$ The Jacobian \(J(u, v)\) of the given transformation is: $$J(u,v)=-\frac{1}{v}$$

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus, helping us understand how a function changes as we modify one variable at a time, while keeping other variables constant. When dealing with functions of multiple variables, such as the problem in our exercise where we have functions for both \(x\) and \(y\) in terms of \(u\) and \(v\), partial derivatives allow us to see the sensitivity of outputs in relation to each specific input variable.

For a function \(f(x, y)\), the partial derivative with respect to \(x\), denoted \( \frac{\partial f}{\partial x} \), shows how \(f\) changes as \(x\) varies while \(y\) remains unchanged. Similarly, \( \frac{\partial f}{\partial y} \) gives us the rate of change of \(f\) with regards to \(y\), keeping \(x\) constant.

In the solution note, these partial derivatives were used to form a matrix that leads to the Jacobian. Calculating partial derivatives is key in determining how mappings behave, crucial when these mathematical mappings represent, for example, transformations of coordinates or transformations in physical systems.

Determinant of a Matrix

A determinant is a special number that can be calculated from a square matrix. It offers us vital information about the matrix properties, such as whether a matrix is invertible and the volume scaling factor in transformations. In the context of the Jacobian, it helps signify the rate of change of variables as affected by the transformation.

To compute the determinant, especially for a 2x2 matrix, you use the simple rule: if you have a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( ad - bc \). For our exercise, this calculation showed the relationship between the partial derivatives of \(x\) and \(y\) with \(u\) and \(v\).

Understanding the determinant helps visualize how transformations stretch or compress areas, indicating whether they maintain orientation or flip it. Hence, in analytical models or graphical transformations, the determinant is a stepping stone in understanding and interpreting data changes.

Transformation

In mathematical terms, a transformation refers to operations that change the position, size, or shape of an object. Transformations can map coordinates from one system to another, alter functions, or adjust geometric figures. In our exercise, we express a transformation through functions where \(u = xy\) and \(v = x\).

By solving these, we found expressions of \(x\) and \(y\) as functions of \(u\) and \(v\). This reverse operation is key in understanding transformations, as it helps in returning to original forms or exploring how variables are interconnected in their new form.

In analytic geometry or other fields, transformations allow us to manipulate expressions for simplification, visualization, or further calculations. They enable conversion of complex relationships into more manageable forms, helping us solve intricate problems by simplifying the mapping between different variables. These transformations also aid in predicting outcomes when variables undergo controlled changes.