Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-16\), find the transformation from the uv-plane to the \(x y\)-plane and find the Jacobian. Assume that \(x \geq 0\) and \(y \geq 0\). $$ u=x^{2}+y^{2}, v=x $$

Step-by-Step Solution

Answer

\(x = v\), \(y = \sqrt{u - v^2}\), Jacobian is \(-\frac{1}{2\sqrt{u - v^2}}\).

Step 1: Identify the Given Transformations

The given transformations are from the \(xy\)-plane to the \(uv\)-plane: \(u = x^2 + y^2\) and \(v = x\). Our goal is to find the inverse transformations that express \(x\) and \(y\) in terms of \(u\) and \(v\).

Step 2: Express \(x\) in terms of \(u\) and \(v\)

From the equation \(v = x\), we directly have \(x = v\).

Step 3: Express \(y\) in terms of \(u\) and \(v\)

Substitute \(x = v\) into the equation \(u = x^2 + y^2\) to get \(u = v^2 + y^2\). Then, solve for \(y\): \[ y^2 = u - v^2 \]Since \(y \geq 0\), \[ y = \sqrt{u - v^2} \]

Step 4: Define the Inverse Transformations

The inverse transformations from the \(uv\)-plane back to the \(xy\)-plane are:\[ x = v \]\[ y = \sqrt{u - v^2} \]

Step 5: Calculate the Jacobian

The Jacobian \(J\) of the transformation from the \(uv\)-plane to the \(xy\)-plane is given by the determinant of the matrix of partial derivatives:\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}\]Compute each partial derivative:- \(\frac{\partial x}{\partial u} = 0\) because \(x = v\)- \(\frac{\partial x}{\partial v} = 1\)- \(\frac{\partial y}{\partial u} = \frac{1}{2\sqrt{u - v^2}}\)- \(\frac{\partial y}{\partial v} = \frac{-v}{\sqrt{u - v^2}}\) using the chain ruleThus, the Jacobian is:\[J = \begin{vmatrix} 0 & 1 \ \frac{1}{2\sqrt{u - v^2}} & \frac{-v}{\sqrt{u - v^2}} \end{vmatrix} \]Calculate the determinant:\[ J = 0 \cdot \frac{-v}{\sqrt{u - v^2}} - 1 \cdot \frac{1}{2\sqrt{u - v^2}} = -\frac{1}{2\sqrt{u - v^2}} \]

Step 6: Summarize the Result

The transformations from the \(uv\)-plane to the \(xy\)-plane are \(x = v\) and \(y = \sqrt{u - v^2}\). The Jacobian of this transformation is \(-\frac{1}{2\sqrt{u - v^2}}\).

Key Concepts

Transformation

Transformations in mathematics help change one set of variables into another. In this exercise, we start with the original transformations:

• The function for u: \( u = x^2 + y^2 \)
• The function for v: \( v = x \)

These equations mean we take values from the xy-plane and transform them into the uv-plane. Each set of variables describes the same space differently. But to work with these transformations effectively, especially in integration, we need to understand the reverse process—finding x and y using u and v.

Partial Derivatives

Partial derivatives are a cornerstone of calculus. They measure how a function changes as its variables change. In this exercise, we calculate partial derivatives to find the Jacobian.Here's how it works:

• When we take the partial derivative of x with respect to u, since x = v and there’s no u term involved, the result is \( \frac{\partial x}{\partial u} = 0 \).
• Similarly, for x with respect to v, it's straightforward since x only involves v: \( \frac{\partial x}{\partial v} = 1 \)
• For y, things get a bit more interesting. The partial derivative of y with respect to u involves using the formula for square roots: \( \frac{\partial y}{\partial u} = \frac{1}{2\sqrt{u - v^2}} \).
• For y with respect to v, a chain rule is applied. It helps find \( \frac{\partial y}{\partial v} = \frac{-v}{\sqrt{u - v^2}} \).

These derivatives capture how small changes in u and v affect x and y.

Inverse Transformation

The inverse transformation is about translating back from the uv-plane to the xy-plane. It allows handling problems that might be more naturally expressed in one set of coordinates but need solutions in another.For this exercise:

• We found that x directly equals v since v = x; reversing this is straightforward: \( x = v \).
• For y, starting from u = v^2 + y^2, we solve for y: \( y = \sqrt{u - v^2} \).

These inverse transformations are crucial, as they help us switch between coordinate systems while solving or interpreting a problem. This ability to move between coordinate expressions broadens the problem-solving toolkit substantially.

Determinant

The determinant is like a magic number that tells a lot about a matrix! For calculating a Jacobian, finding the determinant is essential. The Jacobian here involves a 2x2 matrix of partial derivatives:\[J = \begin{vmatrix} 0 & 1 \ \frac{1}{2\sqrt{u - v^2}} & \frac{-v}{\sqrt{u - v^2}} \end{vmatrix}\]To find the Jacobian determinant:

• Multiply the diagonal: \( 0 \times \frac{-v}{\sqrt{u - v^2}} = 0 \).
• Subtract the product of the other diagonal: \( 1 \times \frac{1}{2\sqrt{u - v^2}} = \frac{1}{2\sqrt{u - v^2}} \).

Thus, the determinant, or the Jacobian, is \(-\frac{1}{2\sqrt{u - v^2}}\).This negative sign often indicates orientation changes. The determinant shows how the transformation "shrinks" or "expands" area between two coordinate frames.