Problem 21 - Calculus Early Transcendentals

Question

Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=(u+v) / \sqrt{2}, y=(u-v) / \sqrt{2}$$

Step-by-Step Solution

Answer

Answer: The value of the Jacobian \(J(u, v)\) for the given transformation \(T\) is -1.

Step 1: Find Partial Derivatives

First, we need to find four partial derivatives; \(\frac{\partial x}{\partial u}\), \(\frac{\partial x}{\partial v}\), \(\frac{\partial y}{\partial u}\), and \(\frac{\partial y}{\partial v}\). Remember that the transformation \(T\) is defined as: $$x=\frac{u+v}{\sqrt{2}}, \ \ \ \ y=\frac{u-v}{\sqrt{2}}$$ To find these derivatives, we will use the following rules: $$\frac{\partial x}{\partial u} = \frac{\partial}{\partial u} \left(\frac{u+v}{\sqrt{2}}\right)$$ $$\frac{\partial x}{\partial v} = \frac{\partial}{\partial v} \left(\frac{u+v}{\sqrt{2}}\right)$$ $$\frac{\partial y}{\partial u} = \frac{\partial}{\partial u} \left(\frac{u-v}{\sqrt{2}}\right)$$ $$\frac{\partial y}{\partial v} = \frac{\partial}{\partial v} \left(\frac{u-v}{\sqrt{2}}\right)$$ Calculate these four partial derivatives.

Step 2: Calculate the Partial Derivatives

Using the definition of \(T\), we can calculate the partial derivatives: $$\frac{\partial x}{\partial u} = \frac{1}{\sqrt{2}}$$ $$\frac{\partial x}{\partial v} = \frac{1}{\sqrt{2}}$$ $$\frac{\partial y}{\partial u} = \frac{1}{\sqrt{2}}$$ $$\frac{\partial y}{\partial v} = -\frac{1}{\sqrt{2}}$$

Step 3: Form the Jacobian Matrix

Now that we have the four partial derivatives, we can arrange them in a matrix called the Jacobian matrix. The rows represent the derivatives of the transformation components, and the columns represent the derivatives with respect to the variables \(u\) and \(v\). So, the Jacobian matrix is: $$ J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} $$

Step 4: Compute the Jacobian \(J(u, v)\)

Finally, to compute the Jacobian \(J(u, v)\), we need to find the determinant of the Jacobian matrix. For a 2x2 matrix, the determinant is: $$\text{det}(J) = \text{det}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$$ Using this formula for our Jacobian matrix, we have: $$J(u, v) = \text{det}(J) = \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{2} = -1$$ Therefore, the Jacobian \(J(u, v)\) for the given transformation \(T\) is equal to -1.