91Ó°ÊÓ

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.

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

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: $$ J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} $$

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.