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Chapter 16: Problem 18

Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=4 v, y=-2 u$$

Short Answer

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Question: Compute the Jacobian J(u, v) of the transformation T: (u, v) 鈫� (x, y) where x = 4v and y = -2u. Answer: The Jacobian J(u, v) for the given transformation T is 8.

Step by step solution

01

Find the partial derivatives

Calculate the partial derivatives of x and y with respect to u and v. We have: $$ \frac{\partial x}{\partial u}=0,~~~~\frac{\partial x}{\partial v}=4,\\ \frac{\partial y}{\partial u}=-2,~~~~\frac{\partial y}{\partial v}=0. $$ Note that since x is given as a function of v (and doesn't involve u), its partial derivative with respect to u is 0. Similarly, since y is a function of u (and doesn't involve v), its partial derivative with respect to v is also 0.
02

Create the Jacobian matrix

Now that we have found the partial derivatives, we can create the Jacobian matrix. The Jacobian matrix is a 2x2 matrix that has the partial derivatives as its elements: $$ J(u,v) = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 0 & 4 \\ -2 & 0 \end{pmatrix} $$
03

Calculate the determinant of the Jacobian matrix

Now, we will compute the determinant of the Jacobian matrix. For a 2x2 matrix, the determinant is calculated as follows: $$ \det(J(u,v)) = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u} $$ Using the partial derivatives we calculated earlier: $$ \det(J(u,v)) = (0)(0) - (4)(-2) = 8 $$ Thus, the Jacobian J(u, v) for the given transformation T is 8.

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