Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The transformation is \(x=\frac{u+v^2}{2v}, y=v-\frac{u+v^2}{2v}\) with Jacobian \(J=2v\).

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

We start with the transformation equations: \( u = x^2 - y^2 \) and \( v = x + y \). We need to express \(x\) and \(y\) in terms of \(u\) and \(v\). First, from \( v = x + y \), express \( y = v - x \). Substitute into \( u = x^2 - y^2 \), giving us: \( u = x^2 - (v - x)^2 \). Simplifying this, we obtain: \( u = x^2 - (v^2 - 2vx + x^2) = 2vx - v^2 \). This can be rearranged to \( x = \frac{u + v^2}{2v} \). Then replacing \( x \) back into \(y = v - x\), we get \( y = v - \left(\frac{u + v^2}{2v}\right) \).

Step 2: Calculate the Jacobian of the transformation

The Jacobian of the transformation from \((x, y)\) to \((u, v)\) is given by:\[J = \begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix}\]First, calculate the partial derivatives:- \( \frac{\partial u}{\partial x} = 2x \)- \( \frac{\partial u}{\partial y} = -2y \)- \( \frac{\partial v}{\partial x} = 1 \)- \( \frac{\partial v}{\partial y} = 1 \)Then, substitute these values into the Jacobian matrix:\[J = \begin{vmatrix}2x & -2y \1 & 1\end{vmatrix} = (2x)(1) - (-2y)(1) = 2x + 2y = 2(x + y)\]Since \( v = x + y \), we have \( J = 2v \).

Step 3: Verify sign based on the given conditions

Since it is given that \(x \geq 0\) and \(y \geq 0\), it follows that \(v = x + y \geq 0\). Therefore, the Jacobian determinant \( J = 2v = 2(x+y) \geq 0 \). This confirms that the transformation is valid for the given domain constraints.

Key Concepts

Transformation

When dealing with transformations, it often involves converting coordinates from one plane to another. Here, the transformation we are discussing is from the \(uv\)-plane to the \(xy\)-plane. The given transformation equations are:

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

These equations help in mapping coordinates from the \(xy\)-plane to the \(uv\)-plane. But to fully understand or manipulate these points, we often need to reverse the process, meaning we express \(x\) and \(y\) in terms of \(u\) and \(v\). This is essential for transforming coordinates back to the original plane or to check the consistency of transformations.

Partial Derivatives

Partial derivatives are critical in understanding how small changes in one variable affect another in multivariable functions. When calculating the Jacobian of a transformation, partial derivatives provide insight into how each variable changes with respect to another. For the transformation from \(xy\)-plane to \(uv\)-plane, the partial derivatives needed are:

• \( \frac{\partial u}{\partial x} = 2x \)
• \( \frac{\partial u}{\partial y} = -2y \)
• \( \frac{\partial v}{\partial x} = 1 \)
• \( \frac{\partial v}{\partial y} = 1 \)

Calculating these derivatives allows us to form the Jacobian matrix, which judges how volume elements deform under the transformation. These values combine in the determinant of the Jacobian to determine the transformation's local behavior.

u v-plane

The \(uv\)-plane is where our transformed points reside after applying the given transformations from the \(xy\)-plane. Each coordinate in the \(uv\)-plane corresponds to a transformation of an original coordinate in the \(xy\)-plane.For our transformation:

• \( u \) gives a measure related to \(x^2 - y^2\), showcasing the difference in squared lengths.
• \( v \) is the simple linear sum \(x + y\), often used to simplify expressions in many analytical contexts.

Understanding the \(uv\)-plane is crucial when reversing transformations or solving integration problems involving these mappings.

xy-plane

The \(xy\)-plane is our starting domain, where initial coordinates \(x\) and \(y\) exist and interact. In many geometric and physical problems, we start with known values in the \(xy\)-plane and seek transformations to other planes for analysis or solution-finding.The transformation equations \( u = x^2 - y^2 \) and \( v = x + y \) express how points in the \(xy\)-plane correspond to another set in the \(uv\)-plane. This plane helps to visualize and understand how initial geometric shapes or solutions might deform or translate when other conditions apply.Understanding the relationships and constraints in the \(xy\)-plane (e.g., \(x \geq 0, y \geq 0\)) is vital for ensuring that transformations remain valid and correctly interpreted.