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Chapter 16: Problem 22

Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=u / v, y=v$$

Short Answer

Expert verified
Answer: The Jacobian determinant for the given transformation is \(J(u, v) = \frac{1}{v}\).

Step by step solution

01

Define the partial derivatives

We first need to compute the partial derivatives. To do this, we will differentiate x with respect to u and v, and y with respect to u and v. $$\frac{\partial x}{\partial u} = \frac{\partial}{\partial u} \frac{u}{v} = \frac{1}{v}$$ $$\frac{\partial x}{\partial v} = \frac{\partial}{\partial v} \frac{u}{v} = -\frac{u}{v^2}$$ $$\frac{\partial y}{\partial u} = \frac{\partial}{\partial u} v = 0$$ $$\frac{\partial y}{\partial v} = \frac{\partial}{\partial v} v = 1$$
02

Form the Jacobian matrix

Now we will form the Jacobian matrix using the partial derivatives computed in step 1. The Jacobian matrix is given by: $$J(u, v) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} \frac{1}{v} & -\frac{u}{v^2} \\ 0 & 1 \end{bmatrix}$$
03

Compute the determinant of the Jacobian matrix

Finally, we will compute the determinant of the Jacobian matrix. The determinant is given by: $$J(u, v) = \det(J(u, v)) = \begin{vmatrix} \frac{1}{v} & -\frac{u}{v^2} \\ 0 & 1 \end{vmatrix} = \frac{1}{v} \cdot 1 - 0 \cdot -\frac{u}{v^2} = \frac{1}{v}$$ So, the Jacobian determinant for the given transformation is \(J(u, v) = \frac{1}{v}\).

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