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Chapter 13: Problem 19

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

Short Answer

Expert verified
Answer: The Jacobian of the transformation \(T\) is given by: $$J(u, v) = \begin{bmatrix} 2v & 2u \\ 2u & -2v \end{bmatrix}$$

Step by step solution

01

Write down the given transformation

$$T: x=2 u v, y=u^{2}-v^{2}$$
02

Compute the partial derivatives

We need to compute the first-order partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\). These are: $$\frac{\partial x}{\partial u} = 2v, \qquad \frac{\partial x}{\partial v} = 2u$$ $$\frac{\partial y}{\partial u} = 2u, \qquad \frac{\partial y}{\partial v} = -2v$$
03

Compute the Jacobian matrix

The Jacobian matrix \(J(u, v)\) is a 2x2 matrix where the partial derivatives calculated above are placed in the appropriate positions: $$J(u, v) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}$$ Substituting the calculated partial derivatives, we get: $$J(u, v) = \begin{bmatrix} 2v & 2u \\ 2u & -2v \end{bmatrix}$$ Hence, the Jacobian of the given transformation is: $$J(u, v) = \begin{bmatrix} 2v & 2u \\ 2u & -2v \end{bmatrix}$$

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