Problem 19
Question
Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=2 u v, y=u^{2}-v^{2}$$
Step-by-Step Solution
Verified Answer
Answer: The Jacobian of the transformation \(T\) is given by:
$$J(u, v) = \begin{bmatrix}
2v & 2u \\
2u & -2v
\end{bmatrix}$$
1Step 1: Write down the given transformation
$$T: x=2 u v, y=u^{2}-v^{2}$$
2Step 2: 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:
$$\frac{\partial x}{\partial u} = 2v, \qquad \frac{\partial x}{\partial v} = 2u$$
$$\frac{\partial y}{\partial u} = 2u, \qquad \frac{\partial y}{\partial v} = -2v$$
3Step 3: 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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